# Show me how to evaluate $\int_0^1\int_0^1\frac{[-\ln(x)]^s}{1-xy}dxdy=\frac{\zeta(s+2)}{\Gamma(s+2)}$

Double integrals

$$\int_0^1\int_0^1\frac{[-\ln(x)]^s}{1-xy}dxdy=\frac{\zeta(s+2)}{\Gamma(s+2)} \tag1$$

$$\int_0^1\int_0^1\frac{[-\ln(xy)]^s}{1-xy}dxdy=\zeta(s+2)\Gamma(s+2) \tag2$$

Where $\sum_{n=0}^{\infty}\frac{1}{(n+1)^s}=\zeta(s)$, valid for $\Re(s)>1$ and $\Gamma(n+1)=n!$ valid for all non-negative integers and rational arguments.

I came across these two double integrals during the time I was on Wolfram integrator, was trying to search for something. It didn't gave me the closed form, just the numerical values and the rest I had to find the closed form base on these values.

I don't know how to prove these integrals, can somebody show me how to prove it, so I can learn from it, so next time I can independently do it myself.

I hope these closed form are correct.

P.s

Please, try not to miss too many steps. Thank you!

• you want to prove that $\int_0^1\int_0^1\frac{[-\ln(x)]^s}{1-xy}dxdy=\frac{\zeta(s+2)}{\Gamma(s+2)}$ ? Commented May 8, 2016 at 20:39
• Yes, please, can you help me?
– user335850
Commented May 8, 2016 at 21:15

For the first integral, we can write

\begin{align} I(s)&=\int_0^1\int_0^1 \frac{\left(-\log(x)\right)^s}{1-xy}\,dx\,dy\\\\ &=\sum_{n=0}^\infty\left(\int_0^1 y^n \,dy \int_0^1 x^n \log^s(1/x)\,dx \right)\\\\ &=\sum_{n=0}^\infty \frac{1}{n+1}\int_0^1 x^n \log^s(1/x)\,dx\\\\ &=\sum_{n=0}^\infty \frac{1}{n+1} \int_0^\infty t^s\,e^{-(n+1)t}\,dt\\\\ &=\sum_{n=0}^\infty \frac{1}{(n+1)^{s+2}}\int_0^\infty t^s\,e^{-t}\,dt\\\\ &=\zeta(s+2)\Gamma(s+1) \end{align}

For the second integral, we can write

\begin{align} J(s)&=\int_0^1\int_0^1 \frac{\left(-\log(xy)\right)^s}{1-xy}\,dx\,dy\\\\ &=\sum_{n=1}^\infty \int_0^1 \frac1y \int_0^y t^n\log^s(1/t)\,dt \,dy\\\\ &=\sum_{n=1}^\infty \int_0^1 t^n\log^s(1/t) \int_t^1 \frac1y \,dy \,dt\\\\ &=\sum_{n=1}^\infty \int_0^1 t^n\log^{s+1}(1/t)\,dt\\\\ &=\sum_{n=1}^\infty \int_0^1 e^{-(n+1)u}u^{s+1}\,du\\\\ &=\sum_{n=1}^\infty \frac{1}{(n+1)^{s+2}}\int_0^1 e^{-u}u^{s+1}\,du\\\\ &=\zeta(s+2)\Gamma(s+2) \end{align}

as was to be shown!

• (+1). Is my first stated integral was wrong or there is a type error?
– user334593
Commented May 9, 2016 at 4:12
• The first integral in the OP is incorrect. Commented May 9, 2016 at 4:27

I can't comment.