Laplace Transform Ultimate Study Guide

okay let's do some math for fun today I have the  Laplace transformations for you guys we have a   total of 24 of them and of course here's the file  you guys can go ahead download that please see the   description for the file and the notes and also  the timestamps for your convenience and here is   the definition of the Laplace transform as you can  see is just an improper integral and you have e to   the negative T times the function right here and  hopefully each converge and to guarantee that this   work here converges you just have to make sure  that F of he has exponential water what that means   is that F of T cannot be too big it is you can't  just think about it if you have e to the positive   T squared in that case that will make this  improper integral diverge right so I just have to   make sure that happens and if you ever wonder why  that last transport can help to solve differential   equations go ahead and check our mu prime masters  video I have that lot I will have that link in the   description for you as well man that was so good  because I actually have never seen that before   when I was taking Laplace transforms backing that  the days I mean taking the differential equation   cast back in the days all right so now here we  go this right here it's the first one of course   we'll be doing a lot of proofs so here we go now  hands transform of e to the 8th Yi well that's   the integral from 0 to infinity and we'll just  use this formula so we have e to the negative   as T times e to the 8th he and then he so you just  apply that a phoenician and in the end we're just   going to be working a lot of integrals pretty much  all right so now this right here and here they are   just Yi so it's good we can just integrate this  from 0 to infinity and we can just come by this   as a - yes yeah so I'll just read yes a minus  s and then what fact how that he right here and   then I will have the TV at you so now to tend out  this improper integral you should technically take   the limit and then you can say B goes to infinity  and then the interval corresponds hero to infinity   all that stuff but in this video I'm just going to  skip that just write down the infinity symbol you   just remember we will be taking limits okay here  we go to do this first of course just note your   integration this is just e to a number times T to  the first power so the integral of this is just   e to the a minus s T but remember the duet the  reverse chandu right the derivative of this is   just that when we are doing the integral be sure  divided by that so we will have to do 1 over a   minus s so that's the integral part and then we  will have to plug in plug in from 0 to infinity   under stuff now here is the deal when we plug in  infinity to the T you will see that we'll end up   with something like this 1 over a minus s and then  e to the a minus s and here you have past infinity   right and then minus and again I know this is  a penetration but as I said it's just a short   notation it's still ok just remember what he can  limit and then you put the wrong here so we have   1 over a minus s and then we have e tired to the  a minus s 2 times 0 like this right so that's the   first part this is the second part and now here is  the deal in water for this right here to converge   we have to make sure each to this power and she  end up to be 0 because we have infinity times   this already the idea is that we need to make sure  this part has to be negative and 1 for this to be   negative of course that means I will have to have  a minus s to be less than 0 and of course we can   solve this inequality move the s to the other side  meaning that s has to be bigger than a we need   to have this condition here in order to make this  negative so that you can see we will have e to the   negative infinity and then that will be 0 times  all this that will be 0 so with this condition we   will see that the first part is just nice T equal  to 0 and then the second part were just - and you   see this is just over a minus s and that e to the  zero power right here is just one so we just have   this right here and finally what we can do is  you can also just distribute the negative here   and here meaning just switch the water so get 1  over s minus a like this so this right here is it   and again until you finish all that you should  technically include this condition so you will   have to say s should be greater than a right so  that's it that's it that's it very good huh that's   the first one not so bad alright so now continue  and we'll be using this for the next few so look   at this look at this ok so again how are you guys  doing let me know leave a comment down below let   me know I'm doing ok and doing that for math  lady how about you guys and hopefully everybody   is doing well of course stay safe stay healthy  all right number two right here and of course   I will be doing this in water in the sense that  one is based on the other blah blah blah right   so here is the second one who are between the  power function instead so we have this our goal   is to get a Laplace transform of T to the N power  and here I'm going to assume that and it's just a   non-negative whole number right so I want to know  what this is I would like to have 0 1 2 so on and   so on in fact later on we can actually extend this  to like some like weird numbers such as n is equal   to 1/2 but that would be for question number 23 so  nah okay this is how you can do it first you can   use the definition so you can put a T to the nth  power right here instead right and you can just   work that out and you can use integration by parts  and times actually technically you have to do that   a lot of times and you how to do a lot of math  or you can do the following check this out this   is how we are going to do it I will use this right  here to help us so have a look right here we know   the Laplace of e to the a T is equal to that and  if you have e to the HT in fact we know the series   expansion for that isn't it so I will read this  down right here for you guys so this is how we are   going to do it so notice that we have this already  a Laplace transform e to the 8th he which we know   this is equal to 1 over s minus a like this and in  this particular case I'm going to choose a to be   greater than 0 I can choose Y for either 1 I can  say love has off into dy over positive a times T   right so of course this is still a yet however you  still have to include this condition so you just   have to make sure that both s then a are greater  than 0 so that's pretty much why I hat and again   I mentioned about the series we know how to handle  the series right here it's already style Rico let   me just put on Rico and I will just put this down  here for you guys e to the X power this is just a   power series expansion for e to the X this is  the series as n goes from 0 to infinity and we   have X to the nth power over n factorial and if  you would like to recall and review your series   you can check out my powers use marathon for you  guys right so that's what we have and this right   here is true for all X so let me also just write  that down right here so what we are going to do is   you see here we have e to the 80s so let's go  ahead and just focus on this part first stuff   when I have e to the 8000 I have to do is just put  the 80 into this egg's right here and this is the   jet because it's true for all the inputs so I can  just open a parenthesis with a t inside raised to   the nth power over factorial like that so that's  pretty nice and of course you can just work out   this right here a little bit and you see this  year's the series as n goes from 0 to infinity   and this is how I'm gonna do it I will put down  t3 and power first like this over n factorial and   then we have a to the nth power like that so this  is what we have all right okay then another thing   I want to do this that's to get this part and I  will just do a little divider here well we have   one over topic so recall that 1 over 1 minus X yes  this is our best spread this right here is equal   to the sum as n goes from 0 to infinity just X to  the nth power that it's however this is only true   if the absolute value of x is less than 1 right  so be really careful on this right here however   it's actually not so bad oh now let's look at  a series expansion for this so we have 1 over s   minus a like this yeah well in order for us to use  this we actually have to have a 1 right here but   this right here is an S that's not good that's ok  because we can factor out the s so it looks that   we have 1 over s times 1 over R factorial yes  so we have one right here and then factor out   the S so we have minus a over s like this and of  course you can put on parenthesis like that now   here is a deal because we do have this condition  have a look as it's greater than a and they are   both positive and right here we have a over s  so of course in this case here notice that the   absolute value of a all 4s the top is smaller than  the bottom right this right here has to be less   than 1 for sure so we'll test said we know we can  use our best friend right here in the front oh we   still have 1 over s and then right here we can  put this into here so we have the series as n   goes from 0 to infinity and we just need to write  this down here to open the parentheses with a over   s and then raised to the and power like that  alright so that's pretty much it and now I'm   just going to erase this and then we will see  how we can make the connection between one and   the other okay oh of course maybe that's simplify  this a little bit early so this right here I will   have the summation n goes from 0 to infinity on  the top we have a two ends power but let me put   that aside like this a to s power on the bottom we  have s to the nth power but we have another one on   the bottom as well so on o we have 1 over s to the  n plus 1 like that ok now have a look right here this is true that we know so here is the deal  let's go ahead instead of looking at the a that   means eat the e to the a T and also this have a  look e to the HT is nicely equal to this isn't   it yes yes so I can just actually put that down  right there so we have the Laplace transform and   actually let me put this down blue we have the  sum from n goes from 0 to infinity of all that   so we have P to the N over factorial a to the  nth power like this and then on the right hand   side we have 1 over s minus a and again in terms  of a series it's equal to this isn't it yes yes so   we can just write that down the series as n goes  from 0 to infinity and we have 1 over s to the n   plus 1 times a to the N power now here is the  thing you had to just believe me in this case   where the plus is actually linear meaning that the  Laplace of a sum is the sum of the other plus and   even though inside it's an infinite sum but let  me just tell you it will still work in this case   all right because this right here converges so  what we can do right here is that we can switch   the water of the laplace and also the summation  and if you want to see a teacher can just change   that to improper integral and what the appa okay  let's just do it right we have the sum as n goes   from 0 to e that he let's put that down first and  then we have the Laplace transport of our we won   inside now here is the deal have a look right here  inside here what I want to do is you see we have   heat and power I'm just going to have that inside  of my Laplace transform so we have a transform of   T to the and power like this right and then on  the outside we have the one over N factorial   let me just put that down right here and then at  the end we still have the a to the nth power like   this so have a look this is a summation of all  that again technically every single being said   the Laplace but I took a 1 over n factorial and  also to put that a to the nth power on outside   and the reason I can do that because in a la  classe world only T matters T is the variable   and the N is just like a constant so just take  that out you can just read out so that's that   here all right and then on the right hand side we  have this is equal to the sum as n goes from 0 to   infinity of 1 over s 2 n plus 1 MJ we have a to  the N power and here is the punchline maybe you   can see already which is very nice have a look  left hand side on the right hand side the above   series and you can just look at this part this  is a 2n power likewise this is also a power so   it's like infinite polynomial right warp infinite  power series namely though you see that this and   that they have to be the coefficients and water  for the polynomials to equal what the infinite   series to be equal to each other the coefficients  has to equal to each other so we can conclude that   this and that must be the same so we started you  see here we actually just have 1 over n factorial   the Laplace transform of T to the nth power this  has to be the same as 1 over s to the n plus 1   so as you can see in the end all we have to do is  just multiply the N factorial ample sites so you   can just see that if you would like this is so  nice they can so nicely and then in the end you   get a Laplace transform of T to ins power that's  equal to factorial over s to the n plus 1 power   so here we have it this is it and technically  right here as I said and is going from 0 to 0   1 2 3 and so on to infinity and that's exactly  how you stress all that good stuff so again this   is factorial over s very nice right so that's  it alright so you can see that this is actually   really useful and then we'll be using that for  the coming up ones as well very cool the next one   is sine and cosine yes actually question student  question poor will be doing two things one so this   is like turning two parts it's with one stone  but we actually happen to kill parts will cure   Laplace transforms all right so here we go this  is question number three I just want to make sure   that I'm still recording so right here question  three we want to have the Laplace transform of   sine D and B is just our number of course so we  need to know what this is and of course when we   have sine that's also talked about the cosine  as well so I'll put down number four right here   number four Laplace transform of cosine of BT and  if I can do this a few ways first way is you can   use integration by parts and again will be similar  things to the things that you have to do with T to   the nth power if you use integration by parts and  you have to plug in the limit and all that stuff   in this video what a [ __ ] I'm just making my  video here I will show you guys how you can use   complex numbers to take your both of them with  just one computations and not a thing you can do   is you can figure out sine of T and cosine T and  later use one of the properties I'll show you and   you can generalize it sound in Campeche by anyway  here is how we are going to do it again I'm going   to write this down here for you guys we are going  to use the fact hat with it earlier the Laplace   transform of e to the a T this is being equal to  1 over what yes s minus a yeah all right so this   is what we are going to do we are going to leave  dangers these sometimes meaning that we will have   to go to a complex world sometimes so have a  look with the a right here right what we are   going to do is I'm going to let a equal to well  we have to have the complex number so we have to   introduce the eye and then Oh buddy I'm using the  P so I will just have I be like this so by using   this you'll see that the next line will have the  following so I will just put this down mmm black   yeah so this is what we have and now we will just  use their so here we have the Laplace transform   excuse my little brace it's really hard for me  to do that by anyway easy and then the a is now   I P and then we still have the T and then this  is equal to 1 over s minus I B like this right   so we can just of course do that and now another  thing that you have to remember is the order spoon   lassoed let me also write it down right here  for you guys so this is the Euler's formula which says when you have e to the I theta this  right here it's nice to equal to cosine theta plus   I sine theta it's amazing formula right and how  we are going to use it is will the left hand side   presents the cosine and sine already yeah and then  the right hand side is a rational expression you   just have to fix that little bit because remember  I don't my to be on the bottom I like to be on top   in that we can just worked out and that's pretty  much the idea so here we go this is how we are   going to do it the left side have a look we have  the theta being the PG right here right so the   left hand size can easily equal to the Laplace  transform of sine well actually cosine first of   PT so we put that down right here and then we  add I