<s>[INST] 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. 
 [/INST] Alternatively, one may follow the same line of thought and use summation by parts formula to calculate the infinite sum. Similarly,
$$\int_0^{+\infty} x\lfloor \frac{1}{x}\rfloor dx =\sum_{n=1}^{\infty}\frac{n}{2}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)=-\frac{1}{2}\sum_{n=1}^{\infty}f_n(g_{n+1}-g_n)$$
where $f_n = n$ and $g_n = 1/n^2$. Now, summation by parts for the last expression gives:
$$-\frac{1}{2}\sum_{n=1}^{\infty}f_n(g_{n+1}-g_n) = -\frac{1}{2}\left(\ \lim_{n\to\infty}\frac{n}{(n+1)^2} -1 -\sum_{n=2}^{\infty}\frac{1}{n^2}\right)$$
But it is well-known that 
$$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$$
Hence, 
$$-\frac{1}{2}\sum_{n=1}^{\infty}f_n(g_{n+1}-g_n) = -\frac{1}{2}\left(\ \lim_{n\to\infty}\frac{n}{(n+1)^2} -1 - (\frac{\pi^2}{6}-1)\right) = \frac{\pi^2}{12}$$
</s>