Supreme Integral with Feynman's Trick

hi - is for you and thank you so much for sending me this also an integral with your station of course and I will present you a solution to everybody right here as you can see here we have sign of air and eggs of course this was the justice to bring the lnx into the sea so let's do that right here we will have sine of X and this is going to be e to the i-x minus e to the negative I and X or / - I like this and as you all know e + Ln however somehow they can cancel each other out just how to do it carefully be sure you bring this to the exponent here first so that you can cancel this out legitimately and right here you do the same bring the negative I to the exponent this X and you can cancel things out legitimately so this is going to be e and L and what cancel by you end up with X to the ice power and then on the other hand here we have this minus and then it's X to the negative ice power or over the - I on the denominator so on all this right here it's pretty much our new numerator oh my god because it's going to get crazy huh so let's take a look of hope exactly that we are having this right here it's equal to the integral from 0 to 1 now and as I said on the top that is just the same as this so let me just put that down so I will write down the X to the ice power minus X to the negative power and then all over let me bring the two I down the Dynamo of course so let me put this down right here and don't forget we still have this additional Ln X so let's just put that down right here and of course we are doing the X world for now so that's what we have at the moment and once again this is the integral from 0 to 1 okay now what can we do next hopefully you guys have seems on like a similar videos in the past because look at this right here we have X to some power and in the denominator we have that Ln X the trouble is this L and X in the denominator that's parting us that's how come that I don't know how to integrate this right you'll be much easier if we can somehow get rid of this arrow next and how can we do that oh well you have watch my other features things in that case you would know how to do this and I will just do a run quick run for you guys first I will put this down this is my other note I'll just put this down right here the secrets that we will be using the differentiation under the integral sign because when we have X for the base if you differentiate X to some variable with respect to that variable you can squeeze out the L and X so Y I mean by that it's a following let me just write this down the DT for example we know if you differentiate T to a third power this right here easy because the 3 print was front and the minus 1 so you have 3t squared but if you differentiate if the base is just a number 3 and they raise to at least power then in this case what you do is you repeat the exponential part and then multiplied by Ln of this base which is 3 like that so if you want to squeeze out the Ornette you just have to you know introduce a new variable and then differentiate that with respect to that variable they can squeeze out the on whatever they want and as you guys all know this is pretty much the finance technique for integration so there we go I am going to just write this down again I will put this down right here so this is the integral that we are looking at from 0 to 1 and we have X to some power times I like this power and then minus X to the negative x power like this or over to I the X well if I have my variable in the exponent I can differentiate this at Kanpur to stop so I'm going to introduce the variable I traditionally if you want to follow the famous technique we use B but you can use whatever that you want you can use T but doesn't really matter so let me put a B right here and we are talking about like a new function because once you have this new variable this new kind of variable is called the parameter then you can defined this as a new function I will just call this to be I of B to be this and of course you can plug in any P for you that you want as known as this legitimate so that you have to make sure this is still going to appear converging all right so we do have to be careful with the choice of the B value but if you notice this and that the difference is just that we have the B and this is B times I this is just I so we are saying is equal to 1 in other word this original integral is nothing but just I of 1 so hopefully we can calculate but I for one place and if you would like to kind of just discuss this first it's like does this even converge in the first place the short answer is just that please trust me and I will tell you guys it will converge because it converges of course anyway a long answer you can just read the comments because I'm pretty sure other people were already anyway so I don't need to worry about that much all right serious business here is my new function I of P that go ahead and differentiate this with respect to B on the left hand side and of course that's do it on the right hand side on the left hand side it's extremely easy because you can just put Prime and can pretend a how you did the derivative I pry off B on the right hand side how do you differentiate Interpol well I said earlier you do differentiation under the integral sign so what you do is you still write on the integral from 0 to 1 and you bring the differentiation inside but inside you have two variables so it becomes partial derivative so I will change the notation to this partial derivative with respect to B and then when looking at this right here namely X to the B I power minus X to the negative I power over to I L and X and if you wonder how come I didn't put a B right here it's okay but I two CDs and negative so I don't like that too much and if you wonder why don't we - why don't we put two like be here and B here you can try that but let me tell you this is enough right okay so this right here and of course we still have the DX all the way at the end technically still in the X world in the dis integral but we had to take up this differentiation first so in that case that's the P world so let me just write down the integral from 0 to 1 the blue part what differentiating this with respect to B namely x 4 pd constant so on the bottom we still have the constant and you know the constant multiple will stay the same so I will just write this down to i/o and I that's just a constant in the B world and then when you differentiate X to the B I power well the X is the constant is just like this tree so you first just repeat this part so I will repeat just a part which is X to the B I power but you multiply by our of the base which is X once again because the X is the constant just ideal situation here alright lastly though we have to do the chain rule because be I and where differentiate this with respect to B so the chain rule says we have to differentiate this and that the ruptor be I with respect to B is just the I you multiply by the derivative of that which is I right here and then for this part look at this is just miners constant constant constant so that the roof feel that it's just a roll