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?

Thanks in advance for your help.


1 Answer 1


$$\begin{eqnarray*}\iint_{(0,1)^2}(1-xy)^n\,dx\,dy &\stackrel{\text{Symmetry}}{=}&2\int_{0}^{1}\int_{0}^{x}(1-xy)^n\,dy\,dx\\&\stackrel{y\mapsto x z}{=}&2 \int_{0}^{1}\int_{0}^{1}x(1-x^2 z)^n\,dz\,dx\\&\stackrel{\text{Fubini}}{=}&\frac{1}{n+1}\int_{0}^{1}\frac{1-(1-z)^{n+1}}{z}\,dz\\&\stackrel{z\mapsto 1-t}{=}&\frac{1}{n+1}\int_{0}^{1}\frac{1-t^{n+1}}{1-t}\,dt = \color{red}{\frac{H_{n+1}}{n+1}}.\end{eqnarray*}$$

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.

  • $\begingroup$ 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. $\endgroup$ Jun 22, 2017 at 19:57
  • 1
    $\begingroup$ 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) $\endgroup$
    – FDP
    Jun 22, 2017 at 20:06
  • $\begingroup$ always right and elegant to the target ! $\endgroup$
    – G Cab
    Jun 22, 2017 at 20:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.