<s>[INST] Double fractional part integral Let $\{\}$ denote the fractional part, does the following integral have a closed form ?
 $$\int_{0}^{1}\int_{0}^{1}\bigg\{\frac{1}{x\,y}\bigg\}^2dx\,dy$$
 [/INST] This is an incomplete answer that only addresses the 1-dimensional case.
We split the integral into continuous pieces:
$$\begin{align}
\int_0^1 \left\{ \frac{1}{x} \right\}^2 \mathrm{d}x
&= \sum_{n=1}^{\infty} \int_{\frac{1}{n+1}}^{\frac{1}{n}} \left\{ \frac{1}{x} \right\}^2 \mathrm{d}x \\\\
&= \sum_{n=1}^{\infty} \int_{\frac{1}{n+1}}^{\frac{1}{n}} \left( \frac{1}{x} - n\right)^2 \mathrm{d}x \\\\
&= \sum_{n=1}^{\infty} \left( n^2x - 2n \ln |x| - \frac{1}{x} \right)\biggr\rvert_{\frac{1}{n+1}}^{\frac{1}{n}} \\\\
&= \sum_{n=1}^{\infty} \left( n - 2n \ln \frac{1}{n} - n - \frac{n^2}{n+1} + 2n \ln \frac{1}{n+1} + n + 1 \right) \\\\
&= \sum_{n=1}^{\infty} \left( 2n \ln \frac{n}{n+1} + \frac{2n + 1}{n+1} \right)
\end{align}$$
With the aid of computer algebra, we obtain that this series converges to $$\ln (2\pi) - \gamma - 1$$ where $\gamma$ is the Euler-Mascheroni constant.
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