we just care about this part right and if you would like a splitted fraction but don't need to do that this is pretty much the result of this differentiation and then you just continue to put on the D X well what do we do really nice the Ln X cancel I are gone and now you pretty much just look at that again do the integration with respect to X now so let me just write this down nicely we have the integral from 0 to 1 and we will let's write this down as 1/2 all the way in the front I was just put it down like one having the front and then we have the X to the B power DX and we on the X world now so to integrate this you just add 1 to the power and then divided by the new power so divided by P I plus 1 but the mere idea is 1 plus B I so this right here is going to give us 1/2 times and let me just put down 1 over 1 plus bi first and then times X to the 1 plus B I power like this well don't forget we still have to do the throw to 1 business so you just go ahead plugging X is equal to 0 up to X is equal to 1 and then just do whatever we need to do right so let's see if I can fit in everything right here let's do something now head when we plug in 1 into X this is just 1 so we just have the first part namely we get 1/2 times 1 over 1 plus bi and then secondly you subtract by if when you plug in 0 into the X you know this is so nice because I don't have to do any indeterminate form because this is just a 0 so you just minus 0 so that's it so this part we just get this 1/2 times 1 over 1 plus bi and if you like and put a parenthesis up you so that's pretty much it but that's that the answer of course so what we have here is just that let me just put this down we know let me just do a little separation I prime of B this is equal to that but I don't really want to know what I pry off B is I want to know I of B because I want to figure I want this is the derivative how can we get back to the original easy you told me yes we can just integrate it when you integrate this time we are we are integrating with respect to B because we differentiate it would be earlier so you ain't doing that so let me just put down DB here and also DB here on the left hand side it's also just as easy you just get the original I of be back on the right hand side this is the constant multiple so let's put that down and remember be it's the variable now because running the B world everything else is the constant so this is just like the usual 1 over 1 plus X to the first power whatever its linear so you get Ln so let me just put out our n of the inside which is 1 plus bi and notice I'm using your parentheses because seriously you have complex number already if you have the complex number you don't need to worry about the absolute for your whatsoever and we don't really worry about negative numbers just just just the same but what's at the root of this the relative of 1 plus bi with respect to B is I when you're doing integration don't forget to divide it by that derivative so we put down 1 over I so this is pretty much it but don't forget to add the plus C for some unknown constant at the moment well but it's that constant this right here by the original definition of I of B is that so let's put that down I want to plug in some value for B so that I can hopefully set up an equation to solve for C what can I use though I want to use I of one that's the things that I don't know if I plug in zero for B what will happen if I plugged in 0 for B then I don't know how to calculate this integral fortunately earlier I said there are some conditions on B maybe but I do notice if B is negative one then look at this is the same as 8 to the negative I and the minus the same thing so the integrand would be just zero so that it's going to be really happy because in that perforce wrote one of the zero function we get zero so let's go ahead and plug in B is equal to negative one I want B is equal to negative one and then this would be zero all right so you write down Ln and then this is 1 and B is negative one so you have minus 1 I and then plus C and then this now is equal to integral from 0 to 1 and B is negative 1 so you have X to the negative 1 I minus X to negative I or over 2 i Ln X DX and the cool thing is that this integral is solid 0 so you're just integrating the zero function and of course this will converge so again other work you have the left-hand side goes to this right here it's going to be 0 so it's equal to 0 so what's C of course this right here is kind of troublesome you can just bring this to the other side so I was just ready stuff from the moment see it's equal to negative 1 over 2i Ln of 1 - I like this that's the C value and then you can refer back to this so this one is I of B which is this and you attach your C value to this now so finally I would just perhaps write this down ok so hope in the end I just have to plug in B is equal to 1 so we can get things done so I won't be is equal to 1 to get the original integral so you at the integral from 0 to 1 X to the power minus X to the negative power or over to Ln X DX this is 1 over 2i Ln of 1 plus 1 I and then minus 1 over 2i Ln of 1 minus I so this right here it's pretty much the answer but what the heck in the world is this it's secretly a real number and you can be sure this has to be real because originally this integral was in the real world right so you should and outside real value I will just do this super quickly for you guys one half I like this I can bring that to the front so we have 1 over 2i and then you have Ln minus another line so I can ready L as Ln let me just write it down like this Ln of 1 plus I over 1 minus i right and I cannot finish this on the 1 port that's unfortunate but this right here what you can do is just multiply by the country can also ever you get Ln and simplify everything I'm young you get out and I write and I this is PI over 2 all right you just take this nice value so what it's the final answer let me just hopefully write this down right here it's of course 1 over 2i times pi over 2 I and then the ice cancel and then you end up with PI over 4 that's it whoo so yeah this part is multiplied by the conjugate which is 1 plus ion pattern on top and you can simplify to get just I and then hopefully you have seen the other video on how to to L and I and you take PI over 2 I for the nice answer and that will be this powerful 4 is this of course is that and then we are done so have a port and that's it open it gets all like this video if you do please subscribe and like you know to reach your comment and you know I tried to solve your questions and things like that but sometimes it takes time and effort in however yeah anyway yeah that's it

Channel: blackpenredpen

Views: 204,063

Keywords: integral of sin(ln(x))/ln(x), integral by Feynman's Technique, integral of sin(ln(x))/ln(x) from 0 to 1, Feynman's Trick for integral, blackpenredpen integral, integral of sin(x)/x, supreme integral, dirichlet integral, balckpenredpen math for fun, calculus, calculus teacher, Feynman's technique for integration, differentiation under the integral sign

Id: E34RfM6wU7Y

Length: 17min 53sec (1073 seconds)

Published: Tue Jun 26 2018