A surprisingly fascinating integral

This is one of those integration results that you'd least expect given the structure of the integral and that makes the solution development all the more satisfying.
Series expansion used in the video for ln(2sin(x)):

[totor]

What i found really cool with this integral is that once it is reduced to -6 time the integral from 0 to pi/2 of xln(sin(x)) if we rewrite it as -6 time the integral from 0 to pi/2 of x(ln(e^ix)+ ln(1-e^-2ix)-ln(2)) using complex exposant form for sin and proprieties of the log then we can split this into three integrals the first and third one being obvious and using the series expansion for ln(1-e^-2ix) witch is - the sum as n goes from 1 to infinity of e^-2in/n and integrating by part we end up with an expression of zeta of 3 for the real part and pi time zeta of 2 for imaginary part however since we are integrating a real function on a real interval the imaginary part must be zero so we can conclude that zeta of 2 is pi^2/6 and we end up with the same result as you. Sorry for my bad english and keep up the good work !

[slavinojunepri]

The result is aesthetically pleasant indeed. Thanks for sharing.

[yoav]

Nice. Btw today michael penn showed today the integral identity 2/(npi) ((-1)^n-cos(nx))ln(2sin(x/2)) from 0 to pi=1/n^2. He proved it in tricky way but now thanks to you, i see that with the series of ln(2sin(x/2)) we can prove it easily😃💯

[patricius]

I appreciate you using ln instead of log, as for some reason, most mathtubers I've watched use log :)

[trelosyiaellinika]

Indeed, it's not an exaggeration when you say "a really cool integral" and "nice to evaluate". Really beautiful. Thanks! Shouldn't we also consider (3/4)(π^2)ln2 as one of the solutions for even values of k? Why did we discard it altogether?

[TheRandomFool]

I've actually solved the integral of x^n * log(sin(x)) from 0 to pi/2 where n is a natural number. It's a complicated finite sum in terms of the Riemann zeta function and Dirichlet eta function. So if that is equal to L(n), then the integral of x^n / sin(x)^2 from 0 to pi/2 is -n*(n-1)*L(n-2).

[zinzhao]

beautiful. can you do more on abreys constant when it comes to these integrals?

[yoav]

Very nice.for the series of ln(2sinx): i+ix -sum e^-2nix/n from 1 to inf.now if we take the real part we find that sum -cos(2nx)/n from 1 to inf is ln(2sinx), and for abonus if we take the im part we find that the sum sin(2nx)/n from 1 to inf is 0.5pi-x (for 0<x<=pi/2).😃💯

[carlosgiovanardi]

Great!!👏👏👏
Can you integrate from 0 to pi/3? (if you have time)🤪

[holyshit]

I trted integration by parts twice and everything would be ok if we have x^2 in the numerator

[giuseppemalaguti]

With bermoulli numners and integration by parts..my result Is pi^(2k)B(2k)/(2k)!(k+1).... Perhaps is correct

[ガアラ-hh]

Note for later let x be arcsin x => Leibnitz => u = x^2 => u = tan x should be doable from there

[alihasani]

Can anyone suggest me an youtube channel which deals with integration but on an easier lever than this😅