I am interested in solving the following definite double integral

$$\int_0^1\int_0^1\Bigl\{\frac{1}{x}\Bigr\}\Bigl\{\frac{1}{y}\Bigr\}\frac{(1-x)(1-y)}{1-xy} dx dy$$

where $\{z\}=z-\lfloor z\rfloor$ denotes the fractional part of a number $z$.

I have been searching for a good change of variable to separate the double integral into a product of integrals. The best option I tried was $r=\ln(1−xy)$, $s=\frac12\ln(\frac{x}{y})$ but I had no success with it.


The given integral equal the sum over $k\geq 0$ of $$ \iint_{(1,+\infty)^2}\frac{\{x\}\{y\}(x-1)(y-1)}{x^{k+3} y^{k+3}}\,dx\,dy=\left[\int_{1}^{+\infty}\frac{\{x\}(x-1)}{x^{k+3}}\,dx\right]^2 $$ where $$\begin{eqnarray*}\int_{1}^{+\infty}\frac{\{x\}(x-1)}{x^{k+3}}\,dx&=&\sum_{n\geq 1}\int_{n}^{n+1}\frac{(x-n)(x-1)}{x^{k+3}}\,dx\\ &=& \frac{1}{k(k+1)}-\frac{\zeta(k+1)}{k+1}+\frac{\zeta(k+2)}{k+2}\end{eqnarray*}$$ which for $k\to 0^+$ equals $-\gamma+\frac{\pi^2}{12}$. The original integral is so converted into $$\left[-\gamma+\frac{\pi^2}{12}\right]^2+\sum_{k\geq 1}\left[\frac{1}{k}-\frac{1}{k+1}-\frac{\zeta(k+1)}{k+1}+\frac{\zeta(k+2)}{k+2}\right]^2 $$ but I am not so sure about the existence of a nice closed form for series like $\sum_{m\geq 2}\frac{\zeta(m)^2}{m^2}$.
On the other hand $\sum_{k\geq 1}\left[\frac{1}{k}-\frac{\zeta(k+1)}{k+1}\right]=\gamma$ by the integral representation for $\zeta$.

  • $\begingroup$ We also have an awkward identity, i.e. $$\sum_{k\geq 1}\left[\log\left(1+\frac{1}{k}\right)-\frac{\zeta(k+1)}{k+1}\right]=\color{red}{0}.$$ $\endgroup$ – Jack D'Aurizio Oct 5 '18 at 21:07

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