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=1∞∑k=1cos(nx)cos(kx)nk. Then, because ∫π0cos(nx)cos(kx)dx=0, n≠k, 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(1−eix)+i2(π−x)

**Answer**

**Preliminaries**

Looking at the real part of log(1−eix), we get

log(2sin(x2))=−∞∑k=1cos(kx)k

Integrating by parts twice, we get

∫2π0x2cos(kx)dx={8π33if k=04πk2if k≠0

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=1∞∑k=11(k+j)2kj=∞∑j=1∞∑k=1(1kj−1k(k+j))(1kj−1j(k+j))=∞∑j=1∞∑k=11k21j2−2∞∑j=1∞∑k=11k2(k+j)j+∞∑j=1∞∑k=11(k+j)2kj∞∑j=1∞∑k=11k2(k+j)j=12ζ(2)2=π472

Explanation:

(3a): 1kj−1k(k+j)=1j(j+k) and 1kj−1j(k+j)=1k(j+k)

(3b): expand the products and separate the sums

(3b): note that ∞∑j=1∞∑k=11k2(k+j)j=∞∑j=1∞∑k=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=1∞∑k=11(k+j)2kj=∞∑j=1∞∑k=j+11k2(k−j)j=∞∑k=1k−1∑j=11k2(k−j)j=∞∑k=1k−1∑j=11k3(1k−j+1j)=2∞∑k=1(Hkk3−1k4)=2∞∑k=1∞∑j=11k3(1j−1j+k)−2ζ(4)=2∞∑k=1∞∑j=11k2(k+j)j−2ζ(4)=π436−π445=π4180

Explanation:

(4a): substitute k↦k−j

(4b): swap order of summation

(4b): (we include k=1 since the inner sum is then 0)

(4c): partial fractions

(4d): k−1∑j=11k−j=k−1∑j=11j=Hk−1k

(4e): Hk=∞∑j=1(1j−1j+k)

(4f): simplify the summand

(4g): apply (3)

(4h): simplify

**Putting The Preliminaries To Work**

∫2π0x2log(2sin(x2))2dx=∫2π0x2∞∑j=1∞∑k=1cos(kx)kcos(jx)jdx=∫2π0x2∞∑j=1∞∑k=1cos((k+j)x)+cos((k−j)x)2kjdx=2π∞∑j=1∞∑k=11(k+j)2kj+4π∞∑j=1∞∑k=j+11(k−j)2kj+4π33∞∑k=11k2=2π∞∑j=1∞∑k=11(k+j)2kj+4π∞∑j=1∞∑k=11k2(k+j)j+4π33ζ(2)=π590+π518+2π59=13π545

Explanation:

(5a): apply (1)

(5b): cos(a)cos(b)=cos(a+b)+cos(a−b)2

(5c): apply (2)

(5d): substitute k↦k+j in the middle sum

(5e): apply (3) and (4)

(5f): simplify

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