a spectacular solution to the Basel problem (sum of 1/n^2 via a complex integral)

this integral it's super amazing in fact not the  real version of it even though everything's real   but rather we'll talk about the complex version  of this and that's the one that's going to keep   us very very nice result (solution to the Basel  Problem) and you guess I kinda liked it so much   I didn't come with the following though I want  to thank my fuher dear viewer dear cv h right he   was the one who show me how to get this awesome  result and we begin by looking at the integral   from 0 to PI over 2 and ln(2*cos(x)) everything is  real and the first step is to make it complex so   let's talk about the following this is the same  as integrating from 0 to PI over 2 ll the tool   tool and we'll be using the complex definition  of cosine for this which is e to the i-x plus   e to the negative x over 2 and if I'm to see if  why this is true you can check out my video the   link is in the description and you see that the  two cancel each other nicely that's good and here   we have the integral from 0 to PI over 2 well from  here to here I will factor out that e to the i-x e   to the pass the IX and we see that from here we  have 1 plus and once we factor out e to the i-x   we will have minus IX from this exponent so we get  e to the negative 2 IX right here if you multiply   this out you see we get that back so we know it  is right and of course we are still have that DX   what's doing the X world and now we see that this  is pretty much the product two things inside of   the head so we can write this down as the sum of  the Ln plus the other and right and we can also   write them as two separate integrals therefore  our just ready style this is the integral from   zero to PI over two first one is Ln of e to the  i-x and then we will just close this integral   secondly we have integral from zero to PI over  two Ln of this one plus e to the negative 2i X   DX and if I have a choice if I'm taking tests  for something I will choose to do this one over   this one of course because Ln and E they cancel  each other out so right here we're pretty much   just integrating my X in the X world and to do  so remember is just like a constant so you keep   it and then you add 1 to the power 1 plus 1 is  2 divided by the power which is over to here and   you have X square right here and you go from 0 to  PI over 2 and perhaps we'll finish this first why   not plug in PI over 2 here you get I over 2 and  then you have PI over 2 squared and then minus I   over 2 and a plug in 0 squared this is PI square  over 4 4 times 2 on the bottom is just over 8 so   therefore we get I PI square on the top all for  8 like this and this is of course 0 so we don't   worry about that and now the harder part is this  right here hmm might want to try integration by   parts I don't want to try substitution I will just  use series for this so check this out which series   of course Ln of 1 plus something right so here  we go let me make a note right here for you guys   well we know that Ln of 1 plus Z and the reason  I'm using these because here we have a complex   number this right here is equal to the following  Z minus Z squared over 2 no factorial by the way   and then plus Z to the third power over 3 and  then minus the 24th power over 4 and then plus   so on so on so on and you can see why this is  true once again can check out my video the link   will be in the description and perhaps that's  not your past I the technical part the radius   of convergence this right here it's only that yet  if absolute value of Z and remember Z is a complex   number so if you take the absolute value meaning  that you're trying to find the distance of that   complex number from the origin if that distance is  and with the exception that Z cannot be negative   one then this right here converges okay and once  again easier explanation the link will be in the   description okay here we have this part e to the  negative 2 IX we can simply first plugging this   into all the Z and then integrate now will be the  strategy so here we go let's take a look of what   Ln of 1 plus e to the negative 2 IX like this ok I  was just first plugging this into all the C's and   the thing I will do is that I will not multiply  out the exponents now just this leave yes how   they are you'll see why anyway first of all I will  do this in two here so which is e to the negative   2i X and then next I will just put down - plugging  this in here we know that we will have to multiply   the exponent with that right because you have this  square I will keep this two in blue and then the   rest of red so we will have e to the negative let  me put down a to first original power right here   and then times 2i X like that and then over to  like this okay and then next we add we do the same   thing plugging this in here I will keep the street  in blue so we pretty much you will have e to the   negative 3 times 2i x over 3 and then the last one  on the list that we have of course this one goes   on forever but the one that we will do is we minus  e to the negative 4 times 2i x over 4 and so on   ok all right this right here is just the function  inside now let's talk about integration and first   of all that's just integrate with respect to X ok  so let's put this down and the cool thing is that   we have e to the a ten times X and this is X to  the first power so we can just do the following   the integral of this is that you keep this part  by you divided by the derivative of the input   function it's pretty much the coefficient so the  first one is going to give us e to the negative   2i x over negative two I like this okay and then  next I'm going to minus here we will have once   again this pretty much stays the same I will write  es e to the negative two times 2i X and I had this   two originally but I will have to divide it by  the coefficient namely