A stellar integral solved using some wonderful complex analysis

[@manstuckinabox3679]

If it was (1+x)^2 instead of (1-x)^2 we could then use gamelin's VII.4.6 which is the integral of x^alog(x)/(1+x)^2 from zero to infinity which is = picsc(pia) - api^2*tan(pi*a)*csc(pia).

[@maths_505]

Ma boi here has spent some real - no wait, complex - time in the temple of Gamelin😎😎😎 Time well spent I must say

[@manstuckinabox3679]

@@maths_505 oh yeah and I think it can be done in the similar method, the only difference being that there are no residues inside the keyhole contour (and for the result just differentiate by a then replace with a = -1/2 (within the domain of convergence).

[@maths_505]

@@manstuckinabox3679 an anime crossover between feynman and complex analysis 🔥🔥🔥

[@HVH_49]

This feels like a complete review on my complex analysis class that I took years ago lmao

[@enderwarrior6185]

So glad i took a course of complex analysis, this is so much fun, great video man !

[@daddy_myers]

Beautiful! I've seriously missed these contour integrals!

[@GearsScrewlose]

Feynman’s Technique works here with odds of zeta(3) canceling out. Kind of cool way of solving it.

[@zunaidparker]

At 11:30 it's not incorrect but it is bad technique to simply assume that the imaginary part cancels out to zero instead of proving it. It doesn't allow you to catch any mistakes. It's worth confirming by checking the symmetry of the relevant functions or by some other independent means.

[@maths_505]

Yeah you have a point but the video is already 20 minutes long.....but yes that would make another cool bonus integral.

[@silvanadamini9122]

4:57 It is unrigoros to pull the -1 out the sqrt. Because let u=-x: sqrt(u)=sqrt(-x)=i*sqrt(x) =i*sqrt(-u)=i*i*sqrt(u)=-sqrt(u). So if sqrt(-u)=i*sqrt(u), then sqrt(u)=-sqrt(u).

[@cpcto945]

Very cool as always

[@maths_505]

Thanks mate

[@SuperSilver316]

I wanna say a contour integral with a branch cut along the positive real access could work to, if you use ln^3(z) instead of ln^2(z).

[@maths_505]

Noted

[@SuperSilver316]

Upon some playing around with the problem, you don’t have to use the cubed logarithm, you can just leave it squared. There’s a double pole at z = 1 underneath the branch cut, but above it it’s fine. The square root gives you this flexibility, but you have to be very careful about how you let the certain radii in the semi-circles go to zero.

[@timemasterdm2462]

Super cool integral. And amazing video as always. May I ask what application you use as a blackboard?

[@maths_505]

Samsung notes

[@user-le1oc9js4h]

I think, in the part where you prove that Г equals 0, you should specify, that not only the integrand goes to zero, it goes to zero faster than the length of circumference grows, that’s why the integral goes to zero

[@zunaidparker]

The integrand incorporates dependence on R when substituting z=R*e^(i*theta). The limit as R goes to infinity is taken AFTER this substitution and shows the whole thing goes to zero as R tends to infinity.

[@cpotisch]

I did it by breaking it up into an integral from 0 to 1 and from to infinity, and then using geometric series!

[@anilrao599]

For anyone looking for the solution to proof that gamma approaches zero here's a brief outline: Call the integral I and then construct the inequality |I|

[@two697]

That's what he said in the video

[@renesperb]

A different approach is as follows : make the substitution x= exp t .Then one gets the integral of t^2*exp (t/2)/(exp t -1 )^2 with limits - inf to + inf .If one splits up into -inf to 0 ,and 0 to inf and uses the fact that 1/(1-q)^2 is just the derivative of the geometric series , one arrives (one can also invoke Laplace transforms ), by integrating termwise, at the series 2*k*( 1/ (k+1/2)^3+ 1/(k - 1/2)^3) ,k =1 to inf.What I was not able to verify yet , that this series gives 2*π^2 , as MATHEMATICA tells me. Maybe someone finds the clue to this.

[@maths_505]

Well.....you could site "by kamaal's video on contour integration...... The series equals 2pi^2" QED😂

[@renesperb]

One finds the following relations: Sum (k/(k+1/2)^3 (1 to inf.) =π^2/2 - 7*Zeta 3 ,Sum (k/(k - 1/2)^3 = π^2/2+7*Zeta 3 ,

[@nicolascurry9520]

Ok. So this isn't an integral but is there some way to evaluate ln(x/ln(x/ln(x/ln(...)))). Its a bit like ur infinite tower of x^x but with ln(x/...) I think it's somewhat related to the Lambert W function as the inverse is xe^x I think? Any help? Love ur videos btw

[@user-dm1tm8uw2o]

Nice work man rekt it.