times sine of BT so this is what we have   on the left hand side and now on the right hand  side to fix that situation when you don't like to   be on the bottom cool I had multiply that top and  bottom by its conjugate so we KS plus IB and then   s plus I like this and like that all right on a  pattern of do this in your head when you multiply   this I'll just get s times s which is a square and  there - and I times I is I squared so it becomes   negative 1 so it becomes a plus and then we have e  squared so this is what we have on the bottom and   on the top you see that we just have s plus I B  just like that isn't it now here is the cool part   again Laplace is linear meaning that the Laplace  of a sum is the sum of the Laplace and when you   have a constant multiple you can also put that  in the front of the Laplace so have a look I'm   going to break this down as the Laplace transform  of cosine of BT now close that and that's added   words I will put the eye to the front and then we  have the Laplace of sine of BT like this right so   I just put you know I just work at them apart  like this and that and again just really excuse   my little brace right all right and I all right  so we can do the same thing we have the real part   and also let me imaginary part the real parts just  this which is the following s over s square plus   B Square and then plus I times B over s square  plus B Square now we'll play the game the game   is called me to match very fun because this and  that yeah both the real part so of course this   has to be equal to that isn't it and then this is  same a generic part though is that so this has to   be the same as that and with that we are done  because you see that the Laplace of Stein PT is   nicely equal to this we have the PIA mater over  a square plus B Square again we are done what's   that and then for this will have the SS on the  table as over a square plus B squared like this   there's no v in my video right so just s this is  now five anyway that's it very cool huh so you   don't have to do the integration by part X so many  times so this is so wonderful it's awful look this   is it alright now we did full already so this  is going pretty well and I have some surprising   intercourse for you guys as well so be sure you  watch that as well I say that will wear will now   what's the next one Oh remember the distinction  cost which is like like the knockoff version of   North here we go number 5 number 6 well here  let me just write that down number 5 we have   the following the Laplace transform of thing  T right since T which is the hyperbolic sine   function well again in this case if you allow you  can go and use the integration by parts which the   definition of the Laplace transform but which end  to the following here we go you have to remember   the definition of the sinc function well this is  equal to the following this is nice the equal to   e to the path the input right here which is T and  a minus e to the negative T all over 2 like that   and we know how to hand off the Laplace transform  into this number T already yes we do so we can't   you just break them apart and let me show you how  first we have the one half let's go ahead and put   that all the way in the front and then again we  can just do the following now class of the first   which is the Laplace of e to the T and there - the  Laplace of the second and then we have the 1/2 all   the way in your front already so this is what we  have yeah yeah put oh I put on the beach heat I'm   sorry I put on the beach things off bTW I'm sorry  so PT is actually our variable of our input right   so I will have to change this this this this to  the PT that's all so easy fix here we have e to   the PT and then here we have e to the negative  PT and then this is e to the PG and this is e to   the negative B okay so just easy fix now we can  continue for this right here of course again the   1/2 it's just all the way in the front like this  and then for this one of course we can just recall   the formula which is just 1 over s goes first -  beat here right here so you actually just put on   be like this and then minus 1 over and we have s -  this is negative PG so we will actually have going   to give that class later on so this is 1/2 and now  of course we can just combine the little fractions   order stuff so right here we will just multiply  the pot top and bottom by s plus B and then as   plus B and I will just multiply this by s minus B  and then multiply the bottom by s minus B so all   that good stuff now we have 1/2 in the front and  then on the bottom you just have s minus B times   s plus B which is s square minus B like this and  I be square sir and on the top though we have s   minus we have s plus B so let me just put on s  plus B minus s and then minus minus becomes plus   so we have a plus B like this and of course just  work that out you can see that the SNS can so and   on top B and BS just go ahead and you have to be  and then in the end of course you can cancel the   2 and the 2 right here so this is very similar  to the original sine function yeah you actually   end up with B on the top as well over the bottom  is s square minus B Square like this so that's   it however do we need to put on any conditions  well Apodaca from here to here remember we had to   put on s is greater than what s has to be greater  than B so I would actually have to say S has to be   greater than B the me at you just write that down  right here has has to be greater than just a hole   we did earlier and now this right here you will  just have to make sure that S has to be greater   than this improve which is negative P and depends  on what B is so you can just kind of include that   and all that stuff but I'm just going to do that  to you guys because I don't like B is negative   or pasty and all that but yeah you can just say  s is greater than B that's actually pretty good   okay all righty anyway this right here yes P over  s square minus B Square like this and number six   very similar that last transport of cosh but the  input here is P again actually this is cosh SM   input now remember the definition of cosh is the  following gosh is nicely equal to e to the improve   of speed he and then plus e to the negative  input which is negative PT o / - like this   then again we can just do what we do over there  so this is the same as one Laplace transform of   e to the PT and then we have to add the Laplace  transform of e to the negative B T like this   thing like that okay let's just go ahead and do  this this is one-half times this is 1 over 1 so   1 over s minus the P right here and in this case  we are adding 1 over s negative and then this is   negative P just like that and again let me just  show you guys this real quick for the first one   we will just get multiplied by s plus B as plus  B and then for this one we multiplied by s minus   B and then s minus B and that's the all the way  in the front we still have the 1/2 on the bottom   again it's going to be a square minus B Square on  the top is s plus B and then this is plus and this   is s and then this is minus B so it's like that  in this case the s I mean the PMB can so SNS is   twist so the 2 and then to cancel so in the end  we end up with an S on the top so this is the S   portion and again very similar to the original  cost original cosine of PG so this right here is   s on top but you have s squared minus B squared on  the bottom like this right and again you can tell   me what the condition is just going to write  that down that s should be equated and yeah because it does already matting yeah because you  square that anyway so I'll just have to make sure   a square has to be greater than B squared then  yeah just alright so that's pretty much it and   continue doing pretty well for this one better  than a differential equation so we have the next   one which is unit step function by this okay okay  so let me erase all this so again as I said who   will be using definitions whenever we have to use  whenever I do two proofs sometimes if we can see   the connection between the things would have done  already you don't have to use the definition all   right number seven I'll put this down here for  you guys seven that last transform and here we   have the unit step function so I'll just put this  down like that like this is the best I can write   t minus some number a like this yeah all right  and right here I have to remind you guys for the   unit step function unit step function is this  right here it's actually a piecewise function   so I will actually 20 for you guys first the idea  of the unit step function is that if you have the   following look at what the a is first so let's  say I'm serious my a and anything before a when   you are doing a when you when you are doing t  minus a anything before a is zero so you have   this and then you put an open circle and then  after a at a you have to jump to one right and   then third open circle as well like this right  so the idea of the unit step function is that   if the input is passed did they will give one if  the inputs naked heat you get 0 and if the input   is 0 some book defines you to be 0 some books  defined it to be not undefined so I will use   the open circle in this case so I already start  here for you guys you are t minus a is equal to   either 0 again if the input so I will write down  t minus a is less than 0 right it's equal to 1 if   the input name eg minus a is greater than 0 and  the reason I want to do this it's because I would   like to tell you if you'd reverse the t minus a  with a minus T which would reverse the picture but   that's happening later on so no so keep that in  mind if the insides positive then you get one if   the incise negative you get 0 if the inside she  wrote some books define it to be nothing sample   defined it to be 0 or 1 depends doesn't really  matter because we are doing integrals so one   point it's not going to change the whole change  anything anyway how can we do this yes definition   of Laplace transform so let's write it down this  is the integral going from 0 to infinity e to the   negative as T times u of t minus a like so and we  have the did he do this so well we'll have a look   this is what's going to happen when you multiply a  function with a unit step function so let me graph   e to the S negatives people you guys which is just  an exponential decay function like this assuming s   is positive so this is e to the negative T right  and originally this integral we have to go from   zero all the way to infinity yeah but now after  we multiply by the unit step function so I will   write this down if you have e to the negative  T unit step function of t minus a what's going   to happen is that we'll pick a where a is that  says right there anything before a this is 0 0   times this that will make everything 0 right so  actually you scared zero right here and you have   an open circle at a there's a jump to one at  one times this is just just that so you will   just jump to here and then you have this okay  so instead of integrating the whole thing have   a look you actually do not have any area from 0  to a at all all we have to integrate is just from   a to infinity just this part and again at the end  points right here it's just the point so if you're   talking about integral doesn't matter all right  so here we go we just have to integrate from A   to infinity and again this right here changed the  zero to a and from a to infinity this portion is   just the same as the original so I can just  write down e to the negative as you like so   and then we have like that and then again right  you can just go ahead and integrate out of this   so integrating this it's just e to the negative  T and then we divide it by this right which is   just going to be 1 over take TS then we plug in  plug in going from a to infinity well well when   you put infinity here we have 1 over negative  e to the negative s and we have infinity right   here and that's the first part and out of - the  second part is 1 over negative e to the negative   and then you have a like right here all right if  this is going to be 0 depends what I see is this   is negative already this is positive infinity s  has to be positive so that we can actually get   a negative infinity out of this right so with  that being said s has to be greater than 0 so   with that condition though we can actually say  the first part has to be 0 and again we need we   need to have that condition and it's in there  right this time right yes that we need to have   that in order to make this work and what are  the rest negative negative becomes positive   and then we just have that that's pretty much it  so in the end we have the plus 1 over s e to the   a so finally we were just ready Stan s e to the  negative as a and some people write to ride yes   e to the negative is it depends on how you want to  ride it but this is okay and then / s yeah again   we have the condition so that's right there down  s has to be greater than zero in order for that to   work just like that right so this right here it's  a unit step function and now let me introduce you   guys something that's very similar to the unit  step function they need a window function so   it's the PI sub apt function I should be able to  do that right here so I will do that right here   number eight so here we have the Laplace transform  and inside the capital PI a comma P this is not   the original type function it's just the I think  Scott window function over oh by the way this is   also called the Heaviside function but it's not  because this sides happier that's why it's zero   happy sighs just a hangover the person right I  thought it was only funny anyway so we have pi   AP and input is just usually with a sheet like  this right so that's what we have and again I   will have to tell you guess what PI ap years well  PI ap it's actually not so bad Oh PI a P P this   right here again as I said it's very similar  to that perhaps I'll show you guys a picture   first well very similar you have to tell me two  numbers a and B suppose P speaker than eight right   so let's say we have a here and B here the idea  of this right here is that anything outside of   a and B is going to be zero and anything between  of a and B is equal to one that's pretty much it   so you can just grab it right here open open like  this anything else a zero anything between is one   and again we do not worry about the end points  right I'll just have that and then here is the   property of the pie a B of T so you can write  it down like this zero should write down one   purse 1 if the input is in between of a and B and  again to not worry about the endpoints 0 otherwise okay so hopefully because it's the problem  zero if otherwise no that doesn't make sense   if otherwise alright so just I'm sorry about  words otherwise meaning that yeah all right so   that's pretty much it however there's a really  nice connection between the unit step function   and PI a B I will also tell you the PI ap  is actually just nicely equal to the unit   step function of t minus a which is this part  and then you - are you - you t - be like this yeah all right so just like that all right we're  stopping said in order to do the Laplace of this   we can actually just use this right here which  is very nice so we actually just have to do the   Laplace transform again Paille be is just going  to be the unit step function of t minus a and   then minus the unit step function of t minus  b like this which is very nice because now we   can just do a Laplace transform the first one -  Laplace transform the second one not part of the   flash transform the first one is what it's kind  of hard no no no it is open here so we actually   just get e to the negative s a over s right that's  pretty much it that's all we have to do and s has   to be greater than zero just like that and then in  between so - so just go and put our - and then we   have this so all we have to do is change the a to  P so we have e to the negative P over s in the end   of course that's right things done together e to  the negative a minus e to the negative P or over s   s has to be greater than zero and with that we are  done Jessica okay so this is the window function   I believe that's pending of that so it's a PI a  business all right next one's going to be pretty   crazy it's still kind of like a piecewise but it's  not as no more looking as the one that we just did   so have a look all right have a look and let me  know if you guys are taking differential equations   if you are man I've missed that class I'm not  teaching differential equation this semester nope   I haven't taught you in three years but you know  I just feel like doing some applause transport so   why not all right so this is the number nine and  this is the Laplace transform of the Dirac Delta   function so you just put down the Delta this is  the Delta and then t