I will have to divide it   by negative two times two I like this and once  again I am NOT multiplying out anything okay   and the next one is plus and on the top were just  ready as e to the negative 3 times 2i x over this   3 and then we divide it by the coefficient namely  negative three times two times I okay so far so   kaha and then minus this is a doing accounting  man's that you keep track of everything in since I   de and ya in fact this is like my fourth or fifth  times recording this problem because I messed up   the organization of my work anyway minus 4 times  2 times I negative 4 times 2 all that stuff okay   that's how we integrate that guy nice huh and then  we will plug in numbers of course 0 to PI over 2   0 to PI over 2 however before we plug in the  numbers I'm going to do some factoring this is   why that I did not multiply out anything on the  bottom whatever because we notice everybody has   negative 2i on the denominator right you see naked  he to I here and negative 2i negative 2i we just   do it this is the negative and then that's why and  negative 2y and also you see this is three and a   motor powers history something is going to be very  pretty okay so I will factor things out into some   algebra before I plug in the PI over 2 and things  like that let me just factor out 1 over negative   2 I all the way to the front ok and then we will  have the first time is just e to the negative 2i   X like this and you know what let me just change  this to red to make me happy a little bit negative   2i because everything was secret yeah including  this one anyway anyway this is that and the next   is - I was still radius e to the negative 2 times  I two times two x IX ok two times two times I X I   have a multiplied or anything yet you don't know  why under dynamometer though we see that we have   two times two lready s2 square ok and then next  I would had on the top we still easily negative   three times 2i X and then on the bottom we have  three times three so that's 3 square and once   again I factor all the negatives right he already  so this size they stay the same and then minus e   to the negative 4 times 2i x over 4 times 4 which  is 4 square and that dot like that and now we are   talking about going from 0 to PI over 2 and I  am NOT going to show you guys the plugging this   because there would be a lot to write so here is  the deal ok notice everybody here is pretty much   having the negative 2i X right 2 X 2 X 2 X and  when we plug in power to here the two in blue   and this 2 in black they will cancel things out  and you'll pretty much be talking about with a   1 left 2 3 4 left and also the negative thing set  up so here is a note F 1 for you guys this right   here it's just another note okay we pretty much  will have the following e to the negative I an   iPad okay everybody is going to have the hi and  also the PI okay and everybody has the negative   exponent now it's just a matter of what this  n is and in fact we only have two choices this   right here will be either negative one what path  do you want this right here will be negative 1   if n is odd and on the other hand this right here  will give us positive what if and is even and now   I am just going to be plugging power 2 into all  the XS right here and this is going to be really   helpful first of all let me just write down the  1 over negative 2i all the way all the way all   the way in the front so open up a parenthesis  ok and let's see our first plug in power 2 into   this X and we see that this 2 and that's what  cancel so we have a 1 right here left so this   is technically negative 1 I tie and that's going  to be this situation so we end outs negative 1 ok   and then I will maintain this - and now we'll  plug in power 2 into this X this true and that   to cancel so we have this 2 right here I'm just  looking at this number which is 2 which is the   even situation which is going to be a 1 on the top  so we have 1 over 2 square and then we'll continue   next we making the plus first plug in PI over 2  into this X the tools will cancel this is a 3 so   that's at a situation that will give us negative  1 on the top over 3 squared and the next one we're   talking this you see the two were canceled but  this is the 4 so that means we will have a 1 on   the top maintain a - stop and now you have the 1  over 4 squared and so on so on so on and you see   that in this parentheses here this is the result  that we will get from we plug in power 2 into   all the XS and now I will subtract perhaps I'll  make this red so that is study more clear okay   subtract next I will part in zero into all the XS  and this is very nice because e to the zero power   is just one so first we get positive one next we  get the - stays and then we still get a one and   over two squared and then we maintain the plus and  this is one over three squared and then minus 1   over 4 squared and then plus da so close that and  then close that ok so it depends on how you want   to draw the parentheses perhaps this is slightly  prettier ok now pretty things happen again ok if   you look at the first term this is negative one  minus one okay leave that but you see negative   one over two squared minus minus becomes plus in  other word this and that will cancel each other   out this and that cancel each other out next this  term stays you are going to combine items in your   second and then this and that will once again  cancel each other and so on so on so on and if   you see that we are going to get we still have  this 1 over negative 2i or the way in the front   and then when we do negative one minus one of  course that's negative 2 and then negative 1 over   3 squared minus 1 over 3 squared that's going to  be negative 2 over 3 squared and if you were like   I could have put down this yes negative 2 over 1  squared right because you could radius 1 square   here to keep the same form and if you