[@niom-nx7kb]

6:48 shouldn’t this second term be a plus sign? The 1/i becomes -i and there’s originally a negative in front of the pi^2

[@niom-nx7kb]

Oh.

[@hsfjw]

Each time reducing log z introduces another Cauchy residue, seems like can generalize as a recursion on log ^N z? 🤔

[@maths_505]

Indeed

[@montreearmy]

Hey guys, I have a question. Can complex's world(Contours Integrals. Sorry if I spell wrong) solve every Integrations?

[@SuperSilver316]

They can solve a wide variety of problems, but are usually only reserved for when the integrand has very nice behavior for large and small x. You wouldn’t want to use it all the time, but it does have a lot of nice utility.

[@amidhmi5243]

If the imaginary part is 0, we can use it to calculate the integral. The first term is exactly the integral we need to calculate. And the second term doesn't have a log in it which makes is simpler to calculate.

[@maths_505]

Not exactly....that log integral has 1+x in the denominator. We needed 1-x. But yes we can count that as another bonus integral. Nd a really nice one too.

[@clayton97330]

Does this integral have a physical interpretation... would it ever be derived from a real system?

[@allmight801]

Why is 0 singularity again and not 1?

[@anilrao599]

Because if you take the lim of ln^2(z) /(1-z) ^2 as z approaches one it is defined

[@maths_505]

I literally explained that in the beginning of the video

[@anilrao599]

😂

[@anilrao599]

@maths_505 can this be solved using Feynman technique?

[@allmight801]

@@maths_505 i know just didn't get it at the beginning i suppose

[@maddog5597]

Why, in the name of all that’s holy, don’t you cross your zs???? They look like 2s!!!!

[@niom-nx7kb]

Isn’t z=1 also a pole?

[@niom-nx7kb]

For the first contour btw

[@niom-nx7kb]

@maths_505 professor I have a question

[@niom-nx7kb]

never mind!

[@holyshit922]

This integral can be calculated without complex numbers Int(ln^2(x)/(sqrt(x)(1-x)^2),x=0..infinity) Let y = sqrt(x) y^2=x 2ydy=dx Int(ln^2(u^2)/(u*(1-u^2)^2)*2u,u=0..infinity) =8Int(ln^2(u)/(1-u^2)^2,u=0..infinity) 8Int(ln^2(u)/(1-u^2)^2,u=0..infinity)=8Int(ln^2(u)/(1-u^2)^2,u=0..1)+8Int(ln^2(u)/(1-u^2)^2,u=1..infinity) 8Int(ln^2(u)/(1-u^2)^2,u=0..infinity)=8Int(ln^2(u)/(1-u^2)^2,u=0..1)+8Int(ln^2(1/v)/(1-1/v^2)^2*(-1/v^2),v=1..0) 8Int(ln^2(u)/(1-u^2)^2,u=0..infinity)=8Int(ln^2(u)/(1-u^2)^2,u=0..1)+Int(ln^2(1/v)v^4/(v^2-1)^2*1/v^2,v=0..1) 8Int(ln^2(u)/(1-u^2)^2,u=0..infinity)=8Int(ln^2(u)/(1-u^2)^2,u=0..1)+8Int((-ln(v))^2v^2/(v^2-1)^2),v=0..1) 8Int(ln^2(u)/(1-u^2)^2,u=0..infinity)=8Int(ln^2(v)(1+v^2)/(1-v^2)^2,v=0..1) Int((1+v^2)/(1-v^2)^2,v)=Int(1/(1-v^2),v)+Int(v*(2v/(1-v^2)^2),v) Int((1+v^2)/(1-v^2)^2,v)=Int(1/(1-v^2),v)+v/(1-v^2)-Int(1/(1-v^2),v) Int((1+v^2)/(1-v^2)^2,v)=v/(1-v^2) 8Int(ln^2(v)(1+v^2)/(1-v^2)^2,v=0..1)=limit(8v*ln^2(v)/(1-v^2),v=1)-limit(8v*ln^2(v)/(1-v^2),v=0)-8*Int(v/(1-v^2)*2*ln(v)/(1-v^2)*(1/v),v=0..1) 8Int(ln^2(v)(1+v^2)/(1-v^2)^2,v=0..1)=-16Int(ln(v)/(1-v^2),v=0..1) -16Int(ln(v)/(1-v^2),v=0..1)=-16Int(ln(v)*sum(v^{2n},n=0..infinity),v=0..1) -16Int(ln(v)/(1-v^2),v=0..1)=-16Int(sum(v^{2n}ln(v),n=0..infinity),v=0..1) -16Int(ln(v)/(1-v^2),v=0..1)=-16sum(Int(v^{2n}ln(v),v=0..1),n=0..infinity) Int(v^{2n}ln(v),v=0..1)=1/(2n+1)v^{2n+1}ln(v)-1/(2n+1)Int(v^{2n+1}*1/v,v=0..1) Int(v^{2n}ln(v),v=0..1)=limit(1/(2n+1)v^{2n+1}ln(v),v=1)-limit(1/(2n+1)v^{2n+1}ln(v),v=0)-1/(2n+1)Int(v^{2n},v=0..1) Int(v^{2n}ln(v),v=0..1)=-1/(2n+1)^2 -16Int(ln(v)/(1-v^2),v=0..1)=-16sum(-1/(2n+1)^2,n=0..infinity) -16Int(ln(v)/(1-v^2),v=0..1)=16sum(1/(2n+1)^2,n=0..infinity) 16sum(1/(2n+1)^2,n=0..infinity)=16(sum(1/n^2,n=1..infinity) - sum(1/(2n)^2,n=1..infinity)) 16sum(1/(2n+1)^2,n=0..infinity)=16(sum(1/n^2,n=1..infinity) - 1/4sum(1/n^2,n=1..infinity)) 16sum(1/(2n+1)^2,n=0..infinity)=16*3/4sum(1/n^2,n=1..infinity)) 16sum(1/(2n+1)^2,n=0..infinity)=12sum(1/n^2,n=1..infinity)) 16sum(1/(2n+1)^2,n=0..infinity)=12*Pi^2/6 16sum(1/(2n+1)^2,n=0..infinity)=2*Pi^2

