Unit square integral $\int_{0}^{1} \int_{0}^{1} \left( - \frac{\ln(xy)}{1-xy} \right)^{m} dx dy $

Question

In an article by Guillera and Sondow, one of the unit square integral identities that is proved (on p. 9) is: $$\int_{0}^{1} \int_{0}^{1} \frac{\left(-\ln(xy) \right)^{s}}{1-xy} dx dy = \Gamma(s+2) \zeta(s+2),$$ which holds for $\mathfrak{R}(s) > -1 $.

Let's define $$I_{m} := \int_{0}^{1} \int_{0}^{1} \left( \frac{-\ln(xy) }{1-xy} \right)^{m} dx dy $$ for $m \in \mathbb{Z}_{\geq 1} $. Then WA finds $$I_{1} = -2 \zeta(3), \qquad I_{2} = 6 \zeta(3). $$ However, so far I haven't been able to find any closed-form evaluations for $I_{m}$ when $m \geq 3$, neither in the literature nor with CAS software.

Questions:

1. What is $I_{m}$ for $m>2$ ?
2. Does this family of definite integrals appear in the literature somewhere?

Answer 1

We can start by considering the following integral: $$\int_0^1\int_0^1 \frac{(xy)^a}{z-xy}dxdy=\frac{1}{z}\sum_{n=0}^\infty \int_0^1\int_0^1(xy)^a\left(\frac{xy}{z}\right)^ndxdy$$ $$=\sum_{n=0}^\infty \frac{1}{z^{n+1}}\int_0^1x^{n+a}dx\int_0^1y^{n+a}dy=\sum_{n=1}^\infty \frac{1}{z^{n+2}}\frac{1}{(n+a)^2}$$ Taking $m$ derivatives w.r.t. $a$ and setting $a$ to $0$ gives: $$\int_0^1\int_0^1 \frac{(-\ln(xy))^m}{z-xy}dxdy=(m+1)!\operatorname{Li}_{m+2}\left(\frac{1}{z}\right)$$ Where $\operatorname{Li}_m(x)$ is the polylogarithm function.
Furthermore, we can take $m-1$ derivatives w.r.t. $z$ (and set it to $1$) in order to find: $$\int_0^1\int_0^1 \frac{(-\ln(xy))^m}{(1-xy)^m}dxdy=(-1)^{m-1}m(m+1)\frac{d^{m-1}}{dz^{m-1}}\operatorname{Li}_{m+2}\left(\frac{1}{z}\right)\bigg|_{z=1}$$ And with the help of OEIS we can obtain the following closed form: $$\boxed{\int_0^1\int_0^1 \left(\frac{-\ln(xy)}{1-xy}\right)^mdxdy=m(m+1)\sum_{k=1}^{m-1}|s(m-1,m-k)|\zeta(k+2)}$$ Where $s(n,m)$ is the Stirling number of the first kind.
In particular, we have: \begin{array}{|c|c|} \hline I_2 & 6\zeta(3) \\ \hline I_3 & 12(\zeta(3)+\zeta(4)) \\ \hline I_4 & 20(\zeta(3)+3\zeta(4)+2\zeta(5)) \\ \hline I_5 & 30(\zeta(3)+6\zeta(4)+11\zeta(5)+6\zeta(5)) \\ \hline I_6 & 42(\zeta(3)+10\zeta(4)+34\zeta(5)+50\zeta(6)+24\zeta(7)) \\ \hline I_7 & 56(\zeta(3)+15\zeta(4)+85\zeta(5)+225\zeta(6)+274\zeta(7)+120\zeta(8)) \\ \hline \end{array}

Answer 2

\begin{align} &\int_{0}^{1} \int_{0}^{1} \left( - \frac{\ln(xy)}{1-xy} \right)^{m}\ \overset{xy=t}{dy} \ dx \\ =& \int_0^1 \int_0^x \left( - \frac{\ln t}{1-t} \right)^{m}\frac1x \ dt\ dx = \int_0^1 \int_t^1 \left( - \frac{\ln t}{1-t} \right)^{m}\frac1x \ dx\ dt\\ = &\int_0^1 \frac{(-\ln t)^{m+1}}{(1-t)^m} \ dt \overset{t=e^{-u}}= \int_0^{\infty} \frac{u^{m+1} e^{-t}}{(1-e^{-u})^m}\ du\\ =& \int_0^{\infty} {u^{m+1} e^{-t}} \sum_{k=0}^{\infty} \binom{m+k-1}{k} e^{-uk} du\\ =& \sum_{k=0}^{\infty} \binom{m+k-1}{k}\int_0^1 u^{m+1} e^{-(k+1)u}\ du \\=&\sum_{k=0}^{\infty} \binom{m+k-1}{k} \frac{(m+1)!}{(k+1)^{m+2}} \end{align}