Computing $\int\limits_0^\infty x \left \lfloor{\frac1x}\right \rfloor \, dx$ This is an integral I computed but can't find the result online or on wolfram. So here's a proof sketch, please indulge this sanity check:
$$\int_0^\infty x \left \lfloor{\frac1x}\right \rfloor \ dx = \int_0^1 x \left \lfloor{\frac1x}\right \rfloor \ dx$$
$$= \sum_{n=1}^\infty \int_{1/(n+1)}^{1/n} nx \ dx =\sum_{n=1}^\infty\frac n2 \left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right) $$
$$=
\sum_{n=1}^\infty\frac n2 \left(\frac{2n+1}{n^2(n+1)^2}\right)$$
$$= \sum_{n=1}^\infty\frac{1}{(n+1)^2} + \frac12 \sum_{n=1}^\infty \frac{1}{n(n+1)^2}$$
$$= \frac{\pi^2}{6} -1 + \frac12\left(\sum_{n=1}^\infty \frac1n - \frac{1}{n+1} - \frac{1}{(n+1)^2}\right)$$
$$=\frac{\pi^2}{6} -1 + \frac12\left(\sum_{n=1}^\infty \frac1n - \frac{1}{n+1}\right) -\frac12\left(\sum_{n=1}^\infty\frac{1}{(n+1)^2}\right)$$
$$= \left(\frac{\pi^2}{6} -1\right) + \left(\frac12\cdot 1\right) - \frac12\left(\frac{\pi^2}{6} -1\right)$$
$$= \frac{\pi^2}{12}.$$
Basically, I used the Basel sum several times, and the fifth line follows from a partial sum decomposition. The seventh follows from the known result for the Basel sum, as well as the fact that the first series in the 6th line telescopes.
I hope this is all correct. 

Use
$$n\left(\frac1{n^2}-\frac1{(n+1)^2}\right)=\frac n{n^2}-\frac{n+1-1}{(n+1)^2}=\frac1n-\frac1{n+1}+\frac1{(n+1)^2}.$$
The first two terms do telescope and the Basel series remains.
