# Closed form for $\int_0^1 \int_0^1 {(1-xy)}^n \, dx \, dy$

I was trying to calculate $$\sum_{k=0}^n (-1)^k \binom{n}{k} \frac{1}{(n+1-k)^2}$$ and I found that it suffices to calculate the $$I_n=\int_0^1 \int_0^1 {(1-xy)}^n \,dx\,dy.$$ Is there any way to find a closed form for this? I tried to use the ordinary and the exponential generating functions of $\{I_n\}_{n\in \mathbb{N}}$, but I couldn't proceed. I also tried the change of variables $$x=\frac{u+v}{2},\ \ \ y=\frac{u-v}{2},$$ but it got me nowhere. Any hints or ideas?

$$\begin{eqnarray*}\iint_{(0,1)^2}(1-xy)^n\,dx\,dy &\stackrel{\text{Symmetry}}{=}&2\int_{0}^{1}\int_{0}^{x}(1-xy)^n\,dy\,dx\\&\stackrel{y\mapsto x z}{=}&2 \int_{0}^{1}\int_{0}^{1}x(1-x^2 z)^n\,dz\,dx\\&\stackrel{\text{Fubini}}{=}&\frac{1}{n+1}\int_{0}^{1}\frac{1-(1-z)^{n+1}}{z}\,dz\\&\stackrel{z\mapsto 1-t}{=}&\frac{1}{n+1}\int_{0}^{1}\frac{1-t^{n+1}}{1-t}\,dt = \color{red}{\frac{H_{n+1}}{n+1}}.\end{eqnarray*}$$
As an alternative: $$\begin{eqnarray*}\sum_{k=0}^{n}(-1)^k\binom{n}{k}\frac{1}{(n+1-k)^2}&=&\sum_{k=0}^{n}(-1)^k\binom{n}{k}\int_{0}^{1}x^{n-k}\left(-\log x\right)\,dx\\ &=& \int_{0}^{1}(-\log x)\sum_{k=0}^{n}(-1)^k\binom{n}{k}x^{n-k}\,dx\\&=& (-1)^{n+1}\int_{0}^{1}(1-x)^n\log(x)\,dx\end{eqnarray*}$$ where $$\int_{0}^{1}x^n\log(1-x)\,dx \stackrel{\text{IBP}}{=}-\frac{1}{n+1}\int_{0}^{1}\frac{1-x^{n+1}}{1-x}\,dx$$ and the conclusion is the same as before.
• Ahh..., I ended up with the integral in my question by the integral of the 3rd line, but I never performed the change of variables $z=1-t$ to the $$\displaystyle{\int_{0}^1 \frac{1-(1-z)^{n+1}}{z}}.$$ Anyway, thanks for your help. Jun 22, 2017 at 19:57
• for $n\neq -1$ an antiderivative of $(1-xy)^n$ is$-\frac{{{\left( 1-x\cdot y\right) }^{1+n}}}{\left( n+1\right) \cdot y}$ (y fixed)