# Double integral involving fractional part

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.

## 1 Answer

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 $$\zeta$$.

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