look at the  team now meet her if you write down more terms you   will really see that the next term is going to  be negative 2 over 5 squared and then negative   2 over 7 squared and so on so on so on so I will  just leave it like this okay and of course the two   will cancel out here here here here right and then  the negative distribute in there you get positive   so that's all good so they meet just TBS 1 over  negative I like this what should never just ready   as one over I the negative will be distributed so  I get positive 1 over 1 squared positive 1 over 3   squared positive 1 over 5 squared and positive  next in extinct terms and all that okay this is   all good but I don't like to be on the bottom I  like to be on the top therefore I multiply the   top and bottom by I real quick so this right  here is once again I squared which is negative   1 so this right here is technically negative  I okay negative I and then you have all that   stuff 1 over 1 squared plus 1 over 3 squared plus  1 over 5 squared at so on so on so on ok and look   at this is this integral and perhaps some of you  guys noticed what we are trying to do and no we're   almost there oh my god I'm going to dry this down  over there and of course you see this integral was   just that I PI squared over 8 so let me just erase  this so we will have more space I will first write   down this right here I PI square over 8 okay and I  can sit by by to this next this whole thing which   by oh that is just this so I will write down minus  I and then we have the parentheses which with 1   over 1 squared plus 1 over and fortunate to have  two squares ok so these 3 squares 3 square and   we don't have the even squares ok so just the 1  3 5 the squares of them and all that stuff okay   so this is what we have and now of course you  notice both terms have the I so of course we   can factor that out and when you factor out the I  we get PI square over 8 and we - this results are   put on parentheses 1 over 1 squared plus 1 over 3  squared plus 1 over 5 squared plus tada and so on   okay so huh it seems like if you change this to  the complex version of it we get this result but   remember originally this is what's supposed to be  a real integral right the real deal at the moment   we see that this is I times the real number keep  in mind everything here is real number so I times   a real number so we can say that this right here  is purely imaginary right this right here has to   be imaginary required for I this has to be just  real so here is the note notice if you take the   imaginary part of this real integral so I'll just  put on I am of 90 bro we will take the imaginary   part of a real integral this right here has to be  0 that means this imaginary part which is equal   to 0 and the only way for us to happen is that  this inside has to be 0 so I will just tell you   guys that this must be 0 we must have I times  0 in order to be 0 so that the imaginary part   is 0 so that when you will end up with the real  value for that and in fact the real value photon   turquoise also 0 all we see that pie square- this  it's equal to 0 so I can say this it's equal to PI   square over 8 so we observe that and just say  observe that here's the next result we have 1   over 1 squared plus 1 over 3 squared plus 1 over 5  squared plus this is equal to PI squared over 8 so   good now of course this is not good enough because  this is it's goodbye it's not famous enough huh   the famous version is what 1 over 1 square plus 1  over 2 squares plus 1 over C square plus 1 over 4   squared and so on so let's see if we can use this  to come with that result I will put this down in   black and a see if I can see everything here  now consider I will just say the following let   s equal to 1 over 1 square plus 1 over 2 square  plus 1 over 3 square plus 1 over 4 square plus   da da ok so let's consider this sum and we'll  see that yes it's equal to I will put down the   terms first so I get 1 over 1 square plus 1 over  3 square plus da da this is the R terms ok and   then plus the even terms is 1 over 2 squared plus  1 over 4 squared plus dot dot okay well this is s   and that's equal to this right here now we know  is PI squared over 8 and we see that this right   here you can look at this yes 2 times 1 in the  parenthesis squared and then can factor out the   2 squared similarly this is that say 2 squared  times 2 squared the next one inside 2 squared   times 3 squared and so on so I can factor out 1  over 2 squared like this ok and you will see that   this inside here becomes 1 over 1 squared plus 1  over 2 squared plus 1 over 3 squared plus da da   in another word we see the equation s is equal  to PI squared over 8 plus 1 over 4 and this is   s that's minus 1 over 4 has done process this  is going to be 3 over 4 s which is equal to PI   square over 8 and of course as you multiply back  for Hopper's to be on both sides so let me put   this down go over 3 tons PI square over a you do  the little thing to keep people happy finally we   see this is equal to PI squared over 6 oh my god  in another word this right here we just proved it   with a very elementary step I will say because we  didn't introduce anything high-tech with the use   since that's complex but nothing really high-tech  we see that this right here it's actually pi   square over 6 and perhaps let me just pop this  famous result and I've also bought this because  

Channel: blackpenredpen

Views: 310,909

Keywords: Integral of ln(2cos(x)) from 0 to pi/2, sum of 1/n^2, sum of 1 over n squared with an integral, complex integral, math for fun, blackpenredpen, a spectacular solution to the Basel problem, sum of 1/n^2 from 1 to infinity, pi^2/6, basel problem, 1+1/2^2+1/3^2+..., 1+1/4+1/9+..., 1+1/4+1/9+...+1/n^2 pi^2/6, series of 1/n^2, calculus p-series, infinte series, integration

Id: 5-pXwWNcsbc

Length: 22min 26sec (1346 seconds)

Published: Thu Aug 23 2018