On the logarithm of the fractional part Integral Let $\{\}$ denote the fractional part function, then does the following integral admit a closed-form ? 
$$\int_{0}^{1}x\ln\bigg(\bigg\{\frac{1}{x}\bigg\}\bigg)dx$$
 A: This is not a complete answer but provides an infinite series expression.
Upon using the $u$-substitution $u=1/x$, we have
$$
\int_0^1 x\ln\left(\left\{\frac1x\right\}\right)\,dx=\int_1^\infty\frac{\ln(\{u\})}{u^3}\,du=\sum_{n=1}^\infty\int_n^{n+1}\frac{\ln(u-n)}{u^3}\,du.
$$
Now using the substitution $x=\ln(u-n)$, the series becomes
$$
\sum_{n=1}^\infty\int_{-\infty}^0\frac{x e^x}{(e^x+n)^3}\,dx.
$$
Evaluating this last integral via integration by parts gives
$$
-\frac{1+(n+1) \ln \left(1+\frac{1}{n}\right)}{ 2n^2 (n+1)}.
$$
Inserting this in for the integral and after doing a little bit of algebra and simplifying, the last sum is equivalent to
$$
-\frac12\left(-1+\frac{\pi^2}{6}+\sum_{n=1}^\infty\frac{\ln(1+\frac1n)}{n^2}\right).
$$
Altogether, this gives
$$
\int_0^1 x\ln\left(\left\{\frac1x\right\}\right)\,dx=-\frac12\left(-1+\frac{\pi^2}{6}+\sum_{n=1}^\infty\frac{\ln(1+\frac1n)}{n^2}\right)\approx-0.754071
$$
where I used Mathematica for the last approximation. I don't think there is much hope for the last sum to simplify because you would have to deal with a sum of $\zeta(s)$ evaluated at odd positive integers, for which there is no known elementary expression.
