How do we prove that $\int_{0}^{1}\int_{0}^{1}{\left(\ln{x}\ln{y}\right)^s\over 1-xy}dxdy=\Gamma^2(1+s)\zeta(2+2s)$?

Question

How do we prove that

$$\int_{0}^{1}\int_{0}^{1}{\left(\ln{x}\ln{y}\right)^s\over 1-xy}dxdy=\Gamma^2(1+s)\zeta(2+2s)$$

Integrate with respect to x first, let $s=1$

$$\int_{0}^{1}{\ln{y}\ln{x}\over 1-yx}dx$$

$u=\ln{x}\rightarrow xdu=dx$

$$\ln{y}\int_{0}^{\infty}{u\over y-e^{-u}}du$$

I don't think I am in the right track here, any hints please.

Answer 1

Hint. One may write, for $s>-1$, $0<xy<1$, $$ {\left(\ln{x}\ln{y}\right)^s\over 1-xy}=\sum_{n=0}^\infty (xy)^n\left(\ln{x}\ln{y}\right)^s $$ giving $$ \begin{align} \int_{0}^{1}\int_{0}^{1}{\left(\ln{x}\ln{y}\right)^s\over 1-xy}dxdy&=\int_{0}^{1}\int_{0}^{1}\sum_{n=0}^\infty (xy)^n\left(\ln{x}\ln{y}\right)^sdxdy \\\\&=\sum_{n=0}^\infty \left(\int_{0}^{1}x^n\left(-\ln{x}\right)^sdx\right)\left(\int_{0}^{1}y^n\left(-\ln{y}\right)^sdy\right) \\\\&=\sum_{n=0}^\infty \left(\int_{0}^{1}x^n\left(-\ln{x}\right)^sdx\right)^2 \\\\&=\sum_{n=0}^\infty\left(\frac{\Gamma(s+1)}{(n+1)^{s+1}}\right)^2 \\\\&=\Gamma^2(s+1)\:\sum_{n=1}^\infty\frac{1}{n^{2s+2}} \\\\&=\Gamma^2(1+s)\:\zeta(2+2s) \end{align} $$ as announced.

The interchange between $\int$ and $\sum$ may be justified by the dominated convergence theorem.