minus a like this right so   again I will have to explain to you guys what the  Delta function is well to a picture tell me where   a is anything besides a yes zero so you have  this open circle so happen at a the function   goes up to infinity all the way to infinity all  the way to infinity you can even see it you can   even see it so just yeah just cannot see it so  it doesn't really make sense for me to graph that   you just have to grab it from the definition and  I'm going to read it down alright so here is the   deal I will just have to write it down for you  guys the Delta t minus a again I would write on   a name later but this is 0 if T is equal to  if T is not equal to a and this is equal to   infinity if t is equal to a so again the pictures  everything is 0 except for a goes out to infinity   right so there's no way for me to give you guys  a legitimate graph of that and again this is Cody   what I should write down the name this is the red  which I think that witty Delta Tom [ __ ] right maybe without he I don't know what city I  didn't ready down by anyway I'm sorry let   me actually just not write down the name alright  so this right here is what we have now this is   how the Laplace transform is going to look this  right here we are continues at the Venetian and   see if Abbas so integral from 0 to infinity  again we have e to the negative T and then we   are going to multiply this with the outer of  t minus a like that like this hmm how in the   world can we make the hospital where I have  to tell you get a few more scenes about the   Delta function right here right so one property  that you have to know it's a following right if   you integrate from negative infinity to positive  infinity because the domain is all real numbers   so you go from negative infinity to positive  infinity it doesn't really matter action and   if you integrate the out of t minus a right  here let me just ask you guys this question   first what do you think what the answer is and  again you can just really imagine that you have   the graph by it's not ideal to have a graph is  right here put the outer is everything zero but   when you have a yes going to be infinity and  remember when you up integral you're trying   to find the area the area right here is zero  the area right here is zero what's the area like let's say this right here is the infinity  this is that infinity up here right we have   one point what's the area from here to here  is there any width no the width is zero but   what's the height the height infinity 1 0 times  infinity technical is indeterminate right however   by definition to solve the the best part what is  right here is that this right here is defined to   be 1 this is the property of reading click with  the Delta function alright so you have to just   kind of take my word for that if you were like  hey so that's the idea that's the idea number   one right and in fact you also have idea number  two right here if you have the following let me   erase my picture and I'm arresting you here idea  number two if you have the integral from negative   infinity to positive infinity if you have a  function let's say f of T times the outer of   t minus a like this well well again let me just  let's just think about as a picture yes a-anything   by a is zero like this so this right here it's  going to give you infinity when X is when T is   equal to a but this time we have infinity times  F of T f of T at a is just F of a right so of   course when we have F of T at a is equal to F  of a that's clear right that's the value of the   function at a of course but you have F of a times  infinity so earlier infinity was right here isn't   it and suppose you have F of a times the Infinity  so of course it's like another motor hole you dam   multiple just multiply by one that's all so  whenever you have this all you have to do is   put a into the function so actually just get F or  hey that's it again it's just like the Infinity   times y but that number is that's it and then you  integrate that so of course you know just take   that number to be the answer so this is like the  craziest part yeah well with this being said in   fact this right here is super easy because what we  had to do is notice this right here is the Delta   function and again this right here even though  it's going from 0 to infinity it's the same as   going from negative infinity to positive infinity  because everything is 0 only when a yes only 20   today right so earlier heart negative infinity to  infinity that's why I said doesn't matter doesn't   matter all you have to do is the following pull  a into the variable T right here and that's it so   finally we'll just have ye to the neck is hey  seriously that's it as crazy as it looks like   and just really hard to take the definition as  how D is because 0 times infinity from your calc   want how to classes you should know that it's  indeterminate meaning you have to do more work   in order to figure out the answer but we take  that to be 1 in this case all right pretty cool okay that's number nine okay now this is somewhat  crazier so we'll be doing number 10 and this time   I will be doing two of them for number ten so here  is number ten now class transform of F of T minus   a sorry wrong color Laplace transform of F of T  minus a and you see usually is just F of T but   when you have F of t minus a that's how you shift  the function in turn left foot right depends on   what a is and then not only this though we also  have a unit step function after that so this is   the first one from number ten meanwhile you might  be thinking that if we just have f of T right so   I will also have to show you what's the Laplace  transform of F of T only and multiplied by U of   T minus eight like this so it's some like the unit  step function business again alright let's figure   this out first to do so let's go ahead and use  our definition so integral from 0 to infinity e   to the negative T oh that which is f of t minus  a times the unit step function of t minus a and   then we have the TD of course again whenever  you multiply this you just have to observe   what's happening right here right so I'll draw the  picture one more time for you guys just in case it   you skipped what we talked about over you so this  right here we are trying to find the area from 0   to infinity under this function let me graph this  right here first and I would do that with flu so   yeah e to the negative T times that let's say the  function looks like this I don't know though e to   the negative T F of t minus a right so let's say  the function looks like that and again you have   to tell me what a is that say it's right there so  this is how the function looks like it's just this   now I will graph it to the negative as t f of t  minus a times u of t minus a so here is going to   be the change after we multiply by that and the  only change is that well anything before a it's   going to be zero so we have this right here like  that and then you jump to 1 and 1 times this is   of courses that so anything after a it's just the  original blue curve like that so it's similar to   what we did earlier and if you want to find area  for o to infinity of this you can just find the   area from a to infinity of the padule function so  that's pretty much the idea so here is going to be   the integral from a to infinity so I will actually  put this time blue why not a to infinity and then   the insides just going to be the following e  to the negative T f of t minus a that's it tgd   that guy yeah now how can we handle this though  we haven't done some the gentlemen integration   for while huh especially for this video let's do  some fundamentals namely u substitution so let's   go ahead and do you substitute but we use the  you already write here so let me use a V right   so I'll say V is equal to the input which is  t minus a and until V is equal to DT so that's   pretty good now take this integral to the world  well keep in mind this right here is T goes from   a to infinity now take this integral to the feel  world he will go to where studying where you put a   right here a - a with your zero and then you pull  energy right here infinitely - the final number   is during infinity so P goes from 0 to infinity  and here we have E and then we have negative s T   is photo well we know P is equal to t minus a so  of course that means P is equal to D plus a like   that so I will just put that down this is P plus  a parenthesis that guy and then we have F of P and   then P P and T G are the same so I can just put  that down right here so that's pretty nice and   now we will finish everything in the world P is  just a dummy variable and we just have to figure   it out well right here of course we can just  work down our little bit this is e to the power   of negative s t minus s a yeah now have a look  into this power it's a same as e to this power   times e to the power e to the negative a so that's  just a constant in the v-world so we can just put   that to the outside so this is going to be the  following let's write down e to the negative s   a first right and then for the inside here have a  look this is going to be the integral going from 0   to infinity and we have e to the negative fee and  then here we have this namely F of P DV like this what's this well here it's something really nice  in the front we still have that e to the negative   si but this ratio is well this is just nicely  equal to our definition of the Laplace transform   of F of T isn't it even though I'm using P V and  V right here but these just a dummy variable as   long as you have this this than that being the  same well is still co-sponsor to infinity this   is still going to be the Laplace transform of  our function which is f of T like that right so   this right here is it so this right here is just  translation formula so all you have to do is when   you have e to the negative a when you have the  function multiplied by the unit step function   if they have the same input all you have to do is  in the S world be sure you multiply by e to the   negative a and then you do the Laplace transform  of the original function and that's pretty much it   and now I just recall that I happened right down  this for you guys usually we can write this down   up is e to the negative a and this right here we  can put it as capital F of s like this right so   this and that just extend the notation the output  of the Laplace transform of F of T tot he is just   F of s I so you can just keep dying mine I'm going  to write this down right here though I think and   I hope this more clear this way so I will put  this down here e to the negative F a and Laplace   transform of better for you guys have he all right  so that's the first one I will show you guys two   examples don't worry but when not on your number  ten because what if the input right here they   don't match no good don't worry I got you so have  a look here well if the input don't match you can   do the following make the match first so I'm just  going to start by saying that so this right here   is the second question yeah kind of this like  camping file let G of t minus a be this right and why do you want to do that because now I can  definitely put this down right here isn't it so   here we go I will just write this down here for  you guys this is going to be the Laplace transform   F of T is going to me just write it down again F  of T and we have the unit step function of t minus   a this now I can just put it down as G of T minus  a so we can say this is the Laplace transform of   can we still see yes G of T minus a multiplied by  U of T minus a like this right and do we know how   to handle this yes we do just that by instead of  using f we have to use G right because I'm saying   G t minus a you t minus a so this right here is  nice to equal to you just do the formula namely   e to the negative s a Laplace transform of G of  T like this yeah hmm so that's pretty much it   but the promise that man I'm trying to show you  guys the function was f and now I have a formula   was G so what the heck is G right don't worry I  got you again if you look at this G of t minus a   which we set it to be f of T well all we have to  do is kind of ship it back because I just had to   figure out what's G of T in terms of F that's all  and to do so all we had to do is change the t to   t plus a you see t plus a minus h of t likewise  we've just had to do the same right here t plus   a so you will see G of plugging you actually  just get T and plugging you just get F of T   plus a so in other word this is the same as that  well finally you can just write this down right   here as e to the negative a times the Laplace  transform of this guy which is our F of T plus a just like this right well here is the power  behind the story if this right here is just   F of T which is not with the same impress  that all we have to do is the following you   still multiply by e to the negative 8 whenever  you have the unit step function just go ahead   and do that let's just go ahead to that but if  they don't have the same clue all you have to   do is do the Laplace transform of F and just  have to compromise it this is minus a you go   ahead and do f of T plus a so this right here is  yet you can we see yes all right so that's that   yeah and that's exactly why I just show you  right here right started two in one and now   I will show you guys two examples for the next  question so let's go ahead and do the following yeah and to be very honest with you guys I do  miss my students haven't seen them for a while and   hopefully everybody is doing well and hopefully  my students you guys are watching and you guys   are doing well too all right so tell number 11 you  haven't you surf a happy face yeah all right so we   have two examples right here for you guys first  one let's see the Laplace transform of t minus 2   square times sorry sorry I forgot the input show  period all right number 11 Laplace transform of   t minus 2 square t minus 2 square times the unit  step function of t minus 2 so in this case the a   is curly Rd equal to 2 right versus the second one  the Laplace transform of just T squared times the   unit step function of t minus 2 right which one  is easier this one slightly easier not so paddle   because again just the computational inside easy  it's not too bad alright all we have to do is the   following they do have the same input and the aid  right here is 2 so we have to do this first you   go ahead and multiply by e to the negative as and  the a is 2 so twist like this yeah and it's really   easy if you ask that rational questions and you  have like t minus 5 because you have to proof 5   right here and becomes e to the negative s and  then 5 so I usually you know try to avoid not   just 5 so this is a stuff too yeah that's pretty  clear anyway you do this part first and then the   next part is you go ahead and do that you shall  not pass of f of T which you focus on this part   just in order input just do can impress t only  so you have to do T squared like that right now   how do you do that well we did that earlier for  question number two already right so you have to   remember this kind of things I will write this  down right here for you guys we call that for   question number two the class transformed T to  the nth power this is equal to n factorial over   s to the n plus one so all we have to do is the  following here we have e to the negative 2 s this   is not 25 this is 2 s and negative 2 s and then  for this and these two two factorial is 2 so we   just multiply by 2 over 2 plus 1 is 3 so you have  s to the third power like that but that's pretty   much it so in the end of course you can register  in either way maybe you can read yes to e to the   negative 2 s over s to the third power like  that so this right here this is s right and   again this is meant to be 2 factorial by 2 Phi 2  is 2 times 1 which is the same as 2 so no worry   about that now have a look here how do you do  it the inputs are not the same anymore but we   can use this right here now so what we have to  do is first notice that our a is again 2 so go   ahead and do this which is e to the negative s  times 2 then what we are going to do is multiply   by the Laplace transform the function here is  T Square but this case you have to do t plus   a of the function instead of the function so we  will have to put t plus 2 so I'll just write on   T plus 2 inside of the function and then we have  to work this out how can we work out a lot faster   and small G + 2 squared oh don't worry just do  the algebra so here we go this thing is just e   to the negative 2 s multiply this out the usual  way so we actually get a lot class transform this   is going to be T square plus 2 of this that which  is going to be for tea and the nasty two squared   which is +4 like this yeah so just like this and  then oh no we can just do this please than that   by