TITLE: How to prove $\int^1_0\int^1_0\frac{\log(x-x^2)-\log(y-y^2)}{(x-x^2)-(y-y^2)}dxdy=7\sum_{i=1}^\infty i^{-3}$?
QUESTION [6 upvotes]: How do you prove that
$$\int^1_0\int^1_0\frac{\log(x-x^2)-\log(y-y^2)}{(x-x^2)-(y-y^2)}dxdy=7\sum_{i=1}^\infty i^{-3}\;\;\;\left(=7 \zeta(3)\right)~?$$
p.s. Mathematica gives a pretty good approximation.

REPLY [18 votes]: Let the desired integral be denoted by $I$. Noting that $1-(2t-1)^2=4t(1-t)$ we see that
$$
I=4\int_{\square}\frac{-\log(1-(2x-1)^2)+\log(1-(2y-1)^2)}{(2x-1)^2-(2y-1)^2}dxdy\quad\hbox{with $\square=[0,1]^2$}.
$$
But, for $t\in(-1,1)$, we have
$$
-\log(1-(2t-1)^2)=\sum_{n=1}^\infty\frac{(2t-1)^{2n}}{n}
$$
So the integrand becomes, for $(x,y)\in(0,1)^2$,
$$\eqalign{
\frac{-\log(1-(2x-1)^2)+\log(1-(2y-1)^2)}{(2x-1)^2-(2y-1)^2}&=\sum_{n=1}^\infty\frac{1}{n}\cdot\frac{(2x-1)^{2n}-(2y-1)^{2n}}{(2x-1)^2-(2y-1)^2}\cr
&=\sum_{n=1}^\infty\frac{1}{n}\sum_{k=0}^{n-1}(2x-1)^{2k}(2y-1)^{2n-2k-2}
}
$$
Thus
$$\eqalign{
I&=4\sum_{n=1}^\infty\frac{1}{n}\left(\sum_{k=0}^{n-1}\int_\square
(2x-1)^{2k}(2y-1)^{2n-2k-2}dxdy\right)\cr
&= \sum_{n=1}^\infty\frac{1}{n}\left(\sum_{k=0}^{n-1}\int_0^1
2(2x-1)^{2k}dx\int_0^12(2y-1)^{2n-2k-2}dy\right)\cr
&=4 \sum_{n=1}^\infty\frac{1}{n}\sum_{k=0}^{n-1}\left(\frac{1}{2k+1}\cdot\frac{1}{2n-2k-1}\right)\cr
&=4 \sum_{n=1}^\infty\frac{1}{2n^2}\sum_{k=0}^{n-1}\left(\frac{1}{2k+1}+\frac{1}{2n-2k-1}\right)\cr
&=4 \sum_{n=1}^\infty\frac{1}{2n^2}\left(\sum_{k=0}^{n-1}\frac{1}{2k+1}+\sum_{k=0}^{n-1}\frac{1}{2n-2k-1}\right)\cr
&=4 \sum_{n=1}^\infty\frac{1}{n^2}\left(\sum_{k=0}^{n-1}\frac{1}{2k+1}\right)\cr
&=4 \sum_{n=1}^\infty\frac{1}{n^2}\left(\sum_{k=1}^{2n-1}\frac{\sin^2(\pi k/2)}{k}\right)\cr
&=16 \sum_{n=1}^\infty\frac{1}{(2n)^2}\left(\sum_{k=1}^{2n-1}\frac{\sin^2(\pi k/2)}{k}\right)\cr
&=8\sum_{n=1}^\infty\frac{1+(-1)^n}{n^2}\left(\sum_{k=1}^{n-1}\frac{\sin^2(\pi k/2)}{k}\right)=8(A-B)
}
$$
where
$$
A=\sum_{n=1}^\infty\frac{1}{n^2}\left(\sum_{k=1}^{n-1}\frac{\sin^2(\pi k/2)}{k}\right),
\qquad 
B=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}\left(\sum_{k=1}^{n-1}\frac{\sin^2(\pi k/2)}{k}\right)
$$
These sums are known, with the notation of The evaluation of Character Euler Double Sums
we have $A=[1,2a](2,1)$ and $B=[2b,2a](2,1)$, and this paper gives us
$$
A-B=\sum_{k=0}^\infty\frac{1}{(2k+1)^3}=\frac{7}{8}\zeta(3).
$$
Thus the right answer is $I=7\zeta(3)$.