Vtyjkyuk

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:

$$int_0^infty x left lfloorfrac1xright rfloor dx = int_0^1 x left lfloorfrac1xright rfloor dx$$
$$= sum_n=1^infty int_1/(n+1)^1/n nx dx =sum_n=1^inftyfrac n2 left(frac1n^2 - frac1(n+1)^2right) $$
$$=
sum_n=1^inftyfrac n2 left(frac2n+1n^2(n+1)^2right)$$
$$= sum_n=1^inftyfrac1(n+1)^2 + frac12 sum_n=1^infty frac1n(n+1)^2$$
$$= fracpi^26 -1 + frac12left(sum_n=1^infty frac1n - frac1n+1 - frac1(n+1)^2right)$$
$$=fracpi^26 -1 + frac12left(sum_n=1^infty frac1n - frac1n+1right) -frac12left(sum_n=1^inftyfrac1(n+1)^2right)$$
$$= left(fracpi^26 -1right) + left(frac12cdot 1right) - frac12left(fracpi^26 -1right)$$
$$= fracpi^212.$$

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

$$nleft(frac1n^2-frac1(n+1)^2right)=frac nn^2-fracn+1-1(n+1)^2=frac1n-frac1n+1+frac1(n+1)^2.$$

The first two terms do telescope and the Basel series remains.

Alternatively, one may follow the same line of thought and use summation by parts formula to calculate the infinite sum. Similarly,
$$int_0^+infty xlfloor frac1xrfloor dx =sum_n=1^inftyfracn2left(frac1n^2-frac1(n+1)^2right)=-frac12sum_n=1^inftyf_n(g_n+1-g_n)$$

where $f_n = n$ and $g_n = 1/n^2$. Now, summation by parts for the last expression gives:

$$-frac12sum_n=1^inftyf_n(g_n+1-g_n) = -frac12left( lim_ntoinftyfracn(n+1)^2 -1 -sum_n=2^inftyfrac1n^2right)$$

But it is well-known that
$$sum_n=1^inftyfrac1n^2=fracpi^26$$

Hence,

$$-frac12sum_n=1^inftyf_n(g_n+1-g_n) = -frac12left( lim_ntoinftyfracn(n+1)^2 -1 - (fracpi^26-1)right) = fracpi^212$$