using that this case is when T is when n is  equal to zero right 20 to the zeroth power all   right so here is e to the negative 2 s see I make  the mistake again this is e to the negative 2 s   all righty square which is the semester I'll put  on the parenthesis for the result of the Laplace   transform here we have two factorial which is  just 2 over s to the third power double check good no it's right here so yeah obviously next  what is just four and then the Laplace transform   of T is going to be 1 factorial over T squared  right and it's 1 so on the top is 4 times 1 which   is 4 over s squared like this and lastly this is  going to be plus 4 and again this is T to this   roots power and 0 factorial is equal to 1 and then  put this over here you have X to the first power   namely for over s like this and usually nobody  oil made you combine fractions for a lot of last   transport so I think this is totally ok right  so just leave it like that if you combine the   fractions I don't think impress anyone anyway so  just leave it like that anyway number 11 that's it okay all right so now we will be doing some more  proofs that kind of things right so continued   number 12 here is the Laplace transform of F 80  F of 8 he please do not call this to be fattens   nanites you say f of 8 yeah alright so here is  the deal I am going to again I will write this   down here for now just make a note that when we  have the Laplace transform of our function f of   T this right here can Gradius capital f of s  right that's what we mentioned the earlier and   for number 12 if we have the Laplace transform  but the inside is not just a stop T but rather   is the Laplace transform of F of a T like this  first thing first yes the increased a T is just   a x of T but right here it's not going to be the  same as f of a s it's not like this this is not   true in general right and this is equal to one  but that's redundant so we actually have to do   this carefully and now here with how we are  going to do it carefully use definitions of   course so integrate from 0 to infinity e to the  negative T of F of T so I'm sorry again I forgot   to make things I read so like this and then we  have the right hmm so again let me just cross   this out no how can we make this happen though  I can u substitution so here I think the best   choice is to have the 8 he to be our u and we  haven't used you in this case over just you so   that u equal to 80 and of course it has to be a  legitimate number on that stuff and oh I forgot   let me just prove the case that when a is greater  than zero let me just say that a case where a is   greater than zero and yeah you see cleaner this  way yeah yeah anyway U is equal to 80 sometimes   it is not greater than zero you can use some  property saying thanks to anyway by the way   D you if you go to a little root of this which is  a DT DT is equal to D you offer a right and again   we have the condition a is greater than zero and  this is why we had to have that condition to make   our proof easier alright for our inter going that  you world it goes from 0 to infinity so that means   T goes from 0 to infinity yeah now let's go ahead  and take this integral to the new world well you   is the same as 80 a is passed deep and when T is  0 we get u being 0 and when he is infinitely you   put here a times infinity a is positive you get  infinity right so that's pretty much it and then   you just continue e and then you have negative as  T and T we have to stop that so let's go ahead to   that justify process by a we care T equals a you  - you obey you operate anyway so s is 2 s but he   is you over a and then here we have F and the  input is that you now and then did he is equal   to this so I will write this down here for you  guys this right here is my U and then a DT is   this so the T you right so now let's just go ahead  and do all the rest so here's the usual business   we have the 1 over a we can put that all the way  in the front and we'll finish everything that you   world so we have 1 over a integral 0 to infinity  they are in terms of you right it's pretty nice   now here is the deal I'm going to rewrite this  as follows is a negative and have a look let   me write this down as s all for a and then times  you right here times you on the side yeah right   and then this right here already has f of U and  then an outside here of course we still harder to   you so this is what we have now have a look one  of race all the way in the front so that's nice what's this dog well you you a new match and this  is integral going from 0 to infinity so this is   almost like our Laplace transform and you have  to remember this right here the definition was   interpoc going from 0 to infinity and we have  e to the negative whatever that we have here s   T and then F of T DT right you see F of s as in  that match and then a GTG match and we have two   negative and also the infinity is 0 on the start  here this right here is our new input in the s   world so all we have to do this put that down  into capital F so we have capital F of s over a   just like that therefore the Laplace transform of  D to add of a T is equal to 1 over a capital F of   s over a like this right and again I only prove  the case for a is greater than 0 you can go and   try it is less than 0 just you know somebody's are  part I could limits of the integration you have to   fix that you'll still work and sometimes if you  don't go through all this you can just use the   property of some function to make the naked eco  where if necessary by the way this right here is   it point number two all right so this is how you  change that oh by the way let me just do a quick   example for you guys how you can make this work  this is actually a little bonus I didn't prepare   data I'll show you guys how cool this is so real  quick example suppose you know this the integral   of sine T is equal to which is with the one on top  1 over s squared plus 1 squared so just like this   s is the sign is which the beyond what happened  this is piece 1 now if you know this this is what   we know and if you want the Laplace transform of  s of T DT well this right here is precisely our   F of s I have a look here is the S here is the  S yeah here we can use that formula this right   here is the number multiple early I used a G but  iam using VT because I can match with the formula   that we got earlier anyway what we can do is just  go ahead and do 1 over Y but this number is yeah   and then you put down s over P in here so just  go ahead to F of s over B so we get 1 over B F   of s over B is just 2 s over B right here so we  multiply by 1 over parenthesis as over B and you   square that guy and you plus 1 after that and  then you will see right here we can just get a   common denominator which is going to be P squared  so go ahead and do that and multiply on the top so   in get actually 1 over B B Square over this and  that is just a square this and that is B Square   and of course this and Arkansas doesn't this  look familiar yes it does it's just the over   s square plus B square so that's how you can use  this formula to generalize sometimes don't worry   about the constant multiple just put on one and  after that you can generalize it by doing that   okay that's it all right so numbers there think  and again the capital F of S is just the result   of the Laplace transform of the need to adopt  number 13 we are going to multiply by something   else Laplace transform of e to the HT e to the  8th so it's eat this is each f of T so here is   the tio you notice that when you have a Laplace  transform of a product two functions in fact   there's not a good formula in general only if you  know a function it's like a special one such as   this one if you have e to the eighty times F of T  this right here it's actually rather special and   I'll show you guys how can actually you know come  with a nice to know for this right so here we go   use definition so here we have integral going  from 0 to infinity e to the negative T yes and   then you just put this down right here e to the  80 times F of T and then we have the TG after   that of course now just do the integral pretty  much these are not we can put them together so   we get integral from zero cannot eat woody and let  me just read yes a minus s a minus s and we have actually no no no no no no well technically this  is negative s plus a the N times T yeah if you   factor out that he and you have like this right  good now this is what we're going to do right here   I'm going to factor out a negative I'll do this  this water right so I have the integral going from   0 to infinity e I was factor out negative and  then input will be the following to the stump   blue stencil it'll be better in Kobe s and then  minus a you can just work out algebra and then   you have that here to that and this is f of T G  do we need to use any use substitution the answer   to that is no we don't have to because again TT  and he match and this is negative and then you   have 0 to infinity already so just to remind you  guys one more time if you have the integral going   from 0 to infinity of e to the negative s T f of B  thought he did he this right here is nicely equal   to F of s yep this match and now you have this  so what's that that's all you include in the s   world that's it so all you have to do right here  is just write down capital F oh - hey like this   very very very nice so this right here is how can  translate in the s world earlier when we translate   it has to be in the T world and use the unit step  function and if you want to make any shipment new   or just have to multiply the F of T with E to the  18 right so that's the idea and now of course I'll   show you guys an example and I'll show you guys  two examples so let's do number 14 so the first   one I have this for you guys and then Val will  just give you guys another one because I think   I would give you guys two examples right first  one the Laplace transform T to the third power   e to the 2t I forgot to write number 30 T to the  third power e to the 2t like this right in fact   you can do this with two ways depends on which  one you take to be the original function f of   T well I'm going to do it with that way so that  means I will take this to be the e to the 8th he   of course I'll take this to PDF of T yeah so this  is how we are going to do it first of all if you   would like you can put this into that form it's  not necessary about to be for you guys anyway this   is the Laplace transform it will be 2 T and then  we have T to the third power right just like this   then what we are going to do is we have to figure  out what's the Laplace transform of T to the third   power first so I will just read this down here  for you guys now flash transform of T to the   third power this right here is you go to what you  two three factorial so that's put on some color   so yes sweet so it will last three factorial over  s raised to the three plus one power and what's   2 factorial 3 times 2 times 1 that's going to be  6 so this is nicely 6 over s to the fourth power   isn't it and in this case this is our s right here  ma attention this right here is precisely our F of   s right so what's the answer for this well this  right here it's just going to be f of s minus   2 all we have to do is instead of the s right  here you just have two stops at us - - because   in here too and all you have to do this six on  the top over s minus two in here and then the   great start to the fourth power so here we have  the s minus 2 raised to the fourth power and you   are done again this right here is precisely our  F or as minus two right here it's regular the two   where F of S is equal to this and then we know  F of S is equal to that so that's all we have to   do very very nice right so again another number  14 as a little bonus because I don't think I am   doing enough for the marathon so number 14 again  so put it a star next to maybe let's do another   one how about let's do a movie star what that  last transformed and let's do this I would like   to do e to the 3t and then let's do a cosine  one so that's the cosine of T like this well   in this case again we do the same thing first we  have to figure out what's the Laplace transform   of the cosine T actually let's do cosine of 2t  why not study more numbers more better all right   so that's going to figure out what's that plus  transform this part which is the last transform   of cosine of 2t and remember cosine is the one  that has the s down the top so this is equal to   s on the top over and you have s square minus P  which is 2 right so you have to square like this   so that's pretty much it two squares for pi i'm  not going to write it down because this is not the   final answer anyway here is yes so you know this  right here is capital F of s yeah and right here   you notice we are multiplying this with e to the  3t so what should the answer be well we just have   to do s minus 3 in the capital F that's all so we  chose how to make sure we end up with capital F   of - three so now here we have s here you have  s so go ahead and just tap s minus three here   over put the s minus three here oops wrong color  and then you square that and then here you have   the - sorry cosines with plus sorry sorry cosine  sweets plus right consensus with plus so you have   this and then in the end you just add two squared  which is four like this and then you are done so   just like that so this is not so bad right one if  you have to go backwards well maybe in the future   not today today is just a lot of class maybe in  the future next time was so I don't think I could   come with that Tony for that class transforms for  the inverse I'm not sure by the way this right   here I see it right maybe you guys can do it or  make that into a contest if you guys would like   to like in first up last transform was on the  other if you're interested let me know and let   me see what you guys related to all my stuff  and I'll try to work things out with you guys oh yeah next one is number 15 Alice are  looking what number what should I be doing number 15 here we have Laplace transform and again  earlier I said a function times f not a function   usually need to have a general formula but today  if you have a function as the f of T so that's a   general case but if you have this multiplied with  just T to the first power this actually gives you   a pretty nice result right this actually gives  you a pretty nice result so let me see how we   can actually make this happen of course this  right here if calling out here we shall use   the definition it's the integral going from 0 to  infinity and e to the negative T and you multiply   with T F of T DT or the stuff inside right okay  but in this case I don't think we can integrate   this nicely though I don't know so we have to  do this from a different point of view right   let me just write this down here for you guys  so let's see this is what we know again if you   just have f of teams that that's capital F of s  so let me put this down like this we know that let me write down which one should I write on  first I want to write down F first what should I   write on the claspers don't just in one stack just  be picky where it's not on other it's just pink oh yeah let me just write on cap that first that's  a really matter that much we know that capital F   of s this right here is just a definition of the  Laplace transform meaning the integral going from   0 to infinity and we will write this down better  ah okay I'm just being picky I'm sorry okay all   right this is just the definition of the Laplace  transform Intel quad coming from 0 to infinity   and we have e to the negative T and we have little  F of T like this right so that's pretty good now   let's make a computer this than that this right  here has an extra T that's all that's the only   thing that's happening right I really want to just  multiply the two inside what's the integral but   we had to do that really carefully because we are  in the key world in the integral right hmm I have   this T right here and this is negative s imagine  if I can differentiate this with respect to s if   I treat this as the variable differentiating e  to the negative T where s is the variable so we   just had to do the chandu and we get negative  T the front that's very nice that's how I can   squeeze out that he instead of the integral  right so that's actually how we are going to   do it we are just going to look at this equation  and you see the left-hand side is a function in   terms of X so we are just going to differentiate  this with respect to s like this right so on the   left hand side of course this is just going to  be the derivative with respect to s of our f of   s like this so that's just going to be some calc  1 derivative or whatever but only right-hand side   this is how we are going to do the derivative  of integral you have to use what we call