Can someone explain this integration trick for log-sine integrals?

I was working on this rather challenging log-sine integral:

2π0x2ln2(2sin(x2))dx=13π545

The upper limit is a waiver from the norm of π2. Anyway, when integrating log-sin integrals one can often times use the famous identity

ln(2sin(x/2))=k=1cos(kx)k

by switching sum and integral, integrating, then evaluating the sums. However, this is only valid when x±π,±2π,. So, just for the heck of it, I decided to do something I thought would not be viable but have done it anyway. I integrated 2π0x2cos2(kx)k2, evaluated the resulting sums and arrived at 41π5180.

Now, take the integral π0x2ln2(2cos(x/2))dx=11π5180. This one can be done by using Cauchy’s cosine formula, differentiating, and so on. Anyway, note that when I add the two results, the correct result for the integral at hand is obtained.

41π5180+11π5180=13π545.

By quandary/query is why does this happen to work?. A fluke? Does that cosine sum represent something that when evaluated and added to the above log-cos integral happens to be equivalent to the log-sine integral in question?.

See what I am trying to explain?. I done something I knew I was not supposed to, but it happened to work out. If viable, what would 2π0x2cos2(kx)k2 represent such that when it is added to π0x2ln2(2cos(x/2))dx=11π5180 equals 2π0x2ln2(2sin(x/2))dx=13π545.

I knew I could not just square the cosine and multiply by x2. But,sometimes it can be done due to Parseval, i.e., ln2(2sin(x/2))=n=1k=1cos(nx)cos(kx)nk. Then, because π0cos(nx)cos(kx)dx=0, nk, one can integrate π0cos2(kx)k2, then sum.

This works for 2π as well, but that x2 term tends to throw a wrench in things But I do not think I can do this in my case because of the x2 term. Or can I?. I was surprised when I saw this. I thought perhaps I stumbled onto a cool way to evaluate this, but I am not entirely sure what I have.

Or, if anyone has their own clever method?.

EDIT: I managed to evaluate this integral by considering the identity

ln(2sin(x2))=ln(1eix)+i2(πx)

Answer

Preliminaries

Looking at the real part of log(1eix), we get
log(2sin(x2))=k=1cos(kx)k
Integrating by parts twice, we get
2π0x2cos(kx)dx={8π33if k=04πk2if k0

In the derivations below, we use the values of ζ(2) and ζ(4) computed at the end of this answer.

Start to evaluate one sum, then get another:
j=1k=11(k+j)2kj=j=1k=1(1kj1k(k+j))(1kj1j(k+j))=j=1k=11k21j22j=1k=11k2(k+j)j+j=1k=11(k+j)2kjj=1k=11k2(k+j)j=12ζ(2)2=π472
Explanation:
(3a): 1kj1k(k+j)=1j(j+k) and 1kj1j(k+j)=1k(j+k)
(3b): expand the products and separate the sums
(3b): note that j=1k=11k2(k+j)j=j=1k=11k(k+j)j2
(3c): cancel the red sums, move the green sum to the left side, and divide by 2
(3d): simplify

Evaluate the sum we started before (using the other sum we got before):
j=1k=11(k+j)2kj=j=1k=j+11k2(kj)j=k=1k1j=11k2(kj)j=k=1k1j=11k3(1kj+1j)=2k=1(Hkk31k4)=2k=1j=11k3(1j1j+k)2ζ(4)=2k=1j=11k2(k+j)j2ζ(4)=π436π445=π4180
Explanation:
(4a): substitute kkj
(4b): swap order of summation
(4b): (we include k=1 since the inner sum is then 0)
(4c): partial fractions
(4d): k1j=11kj=k1j=11j=Hk1k
(4e): Hk=j=1(1j1j+k)
(4f): simplify the summand
(4g): apply (3)
(4h): simplify


Putting The Preliminaries To Work
2π0x2log(2sin(x2))2dx=2π0x2j=1k=1cos(kx)kcos(jx)jdx=2π0x2j=1k=1cos((k+j)x)+cos((kj)x)2kjdx=2πj=1k=11(k+j)2kj+4πj=1k=j+11(kj)2kj+4π33k=11k2=2πj=1k=11(k+j)2kj+4πj=1k=11k2(k+j)j+4π33ζ(2)=π590+π518+2π59=13π545
Explanation:
(5a): apply (1)
(5b): cos(a)cos(b)=cos(a+b)+cos(ab)2
(5c): apply (2)
(5d): substitute kk+j in the middle sum
(5e): apply (3) and (4)
(5f): simplify

Attribution
Source : Link , Question Author : Cody , Answer Author : robjohn

Leave a Comment