[@maths_505]

Okay bro that is just boring af

[@two697]

There's no way you just typed all of this out

[@JO06]

Why would u write this out…

[@holyshit922]

@@JO06To show that complex analysis solution is not easier nor even faster than real analysis in this example

[@tueur2squall973]

6:41 for the 2nd integral , this should be a + sign instead no ?

[@tueur2squall973]

Aight' you corrected yourself it's all fine

[@maths_505]

Ahhh yes....my favourite algorithm bumper 🤣🤣🤣🤣

[@pratimguha5905]

If you don't mind , can you please share me some of the names of books from where I can get these integrals ?

[@maths_505]

Won't find most of these in books. Alot of the integrals I've solved are ones I've made up. This one I got from Michael Penn. He solved it using real methods.

[@maths_505]

Though there are some great books Paul nahin's inside interesting integrals There's this book called almost impossible integrals sums and series and that has lots of cool integrals and tools. Then ofcourse there's a wealth of integral calculus knowledge on math stackexchange and KZbin

[@pratimguha5905]

@@maths_505 ok sir , thanks a lot

[@NC-hu6xd]

@@pratimguha5905 Pls sir do not redeem the integrals bloody basterd

[@tolberthobson2610]

Wouldn't the key-hole contour have made this so much easier??? Rather than an upper-semi-circular contour.

[@maths_505]

Give it a try.....

[@SuperSilver316]

It’s actually a lot harder with a keyhole contour believe it or not!

[@Akenfelds1]

Fascinating but I have a bit of constructive criticism: This video needs to be cleaned up/remade. To much backtracking on errors, etc..

[@maths_505]

Thank you

[@andrewneedham3281]

The lack of being able to differentiate z's from 2s led to 5 minutes of frustration trying to follow one of your steps when calculating the residue. Just ... no, thanks.

[@cycklist]

Zed

[@thefirminator]

just start the vid with i = whatever instead of doing it everytime

[@maths_505]

Already started doing that bro. Damn I miss bobby at anfield.....

[@thefirminator]

​@@maths_505me too man

[@holyshit922]

You really made it complex . Especially when original integral can be calculated with techniques accessible to beginners in integral calculus One more thing There are many integrals on your channel and no other math stuff

[@maths_505]

Beginners??? Beginners only use antiderivative.....if you mean people with some knowledge of real integration methods then it's no harm to try new things......its often fun and opens up new insights.... I get it that you can evaluate the integral using real methods but some solution developments are just more fun than others.....

[@maths_505]

Nd my channel is mostly integrals and DEs cuz I like them the most..... Although I am preparing lecture notes for formal courses that I plan to upload in a couple months.

[@two697]

​@@maths_505 for what courses if you don't mind sharing?

[@maths_505]

@@two697 complex analysis and ODEs