the   Leibniz rule and sometimes that's also called  the famous taking of integration but anyway if   you do the derivative of this you will just have  to do the partial derivative of the inverse that's   okay that's true so let me write this down here  for you guys where you will first write down the   integral from 0 to infinity instead of putting  this on outside we are going to put this inside   the integral and we will do the partial with  respect to s and the reason will change to this   to the partial notation it's because the inside  here technically we have two variables by anyway   to the partial of this expression with respect to  s so I will write this time red we get e to the   negative as T and then we have F of T in order  to focus on that and then on the outside here   was to half that eg another focus on this part so  do this first that means we can write an integral   from 0 to infinity and then we still have to did  heat on the very outside now here we are taking   the derivative of this expression in terms of  the s in the S world so as is the constant T is   just the variable S is the variable XI is just  a constant there we go I'm sorry all right so I   can just write down the constant multiple first  which is the F of T no BTO and then Teddy the   Rope T of e to the negative T well me the root  of this always repeat part so we have e to the   negative T then use the chandu as how efficient  the area multiplied by DT root of this the review   of negative as T in the s world is negative T  so you just multiply by negative right here and   that's not wonderful isn't it because this matchu  is that very nice except for the negative sign   well what can we do with a negative sign this is  just a negative 1 we can kick that to the other   side right so happy dog this is showing us the  integral from 0 to infinity of positive T and   our actually just weather everything right this  e to the negative T to the T right here he F of T   DT this right here again we can keep the negative  to the other side so just have to multiply the   negative with the derivative in a swirl of the  capital of F of s like this so that's pretty much   it all you have to do is look at what f of S is  look at that first and then go ahead differentiate   that with respect to s right sorry let me just  write down better for you guys look here what F   of s is first and then depreciate is with respect  to s but not thanked yet because you still have   to multiply by a negative so that's it just like  that very nice now I would demonstrate an example   for you guys that's number 16 I'll show just put  it down right here should work okay number 16 is   equal to the follow a Laplace transform of T times  sine oh like this well if you're outgoing now we   can just run through the Laplace the top finish it  again but we will just use the result now here is   the deal you see we multiply this with t to the  first power so I just have to figure out what's   the Laplace transform of the sine of BT so I will  just rewrite we know F of s which is the Laplace   transform of sine of BT this right here is equal  to B over s square plus B Square I am NOT making   a mistake like earlier so this is good so this  is our F of s well this right here is equal to   the following we just have to go ahead and do the  negative and differentiate this with respect to   s of the function which is f of s which is  just this I'll just put this down half of s   yeah and of course this is the same same negative  differentiating and to differentiate F of S which   is that remember P is a constant is this the  variable we can write this as the following put   this up here with a negative exponent the clock  will start right so we can write this as B times   s squared plus B squared to the negative 1 like  this pretty nice huh now of course bring the power   to the front minus one and then that's ready  down negative times negative is going to give   you past if so that's good and then we'll just  have the parenthesis a square plus B Square to   the negative two power and of course don't forget  your chandu multiplied by the derivative inside   the routier of two that's the relative a square  plus B Square is 2's like this and now we can of   course just put everything down together this  done with ha so we have 2's with the be on top   again this is the S right over this right here  is the parenthesis a square plus B Square and   then raised to the square power on the bottom  like this so this right here is the Laplace   transform of T times sine of BT right just like  that very cool huh and now as an extension what   if you motor perfectly square yes I said just go  ahead just stand the idea if you want if you have   G Square here you to the second derivative and  you multiply negative 1 twice if you have t third   power just go ahead and do what differentiate the  three times and then you multiply by negative one   to the third power and all that good stuff  alright so that's pretty much it for number   16 and I were actually like to go back to  number of working for you guys a little bit arrested that not good actually let me just put  down what I have right here on the notes I wrote   this Tanis honorable mention so for the idea so  this is just the honorable mention well the idea   is that if you have the Laplace transform and  originally if you have the function right here   already yeah but if you have to multiply this  with T to some power and assuming any just from   0 1 2 3 4 5 and so on maybe they talkin to Sun  like fractional to row that here but that they   turn anyway though this right here all you have  to do is the following first do care what Laplace   transform of F of T s as a result just ready as  F of s yeah this is T 3 and power all you had to   do this duty and the rabbit here in the s world  like so I to the nth derivative in the s world so   let me just read it some nicer to you guys the  power over like that but you see each time to   knit a row but he not only you to need a robot  heap I also had to multiply by negative 1 yeah   so you actually have to do negative 1 raised to  power just make sure that you have the right sign   in all that stuff I saw this is my honorable  mentioned for you guys this is just doesn't   look nice sorry one better maybe alright so this  right here is just an honorable example no no no   no I'm not doing that I wrote to you I'm going  to do the following the Laplace transform of T   to the third power e to the 2t does this look  familiar he should that's question number 14   anyway earlier the way that we did it is we took  this as F of L F of T right that was actually easy   way now I'll show you gets the hardware to do  this now I would have to take this as my F of   T the little F of T and this is T to the third  power because here we have T to the third power   so that means I will have to multiply the negative  one three times and then I will have to find his   third row but here which is the jerk in this  case the third row but here oh the following   well let me just remind you guys let me write  it down right here if you look at the Laplace   transform of e to the 2t if you just look at  this party the 8th values right here is 2 so   this right here is 1 over s minus 2 s has to be  greater than 2 pi don't write that down too much   and this right here is precisely our F oh right so  to desert the rubber tip of F of s which is just   a 1 4 s minus 2 like this but again if you want  to to deter operty of this it's easier if you'd   write es s minus 2 to the negative 1 power and  then just enjoy your derivative all right here   we go the trick is to do this carefully with  the negative signs here you can negative 1 to   the first derivative so I will have just D to D  s to write to the first erupted heat right here   I'll bring that one negative 1 to the 4 minus  1 the Chandu is just 1 so doesn't really matter   so you get negative s minus 2 like this when  I minus 2 yeah like that so that's the first   derivative and then let's just have the negative  1 all the way near at next you do this again put   this to the front minus 1 this kind of fund you  just wander over here because you just did want   a rubber tip already here you again negative  negative becomes positive 2 and then you have   s minus 2 and that's negative 3 and then what  do you do next yes you do you again that's the   last one put this to the front minus 1 so that's  going to be negative 6 times negative 1 oh no it's   going to possibly 6 and then you have parenthesis  s minus 2 to the fourth power and of course we can   write this down as 6 over s minus 2 to the 4th  power like that right just like this very good   huh and of course don't do it this way don't throw  it this way this longer this way you should take   to the 3rd power as you are f of T and then you do  the shift right that's much better but you're not   sure you guys both right so no complex alright  so that was number 16 it's my honorable mention earlier we multiplied by T this time we are going  to fresh welcome to / T God and you can expect it   we will be getting into gross of course so let  me just write this down right here for you guys   alright so number 17 but I mean you actually  register reversed again Laplace transform he   actually let me register because it's really cool  you put it on like this first Laplace transform of   F of T this is just f of s good and as we all know  if you have the Laplace transform if you multiply   by G of f of T this right here all you have to  do is you depreciate in the S world of that so   you have the capital F of s right here but don't  forget you multiply by negative one and now have   you thought about what if we have to divide it by  T so let's go ahead and look at question number 17   here is the question what's the Laplace transform  of F of T let me put it down on the top and then   with the PI D PI instead so how will we get if we  do like that well this is like T to the negative   one power it's just that you do the integral huh  yes I agree but we have to do it carefully because   you cannot just integrate f of s because if you  do this you forgot the constant you forgot that   yes well no joke aside if you do this you may  end up with some constants yeah because this   is a definite integral and maybe you say maybe  just go from zero to infinity that's actually   not true so you have to do this one carefully and  this is how so have a look let me write down the   definition first this right here is the integral  going from zero to infinity and we still have e to   the negative T and O this I will just put it down  right here for you guys in red I have up T over T   like so and then we have the DT so that's exactly  what we are right hmm let's take bow it this is   the F of T and then we have the F of T here how do  we prove this earlier we differentiate this yeah   we differentiate that with respect to s where we  use the fineman's technique or the librarians rule   derivative of e to the negative as T that's how we  can produce an extra G that's how we get out right   this time we want to divide so maybe this guy  okay I'm not going to differentiate I should be   integrate because if I integrate e to the negative  T I will have to get what I will end up with D by   D by this right so that's the hint and I will  just have to make some observation and perhaps   that do the following let me put this down right  here so first put on some observation right if we   integrate e to the negative s just integrate this  part well what's the answer for this this is going   to be e to the negative as T that works that's  great but the reason I put down dgod forever it's   because of the following I'm trying to produce the  over T on the bottom so I'm not going to integrate   this in the T world instead I will be integrating  this in the s world so T is that why because in   the s world T is the constant so when I do this  integral in the S word we will have to divide it   by negative T so I actually end up with 1 over  negative T like this so we had to integrate this   in the s world really carefully we really kept on  that so that's nice though but the promise that I   was to end up with some bizarre things because I  will have the constant plus C for example right   so that's no good so we should somehow use a  definite integral to help us out so maybe it's a   good idea to still have infinity because he popped  infinite he put it here and you just need to write   down s has to be greater than 0 and that would  become 0 that's nice but I want to end up with hmm well let's see so I just integrate this from  0 no actually I'm integrating this in the as well   I get that and then he no cannot inter cannot  put this down this is 0 if I put on 0 again were   integrating in this world I will have to put a 0  in here sorry I should have said the infinity is   for the S right here because what we in this in  the S world if I put a 0 in the S that part goes   away I don't have to eat with negati as T anymore  so be really careful I actually have to integrate   this from s to infinity check this out right so  again let's integrate e to the negative T from   s to infinity of in the S world so of course you  draw into equation and we got that and then what   we do is we have to plug in plug in I will just  go ahead and plug in this in black I'll say plug   in plug in as two but the idea is that again we  are doing the s world okay now here is the piece   of art you can plunking s is equal to s s is equal  to infinity the promise that nobody does that some   people are okay with that but some people are  not what you had to do is right here instead of   using the s you will have to change that to a  different variable because the dummy variable   and that cannot be the same and the promise that  only at the end you want to end up with the S so   you just have to change the S earlier so earlier I  will just have to change the OD s inside there to   be you just say that you and the same right here  this is still going for ice to infinity that's   good but now this s it's actually you and again  this is just how you can be savvy technical and   this is again just the observation that were  making I don't want to tell you guys just do   that and you'll get the answer but this is the  thing that you have to do a little bit on your   sometimes to make sure that you can actually get  this to work my duet not continue you ghost use   infinity here you put it here and of course T is  greater than zero because originally this right   here T is going from zero to infinity so that  will take care of the business you push this   right here the first part will be 0 and then  you will have to - you put s right here so and   outwards 1 over negative T and then you have e to  the negative 2 put s right here ok s T like that   and all you end up with each will leave negative  as T over T this is so wonderful isn't it because   now this portion you can just have a intercooling  set that's the idea all right so why exactly do   we have to do with this though let me show you  so this is the observation it took us a while   but now this is what we know again let me just  write down the original let's see where is that   thing we know this yeah so we know the integral of  the Laplace transform of F of T is equal to f of   s this is what we know now for this right here  I can change that to the definition namely the   integral from 0 to e e to the negative s right  T and you have F of T and like this and this is   equal to the integral and you can have that to be  - you know that but really we want to integrate   and what to change the time we variable to be  you so again this variable and that they will   match so instead of putting down s right here I  will actually use U and then right here I will   actually have F owe you and this right here  it's actually tu because what integral is in   a new world and then no sorry not integrating  yet because the Laplace it's already a let me   do again all right this is probably not yeah so  let's put on the definition of Laplace transform   of papers which is the integral going from 0 to  infinity e to the negative s T and then you have   F of T DT and that's equal to F of s yeah so this  is just a definition and of course the right hand   side to s now check this out this s and s are the  same but in fact we are about to integrate and we   still want to end up with s to be nice yeah so  in started having the S right here I will change   that to be you to be you and again this is just  a dummy variable you could have used to fire you   one other stuff right now what we are playing  to is we are going to - what we did right here   we will integrate both sides from s to infinity so  right here I'll integrate this from s to infinity   and then from here are also integrate this on  s to infinity and again you see the output of   the integrating plaque would be in terms of U  so I will have that EU right here