Compute the integral $$\int_0^1 dx\,dy\, \frac{\ln(1+y(1-x))}{1-xy}$$
I was just wondering if there is a way to convert the integrand into a polylog? This comes from a tutorial following a lecture on polylogarithms and co-product expansions so I am thinking this is the desired method of solution. Also, what are the limits of $x$ and $y$ here? I have never seen a double integral expressed like that. Thanks!
The path outlined below is the following: we exploit the Taylor series of $\frac{1}{1-z}$ and $\log(1+z)$ in order to write the double integral as a double integral of a double series of monomials. Then, exploiting the Beta function, we integrate each monomial over the unit square, converting the last expression in a double series over $j$ and $k$. We take $n=k+j$ as a new summation parameter, computing $\sum_{j=0}^{n}(-1)^j j!(n-j)!$ in order to remove a summation sign. After that, we are left with a familiar series, related to $\zeta(3)$.
$$\begin{eqnarray*}\iint_{(0,1)^2}\frac{\log(1+y(1-x))}{1-xy}\,dx\,dy &=& \iint_{(0,1)}\sum_{k\geq 0}x^k y^k \log(1+y(1-x))\,dx\,dy\\&=&\iint_{(0,1)^2}\sum_{k,j\geq 0}(-1)^j \frac{x^k y^k y^{j+1} (1-x)^{j+1}}{j+1}\\&=&\sum_{k,j\geq 0}(-1)^j\frac{k!j!}{(k+j+2)\cdot(k+j+2)!}\\&=&\sum_{n=0}^{+\infty}\frac{(1+(-1)^n)\cdot(n+1)!}{(n+2)^2(n+2)!}\\&=&\sum_{m=0}^{+\infty}\frac{2}{(2m+2)^3}=\color{red}{\frac{\zeta(3)}{4}}.\end{eqnarray*}$$