likewise the   output right here will be in terms of U so this  is again in the new world like this right so now   here is the deal we end up having to grow but  not so bad at all because if you look at that   he and also that you I will write this down here  for you guys let's say this right here is the T   axis and this right here is that you access yeah  can we see okay okay first of all T goes from 0 to   infinity so it's just this portion yeah I then u  goes from s to infinity and I say s is somewhere   right here so the region that you're talking  about is just this part so it's just like a   rectangle yeah the best part is that we can just  switch the water of integration and that's very   nice let me do that for you guys let me do the  integral from 0 to infinity and then the blue   integral from s to infinity and we'll have the  tu first and then we have the TT after that and   for the inside well I really want to have this  to happen so what I will do is let me just read   it down again let me just read on everything again  so inside here is e to the negative u and we have inside here we have e to the negative and then  we have F of T like this now the right hand side   is still equal to the integral from s to infinity  f of U T you okay only that hand side have a look   we are looking at the world first that means f of  T is just a while it's just a constant so you can   put that right here can I put it all the way in  a bundle because not to deal with time that's the   world so have a look all right so let me put in  straight here so we have two integral from zero   to infinity and then the F of T right here and  then we have the actually let me just put the   F of T in black so we have the F of T right here  and then we have the integral from s to infinity   and then we have e to the sorry e to the negative  UT e to the negative u T just copied it wrong e to   the negative T and then T U and T right and now  have a look we know this right here is equal to   this is the integral from s to infinity of f of  U the U what's this in blue this is exactly what   we came up with earlier so I can just put on  a result right here I'm not going to work that   out yet you can just you know look at that so  we actually have the integral going from 0 to   infinity and this is f of T and then we have this  put the whole thing so we have downwards e to the   negative T right so now we actually end up with  this so that's why it was really sorry about you   know that's one of the things what we used to  anyway this is equal to that we had a TT what's   this this is exactly what we're trying to find  yeah so I'll just put this tongue I know it's   like a triangular equal sign by the way finally  I'll write this down right here for you guys this   is telling you that the Laplace transform of F of  T if with the party pide here right if we divide   it by G because this pressure represents that  this is equal to the integral going from s to   infinity of F of U T U and this right here is  it very nice huh so you can see our assumption   was correct earlier when you multiply it by T  you differentiate x negative 1 and just how to   do it carefully when we divided by T we actually  integrate but you had to do it from s to infinity   and you don't actually multiply by a negative  whatsoever so have a look have a look have a look   pay attention to the observation especially that  part and it's just how we can make things happen   that's wanted to tell you guess all right so I'm  going to leave this on the board just remind you   guys because for the next one it's going to  be an example I can figure out crazy now class   based on that all right so of course that's  copper to black pen where pen events I used   to do sometimes all right number 18 right here  for you guys I'm pretty here let's look at the   Laplace transform of sine G just sine T right  and then I want to divide it by T oh my god first thing first perhaps I will write it down  right here we know that we know that what's the   class of sine T this is just sang of 1p yeah so  it's 1 over s square plus 1 that's all we know and   again we are about to integrate so I'm not going  to use s instead of the s right here we'll just   change to you right so here we go because we're  dividing by T right here all we have to do is we   will be looking at the Laplace transform of sine  T which is this let me write down 1 over u square   plus 1 for you guys so this right here is our f of  u yeah and then you go ahead integrate from s to   infinity so you look at this and you integrate  that from s to infinity and you are in the new   world and how can we continue just do the calculus  of course all right so what's the integral 1 over   u square plus 1 in the world in first tangent  so for this we just have the inverse tangent   of U and then we're parting as an infinitely  other stuff cloudy infinitely inverse tangent   of infinity is precisely power to let me let me  just write down everything for you guys inverse   tangent of infinity and then - just like this and  then - inverse tangent oh yeah this right here is   what PI over 2 as I said the earlier so we get  PI over 2 and a - inverse tangent of s look at   that how cool is this yes this is the Laplace  transform of the sign T over T very very nice   thanks to this formula some honorable mention  right this is number 18 and I'm almost done but   some honorable mention when you have that you can  actually end up with some very nice integrals let   me show you remember what's this this right  here it's a cemetery integrating and just to   cut the whole thing inside as your function this  represents the integral going from 0 to infinity   and we write down e to the negative T and we  have sine T over T like this right now with   this being done you just have to tell me ask value  you can put it here it can come with a lot of cool   integrals so let's see what's the value of lemon  juice tear gas on abroad mention number one what's   the value of the integral from 0 to infinity of  sine T over T debt well this computes that this   is the case when s is equal to 0 so I have to do  is tell you guys this is equal to 0 and then put   0 into here and work that out so this right here  it's going to give you PI over 2 minus inverse   tangent of 0 and that's 0 so on y'all we end up  with PI over 2 very nice isn't it and of course   you can come with another one such as what's the  integral from 0 to infinity if you end up with   something else that's crazier let's say let's  do this e to that he to the negative T e to   the T actually let's put that that's just Twitter  and then we have sine T over T tt what's this dad   well in this case s is equal to one and can s be  negative one no P capital right if s is negative   one this won't converge so this wouldn't make  sense so you still have to do this with culture   if you do that be sure you have to make sure that  s has to be positive and so they mean just credits   down here for you guys s has to be positive by the  way that's not ruin the fun s is one in our case   so we're just plugging one into there so we get  PI over 2 minus inverse tangent of 1 what's the   inverse tangent of 1 well here we have PI over  2 this right here is minus inverse tangent of 1   is just PI over 4 and get our common denominator  2 & 2 so you know it's 2 pi minus pi is PI and 2   times 2 is 4 and now match so it's just PI over  4 so yeah we have another one very cool huh so   thanks to this formula installed cool isn't it I  have another one for you guys number 3 on there   Paul mentioned so let's look at the integral  from 0 to infinity actually let's do it like   this let's do the negative infinity to positive  infinity and here I would like to ask you sine   of e to the X Oh God hey this does not look that  any of that how can we do this well do our usual   calculus stuff let's do some substitution yeah so  have a look here maybe that's do some use up that   u equal to the input function which is e to the X  and you see T you will see the same as e to the X   DX so that means DX is the same as T u over e to  the X don't worry you see after the substitution   you see that G u over u so if you put it back have  a look in the X well X goes from negative infinity   past infinity take this integral to the New World  well when X is negative infinity put it here e to   the negative energy is nice to equal to 0 to  infinity here e to the infinity years so you   goes from 0 to infinity like that ok inside  here we have sine this is our u so assign u   DX is T u over u so I can put down tu over you  does this look familiar now yes it does it's   just nicely equal to PI over 2 isn't it so as  you can see this is the reason why we say the   10-meter robot as a matter because the outputs  just contain number so is it TT and he you you   and you and that's not so let me match this  right yes the same as that which is PI over 2 PI over 2 very very very very very nice right  so some really cool stuff if you know some up   last transport some very nice intercourse I can  also do for free very good so let me finish the   things and then perhaps Aki because my other  unwelcome mention later on but processing tell   you see if you have co-signed oh right if you  have cosine T over T in fact that the past does   not just because the integral is actually type  project by if you have one minus cosine T over   T that actually will give you a convergent  integral and you actually get nice one as   well I'll do that all the way at the end just  a little pompous but this is really cool right all right we are going to be taking Laplace  transform of a derivative so have a look here   number 19 on the board for you guys number 19  here we have the plas transform of F prime so   it's no longer just the original yes DITA robot  here all right all right all right use definitions   of course so here we go this right here by  definition is the integral from 0 to infinity   e to the negative T times F prime of T and then  yep very nice persona it's not so nice because   how can we deal with this empowering to code us  that I wrote it well I want to have the original   because you have the original we know that's just  the original capital F of s this is F prime so   how can I give to F of T from F prime will have to  integrate this guy yeah you see where to integrate   just one part inside the integral so what should  we do with that go ahead and differentiate one   part to be integrated the output part the other  part would be differentiated so what should we   do yes integration by parts and yes di method so  have a look oh my god this is so much fun d and   an i and now put on the plus minus just two rows  is enough don't go too crazy of course we want   to integrate F prime of T because if we do that  sorry that's a horrible F because if we do this   we'll just get the original F of T very nice and  of course in the meantime that differentiate e   to the negative T and don't forget we're in the T  world and don't forget the chandu the derivative   of e to the negative T is e to the negative T and  the channel says x negative s because negative is   the constant with the teeth just like that all  right now first part is just this times that so   this is equal to this times that which is e to  the negative T times f of T this is the first   part of the answer but remember this is a definite  integral so go ahead and plug in plug in from zero   to infinity yeah and then we are - - we still have  to do the part right here and this is still going   to give us and integral so we have to actually  do - times - oh it's actually a plus so I just   put down right - times - this - ingre first times  that - simplify becomes a plus very cool anyway   and I'll put everything except the integral so you  have the integral and then I'll put this and that   is e to the negative T and then we have F of T DT  and this is the comment from zero to infinity and   you might be you know worried about that didn't  put on yes well I can put on s here passing see   asymmetry world s it's just a constant allow me  to put on s in the front right so have a look   the hardest part to teach the past tense point  is - radius anyway could I ask in the front so   that's the s in blue now we know a few things  first when we plug in infinity to this right   here because in the beginning of the video which  is the pollak towards our course all assume that   f is with exponential order so that means if  you put infinity to this it would be zero so I   will just have to write this down again for this  down f of T suppose it has exponential water so   when we do that the first part will be 0 right so  that's nice then of course that's the first part   we are going to - put a 0 into here and there  so we have the rest is just e to the negative   we are doing that in the humor so you put the  zero into the T so it's negative s times zero   and then times f of zero like so right and then  we add again here is dice but what's this part   actually although this is yes here we go what's  this part this is our definition of the Laplace   transform of D to F very nice huh this right here  is nicely equal to capital F of s of course you   can if you like if you'll read a Laplace of F  of T but yeah you prove that anyway and we're   almost done just like clean things up notice  this is just what one right and this is minus   F of zero so be sure that you do the following  and how do I write it or I would write it down   like that so okay so our fitting everything  here finally we have the F oh yes - E - f so   you have to refer back to what's your little F  it's alright so you just look like a functions   and usually you do this when you're solving  a differential equation initial value problem   actually so usually this would be given and you  have F of 0 that it's for this and we are done   very nice right and again F of s if you like this  is the Laplace transform of little F up like that all right so that's it now after the first the  robot here what should we do sure let's go ahead   and do our second derivative hmm the C number 20  a plus transform I want to figure out what's the   second derivative we can do this again and just  have to integrate No twice so you have three rows   of the stuff but life doesn't have to be that  hard sometimes we use we have that already so   use that make good use of our results so this is  how we are going to do it so I put this down in   blue first this is what we know right so have  a look here we know that the Laplace transform   and F prime of T let me write it down like this  this is the same as saying we have F of T right   and then we differentiate that that's how we get  priority good and this is equal to well you see   F of S is equal to that so let me put this down  right here for the F of s so this is equal to s   in the front and then F of s is the same as  that so that's why properties wrote it down   over there and then the input here is f of T very  cool and then in the end here we have little F of   zero that's a - little F of zero just like this  this is what we know and now let's make the best   use of this now what's the connection between  second truck in the first derivative yes you   just differentiate the first derivative again so  have a good happy dog so now if you want to get   the first if you want to get a second term over  here let me write it down better for you guys sorry now so let me just literally put on now  if you have Laplace of the second the reality   of T like this right this is nicely equal to the  Laplace transform you just have to be appreciate   the first derivative so as you can see all I did  is just rewrite this f double Prime the same as   that Prime and the Panda again yeah now check  this out I thought he little puppy little T e   to F of zero this this and now they are the same  time now my kind right here is f prime so next oh   I have to do is the following here as two FS and  then I will just do the Laplace and then this is   just going to be this yeah so put down F prime of  T very nicer and then minus instead of F will put   on F prime so we have F prime of zero like that  and this is so wonderful have a look this this   and that are the same very very nice isn't it  now do we know what this is of course we do is   the same as that so we know the following I'll  just put this down for you guys this is equal   to the following s in the front but for all this  part we can just write it like that which is with   a parenthesis and then we have the s capital f  of s minus zero so this is for that so f of zero yeah and then in the end we have the minus F  prime of zero now finally we just do some work   and usually we put down the half point first  before that and that's why I did actually I   didn't do that nevermind so I'll just write  down everything for you guys you distribute   the s you have s square f of s minus s little  F of 0 and then minus F prime of 0 like this   Congrats this is our Laplace transform of the  second derivative of some motion very quiet okay so twenties are us transportin  and then four more to go so earlier we did a first and also the second  derivative and now what should we do yes we   shall do the integral so number twenty-one right  here the best for right here so number 21 let's go   ahead and do the Laplace transform and instantly  recap oh is the integral going from some number   to T right some was responsible to T so you have  to tweak a body this is the integral going from   zero to T and then here I'm using C as my Tommy  variable so I have F of T DT and then that's   something cool to what I don't know yeah can  we use this again this again I think so because   the rub t think the cooking cancel half a dog so  that's to what we did earlier so let me raise down   here for you guys this is what we know earlier  when we do the Laplace transform of the derivative   of some function and I wrote it down like this  right this right here is equal to s times the   Laplace transform of F of T and a minus F of 0  like that yeah now have a look this is my input   let's go ahead and put this into here here and  also here how do I do that I will show you don't   worry so our function here I will change a little  bit because the notation is going to be this is   the F already so that's not good I mean actually  put down G if you don't mind right because in our   case here this right here I'll just cut out  to be G of T and again this is not a capital   F whatsoever because capital F is reserved for  the output of the Laplace of the two F so that's   why I'm using G so in this case I will just say  the missive I can put that down so in this case   I'll just say that right G of B that and again you  into a quick wipe a little F is plugged into your   parking key the output is it with T so G of T and  you get the integral from 0 to t f of v TV and of   course that go ahead and plug in to everything  so we are looking at this right here G of T is   just an integral so it's just what we are going to  do integral from 0 to t f of P P P prime that and   then close that and then s naught plus transform  of the integral going from 0 to T and then we have   F of V DV good - and I'm sorry this right here  I'm using G G so this radius also be G and should   be G of 0 ok the reason now try this down I'm per  proton G of T like this on purpose it's because G   of 0 it's this well all we have to do is putting 0  into the T you see here is where that he is I just   have to dance you were right here so G of 0 it's  actually integral from zero to zero of that F of   P TV and that's nice T equal to zero so this is  the gentleman D equal to zero which is very nice   right so just wanna show you that this part goes  away because of how we will how do you take it a   review of this the rub deep and integral cancel  by the fundamental theorem of calculus part one   so I have a look the left-hand side they can so  you put a T right here though how can you put it   inside here though so we actually end up with  F of T and that's it right that's it and then   on the right hand side of course you have the  s times the Laplace transform of the thing that   we are looking for so we have the integral going  from 0 to t for P P like this and finally what can   we do great onion so of course I want to figure  out this all i have to do is divide the s sample   size so finally Laplace transform of the integral  from 0 to t f of T DT this is equal to the Laplace   transform of F of T and then on the outside right  on the outside the tip ID pass so you can put this   on OS and of course if you ally you can write  us f of s divided by s after you whichever one   that you like so one way or the other just like  that very cool now next it's another Laplace of   intergroup bias with a star that's a new kind it's  a special kind of integral how the tongue fiducia pop quiz for you guys do you guys know what does  the word Laplace me if you know leave a comment   down below right and question number two what does  do a composition me if you know if I comment down   below alright so number 22 right here here we go  here we have the Laplace transform and input here   is what we call the it convolution namely F of T  and you do the star right here and with another   function GOP like that now here is the idea this  is actually now regular multiplication if you have   a Laplace transform two functions the functions  has to be special enough like the ones who have   done either e to the AG were T to some power  so you can doubt with nice formulas but if you   just have a general function stack the plastron  spoon of L and T time Sankey whatsoever there's   no proper rule for the Laplace unfortunately  however if you have the composition inside   it's a different story you actually end up with  something extremely nice in this case not sure I   guess what is but of course the first thing is to  tell you get the definition of the convolution so   write this down right here this is Cody come for  just this is called account for Lucian and again   F of P star GOP by the way that's called a star  that's why I say stop this right here thank you   going from 0 to T so it's kind of similar to the  one with the early ok but not exact inside yeah   F of T minus the dummy variable T and then motor  time this is the regular multiplication and then   G of B TV yeah so just for fun for you guys right  just for fun question for you guys what's one dot   one question - what's one start one huh leave  a comment down below and let me know if you   know yes for that in fact yeah if you have seen my  videos in the past thank you so much alright so I   did that in the past off I would include a link to  the video in the description for your convenience   oh my god man it's not funny because Laplace is  integral and now we're putting it not interpret   there we'll try our best first Laplace transform  is the integral going from 0 to infinity e to the   negative as yeah and then we'll put this inside  and I might as well just put on the definition   right here already so I will have integral going  from 0 to T and then we have F of t minus V and   multiplied by G of T DT so this is the composition  and then on the outside we still have the TT it's okay it's just stopping they're gonna see  what we can do first thing first when you go from   zero to T about Hyup idea we don't we don't like  that we prefer to have red handles right so now we   can just change the water of the integration  that's much better if you want to change   integration right now leave a comment down below  how you care but I'm not gonna do that now I will   lie to her this going from zero to infinity as  well because though that can actually end up with   a really nice that plus for the function f and  also for the function G you will see now how can   I make the happen though hmm one hint right here  is that we have f of t minus V have we seen that G   minus V s input already yes we have where yes the  unit step function so the see what we can do with   unit step function right well in our case here  if you would like you can let's see how should I   let me just focus on the red part right here so  I am just going to graph this power here just a   report I so I will just do it in Graham blue okay  so I will keep it as a picture right here suppose   here we have f of t minus T times G of P so this  is our function in red and here remember we are   in the v-world sophie is the fee is the variable  i don't know how it looks like let me just give   you a sketch so maybe it looks like this I have  no idea really all right so let's say this is how   the function oxide now here is the deal I would  like to multiply by the unit step function well   V is the variable t is the constant and let  me just remind you guys when we have the unit   step function this time table because they both  look like they're opposed to me so I've already   done like this when we have number sorry let me  do it like this first home we have a variable - hashtags the number I just easier that way  right this right here produce the following   why ever done and number is right this is why I've  read that number is anything before the number is   zero anything after that it jumps to one I thought  this is how it goes if the in size path deep is   one so variable greater than the number is one so  just like that however if you have the unit step   function if you have the number minus the variable  this will keep you the other direction so what   that means is that anything so that's a here's  the number here's a numbering read anything before   here's one add anything after that is negative  one so be really careful with this now here is the   deal pay attention to the red integral we are just  going from zero to T so with this picture right   here we just want to go from zero to T and again  T is our number in this situation and that's the   integral that's the area for the integral that's  all we need with only anything for work so we   actually had to multiply by this kind of situation  but it's actually so wonderful because you notice   that he is the number B is the variable in the  v-world like you're right so have a look if we   have this so I'll just put it down blue if we have  F of T minus V G of B and we multiplied by U of T   minus D and again this right here is the variable  and this is the number so when I multiply this   with the function here what happened what happens  is that anything before in main tanks so the blue   portion is no domain names and and you see after  that becomes zero and zero times that will just   keep your zero like this so the blue portion is  for the blue function right here so what I'm gonna   do is I can multiply by U of T minus V and then to  get the same area we can't you just go from zero   to infinity everybody happy so now this idea right  here really it really well okay so let me erase   this because we will push any more space all right  first thing first copy down the things that we   have so this is integral 0 to infinity here is e  to the negative T and then this is the integral of   going from zero to something let me not put on yet  I already start get F of t minus V G of T and then   let me multiply this by U of T minus V because  I'm doing this from you front because I multiplied   by U of T minus V front here none worth zero so I  will change that to infinity so that's really very   nice and then just continue we have the TV and  then we have that TT and everybody happy right now perhaps we'll just clean things up a little bit  I'll do this in black and red I will first put   this inside and then we'll switch to water in the  gross I will write on this two steps right first   integral in black and the integral in red they are  both going from zero to infinity I'm just going to   put this inside so we have the e to the negative  as T and then we have the red part which is the F   of t minus V G of T and then of course that keep  the blue blue of course you of t minus V and then   we have the TV and then we have that all right  I'm just putting everything inside and now I will   switch the water of the integration well what's  that being said the Ranger cuoco first bottom that   heat at the integral in black will call next and  then what still write down everything in second   actually no let's let's do this in your head  right so let me put the brain tech one outside   so we have this and we have the BP on the outside  right and then I will have the T T so we have the   integral from zero to this now we have everything  right here you can put it down right here but in   the t world hey gia V is a constant so we can put  that to the front this one has to stay because he   has that he this and that they also have to stay  because he has that he so I can put a G of V in   the front right here sorry G of T first to your  fee and then the integral from zero to infinity   and then the rest I was just ready time black e  to the negative T and then this is f of t minus   V and then u of t minus V like this then I'm  doing the TT first and then the P after that   just like that very cool unit step function and  I think we have done this already yes have a look   here we have the integral in red yeah that's zero  to infinity and this is G of P good what is this this is the Laplace of this function isn't it yes  yes yes so I can write everything down right here   as the Laplace transform of F of P minus P U of  T minus P and then on outside we have the T P I   should stop using my pen otherwise my hands are  just going to get really dirty so this is what   we have very nice huh all right so in the front  we still have the bread integral going from zero   to infinity and we have G of T do another class  of this yes we do well in this case though V is   a constant black in the black right here T is the  variable so V is the key is the constant so it's   pretty tricky so be sure you just kind of keep  all this little things in mind so we have the   F of T minus P yeah it's the constant earlier we  were doing f of t minus a times U of T minus area   and how it get is the following we get e to the  negative P and then you have the s yeah and then   we have the rest it's just F of s right because  the input match so just have this right here and   again I'm just using feet in 38 so same thing  and then on the outside we still have the TV   so nice huh because now F of s is just a constant  in v-world I know it's just really crazy but it's   really fun you can put the F of s all the way in  the front and then you look at that real part so   here we have the integral going from 0 to infinity  and allow me to write this down first which is e   to the negative V s and then this right here  is actually the mean red yes Espie yeah let me   read yes sv and then write down G of P and then  we have TV and do we recognize so this is V V V   match and then it goes from 0 to infinity this  right here is precisely the Laplace transform of   the G function so finally we can just write down  F of s and regular multiplication of capital G of   s like this this is so so so wonderful and of  course if you like and ready dance the Laplace   transform of D thought he little F of T did hold  multiplication right just a regular modification   and compares your type you like and this is just a  plot transform of little G of like that so one or   the other right just like this so so so cool this  is one of the most wonderful Laplace transform   and yeah so so so nice right all right and again  don't forget to comment what's one star one right all right so let me show you guys eat samples  okay I'll just let me finish number 23 I'll   keep you guys tweet samples okay what's the  Laplace transform of sine Riggleman regular   multiplication 1/2 cosine T versus the other one  Laplace transform of sine T star which is the   composition and this is cosine T right just like  that in fact I'm not going to say this is question   number 23 this is just a Honorable Mention right  so let me just say this is the Honorable Mention   right so Laplace transform sign he co-signed  this is so not pretty sorry sign he co-signed   he / sister all right which one is easier this one  they mean - this one first this right here it's a   convolution we can use the convolution theorem  that we proved the earlier this is the same as   a to the Laplace transform of the first function  and then you multiply by the Laplace transform of   the second function which is cosine T like this  what's the first one well well this is just 1   over s square plus 1 squared what's the second one  it's just estimate op over s square plus 1 squared   final answer s on the top we are multiplying so  it's parentheses s square plus 1 and then square   that one square is 1 so that's right at the end  put on square then how can you do this this is the   regular mode two-pager it's not easy but it's also  still doable so notice that this is just a sound   right here for you guys when you have sine cosine  you can look at the top angle formula sine of 2t   this is just to sign cosines right just like that  very good now here we have just one front so I can   justify deposits by two so here is the deal I will  just write it down so this right here it's a same   as saying the Laplace transform I would divide it  by two so I will have 1 over 2 and then we have   sine of 2t just like that it's the same so use  identities sometimes now 1/2 is 1/2 good and then   sine of 2t well we do this a few times already  yong-ho top right so we have the two on the top   and then over s square plus 2 square like that hey  guess what listen that can so pretty cool so final   answer we get 1 over s square plus 4 cool all  right so this is how you could use the convolution   and you have to just really make sure if your  competition will regular multiplication and all   that good stuff now just two more to go so let's  go ahead 23 Laplace transform and now here I have   square root of x okay this right here it does have  a lot of fast transform and I'll show you guys   what the answer is you can use the definition and  then you can do some use substitution and then you   can use the Gaussian integral and so on so on so  on but I will take a shortcut right here right so   change I can just show you how can extend the idea  of the factorial so know the following remember we   know that the Laplace transform of T to some power  n this right here it's equal to well pay attention   to what n is this right here is equal to n  factorial that divided by as to the plus one power   like this and the truth is this right here doesn't  have to be just 0 1 2 3 4 and so on you can   actually have a lot more numbers this right here  if you like you can have the Laplace transform of   T to the Arts power I'm using all right here and  we will say this is our factorial over s to the R   plus 1 in this case though R has to be greater  than negative 1 and what I'm going to do is   actually the following not yet I'm sorry all right  that's that's that's put atoms are it or has to   be negative 1 greater than negative 1 right so no  no no no all right these are not almost identical   that for we are going to say we can actually use  what we call the gamma function to extend the idea   right here so that's what we have and another  notice let me just show you this right here oh   yeah so this is called a gamma function let me  just put this down gum function and it's like   hangman of course like this gamma of some input R  this right here is equal to the following a plus   is what is integral what gamma it's also integral  you have the integral going faster protein nothing   I'm not making this up I don't know who did well  anyway nobody did it legitimate math so you start   with T / stop and then you to R and then you have  minus 1 and then you do e to the negative like   this so this is what we call the gamma function  and the good thing is that you can say the gamma   of and you can also have the PI function but let  me just use comma right here the gamma of our   plus 1 which is precise you put our plus 1 here  and just work that out and you can just say this   right here is equal to n factorial R factorial  the mere creature spread is done 0 to infinity   and let me just you just triggers T to R plus 1  minus 1 just are ya and then e to the negative T   DT and again this particular one it's also called  a PI function and just cap the PI of our right so   I just don't know it's accounting class the PI  but this is not a PI ap it's just a capital PI   function and the very nice thing is that you end  up with r factorial but is very nice and again in   order for this to work you will show technically  say or has to be greater than negative 1 ok so   just to make you slightly more legit by the way  you can do the following our factorial is equal   to this so you can register as gamma of R plus 1  and then over as to the R plus 1 like so very nice   right and the best part is that you can use this  as your new definition to extend the concept of   factorial and you can end up with the following  if you have RS equal to 1/2 right then you get   1/2 factorial and again you can just put a 1/2  in here and that's pretty much going to be that that's pretty much how you're going to start with  the Laplace transform the square root if you want   to do the definition but I have to have a video  on this so you can go ahead and check that out   it's a famous result if you have seen me already  1/2 factorial is nicely equal to square root of pi   over 2 so this is a very nice result it used the  gamma function so with this being said this is how   we can do it now I'll just put it down here the  Laplace transform of the square root of x sorry   no not squared of X I'm not using X I'm sorry I  squared yeah this right here of course it's the   same as saying the Laplace transform of T to the  one-half power and we can just put a 1/2 into here   if you wish so in other world can say this is  the 1/2 factorial again 90 you choose cents you   have to use a gamma function to actually work that  out divided by dou s xx 1/2 plus 1 and now based   on what I told you guys right here this right here  1/2 factorial is just nice to equal to square root   of PI over 2 yeah so cool PI over 2 and then s 1/2  plus 1 it needs no introduction is just s to the 3   half like this so this is it yeah this is it so  for this one let me just show you guys what the   gamma function is and then it actually works out  very nicely pretty cool stuff just like that 3   hours not so bad oh my life goes to run a marathon  26.2 miles under 4 hours and my latest one which   is which happened in two weeks ago right there a  marathon the meadow I showed you guys last time   take a guess my time that take a guess my mirrors  on time I post my I post a picture on my Instagram   all that so you guys know you know it but if you  don't you can take this all right question number   20 what is this really question number 24 and  then Laplace transform off I start with an L so   of course now and haha an actual log of T crazy  stuff huh so let's have a look this is going to   look like the Interpol conference row to infinity  e to the negative T and then we have to have   the Ellen you might do it a few ways you can do  this from scratch with integration like this and   working with you working out power to it with the  gamma function that I just showed you guys earlier   and man I forgot a comma function it's somewhere  anyway that here we go so this is what we know so   they mean to us put this down right here and let  me actually just start with what we had earlier   we know that let me put it down right here any  more space it should be okay we know that just   I vomited earlier the Laplace transform of T to  the art power I will put this down in red why not   I mean black why not so like this right this right  here is our factorial over s to the art class one   power but our factorials SMS gamma of R plus 1 so  Buddhist honest gamma of our plus 1 over s to the   R plus 1 isn't it nice and right here of course we  can recall the definition of the class transform   this right here is the integral going from 0 to  infinity e to the negative T and then we have P to   the earth power and then we have and that's equal  to this of course so gamma of R plus 1 over s to   the R plus 1 power good stuff now compare this  and that and to squeeze out an LNG the thing about   the relative because with the a temp out really  similar to this already right earlier if you want   to squeeze out at he we can just differentiate  this with respect to s so can squeeze out a G but   now I'll just squeeze out there and T I'm not okay  and this part anymore actually have to look at   this part because when we have T to the art power  if we differentiate this with respect to R then we   can squeeze out Ln T and that's very very very  nice so let me write it down for you guys then   just put on some note right here yeah so recall  that if you differentiate your function as a DDX   some number for the base that jet pays to X power  if this is the variable then you get B to the x   times and B so just keep that in mind so it's  tying my I will go ahead and look at this and I   will go ahead and differentiate this with respect  to R yeah just like that and training for this   show put on the parsha OPA I will do that inside  anyway so here we go I'm going to differentiate   both sides this is the key function in terms of  our Isis the constant anyway so when I put a sign   inside that's the time change that partial zero to  infinity and then we have the partial with respect   to R and then the inside is e to the neck yes t  e to the R and then we have the T and then again   the right hand side were just go ahead and do our  usual derivative let's say I do s is the constant   we know already anyway here we have the gamma of  R plus 1 over s to our class 1 all right here we   go this is just a constant so it will stay we will  have the integral going from 0 to infinity this is   staying so it's just e to the negative T yep then  literati with respect to R of T we are we get this   so I'll put everything down red we get e to the R  times Ln P like that and then we have the PT very   good now the right hand side just go ahead and do  the derivative by using the quotient rule because   I don't the party on the top they both have R so  now here's the deal on the bottom I will have to   square it so as to the R plus 1 and then I will  square that and then on the top I will put on the   first function on the bottom function so it's s  to the R plus 1 times the derivative of the top   and for now I will just write down gamma prime of  R plus 1 what's the chendo says the channel says x   dt roof inside but the derivative of our class was  just 1 so it doesn't matter then you continue -   the top function which is oh man gamma of R plus 1  times the derivative of the bottom with respect to   R yeah so you first repeat as to the R plus 1 and  times the root of this which is again just 1 so it   doesn't matter so this right here is what we have  this is what we have now compare this and that   what can we do I can just make R equal to 0 don't  want your by yet yeah so I would just say that R   equal to 0 in that case the left hand side give  us the integral from 0 to infinity just e to the   negative as T and then Ln T DT which is exactly  what we are trying to get now on the bottom here   we have s to the 0 plus 1 to s 1 and then to a  second power oh no we have s square on the bottom   good put your own here we have s to the first good  and then we have the the relative of the gamma   function at 1 and then minus gamma 1 oops just  make it one more kiss tick and then over n times   s to the first power like this so that's pretty  good now what's gamma prime of 1 and what's gamma   1 okay so let's see here note so remember gamma  off and because we're talking about functions so   perhaps I'll talk about X yeah our wiper actually  that's just put the arc because that's what we do   so gamma of our car yeah earlier though we said  gamma of R plus 1 is equal to R factorial right   and you can see that I can just push the Rho into  that so this is actually pretty that's that's to   this first no let's just do this first why not  sometimes anyway I'm sorry note gamma of R the   definition of gamma is going to be the integral  from 0 to infinity T to the R minus 1 and then oh yeah and then e to the negative T DT like  this all right so this is what we have a   belief that's yes good all right so gamma of 1  in fact this is how we can argue that this is   zero factorial how can a good stuff actually one  you can put on one right here this is integral   from 0 to infinity T to the 1 minus 1 power so  that will be 0 actually and then you do e to the   negative power and you can do this in your head  integrating e to a negative Y because this is just   one anyway and then you could negative e to the  negative T plus infinity plug in 0 worked out you   actually end up one so this right here is nice  - equal to one and in fact that zero factorial   now let's look at a gamma prime of R so to do  that I would actually have to do the derivative   right here and again we will technically how to  differentiate this with respect to R so use the   like nice rule right so we actually end up  with the following so the action went down   here so gamma prime of R equals okay still have  the integral going from 0 to infinity this right   here stays because teachers are constant in the  art world we are doing this with respect to art   so you have this and we did that earlier so if  you depreciate that you actually get this repeat   which is T to the R minus 1 times L and T you  actually have this and then you have this right   here stays which is e to the negative T like  this right so if you have the gamma prime of   1 when R is equal to 1 you will end up with  the following put a 1 in here G to the 0 is   just 1 so it's nothing there and you actually  get integral from 0 to infinity I'm so sorry gamma prime of 1 this right here is the integral  going from 0 to infinity this right here is just   hit with 0 which is 1 and you end up with  L and G times e to the negative T and you   just want to know what the answer to this  is this right here so cool this is also   called gamma this is the oldest constant  gamma right and let me just show you this   is approximately 0.577 and in fact I'm not  going to do this integral right here for you   guys I'm sorry but of course you guys can go  ahead and check out dr. PI and check out dr. Kyon's video I will have this video in the  description for your convenience so although   this right here it's actually a special  number called gamma if you haven't seen   that before very cool stuff so right here  I will actually come here and tell you this   right here is equal to gamma like that very  very good stuff right and that's pretty much   it I'm afraid I'm making a small-time  mistake let's see okay I should have hmm Ahana this rainy should be negative karma akarma  that's see see this is what I have to look at   doctor pants video sometimes that all right let's  see if it makes sense if you have zero this is   negative infinity times one so it's negative and  then went to infinity does zero so when he is one   that would be zero so you have negative part and  then you have some not so big may pass the parcel   I believe okay I believe this right here should  be negative gamma otherwise it won't work so it   has to be negative gamma let me just double check  to see if I have any mistakes or not yes I really   believe this ratio pian negative gamma and make  a gamma can we see okay so this negative gamma is   just like a constant just like pi water II all  right or Euler's constant yeah and I will come   back here and then actually this is negative  gamma right finally ladies and gentlemen as   you can see the left-hand side is equal to the  hands of obviously but of course we have the s   s then s we can cancel them out so you can see  that on the top you have negative gamma minus 1   over cancelled out so you have s like that oh  one more thing I forgot I'm so sorry I'm very   sorry I'm very sorry all right so you have all  this to little row but if that's too easy with   quotient rule the bottom function square and the  top is the bottom function times the derivative   of the top minus the top function times DT really  at the bottom derivative of s to the R plus 1 with   respect to R as our plus one times Ln s times s  why is that what we did earlier test just about   with you are you so in firms re this right here  which will have another L&S after that yes so   I'm missing that the own s well anyway the rest  is OK right here and again this s dies and that   s quick can reduce you finally care negative  gamma minus 1 Ln s sauce - yes all for s side   uh-huh so already done negative gamma minus 1  s oo / s like this tan tan tan tan tan so done   yes not so bad all right so this right here is  it $24 transforms for you guys right so that's   pretty much it and in fact I will have to ask  you guys the following now so problem just as   we got this so I leave this to you guys question  for you guys this is again the Honorable Mention   question what's the Laplace transform of 1 minus  cosine T over T right what's Laplace transform   of that and then after this can you also find  me the integral going from 0 to infinity of   1 minus cosine T over T e to the T all right so  please do that and let's see if you guys can and   hopefully this right here 3 T negative gamma and  if so that would be great all right so that's it   and hopefully you guys all at this and by the  way I should say s is greater than 0 somewhere but if because otherwise you will converge  fine so anyway this right here is it and   as usual I wanted to put on my marathon medal  this is the one I got from 2017 this register   both sites by sketcher in my hand it's  dirty very dirty dirty dirty right okay   so and you're stuck every time I put on  my metal so that's it hopefully you just   find this video to be helpful and yeah let  me know how you guys to doing other stuff   which you get the best which you guys safe  and healthy right so as always that's it

Channel: blackpenredpen

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Keywords: Laplace Transform Marathon, laplace transform of ln(t), laplace transform of sqrt(t), Laplace transformation, Laplace transform of t^n, Laplace transform of Dirac Delta function, Convolution theorem, Laplace transform of ln(t), laplace transform study guide, ordinary differential equations, laplace transformation, calculating laplace transform, blackpenredpen laplace transform, laplace transform properties, laplace transform of derivatives, Laplace Transform Ultimate Study Guide

Id: ftnpM_RO0Jc

Length: 190min 50sec (11450 seconds)

Published: Wed Mar 25 2020