<s>[INST] Prove that : $2\int_{1}^{\infty}\left(\lfloor x\rfloor-(\lfloor x \rfloor)^2\over x^4\right)dx=-\gamma$ I saw this $\gamma=\int_{1}^{\infty}\left({1\over \lfloor x\rfloor}-{1\over x}\right)dx$ on wikipedia so prompted me to investigate.
I used wolfram integrator to look more of these types.
I found one but wasn't sure it is correct,
$$2\int_{1}^{\infty}\left(\lfloor x\rfloor-(\lfloor x \rfloor)^2\over x^4\right)dx=-\gamma\tag1$$
So can anyone please help to verify (1)?
No ideas where to start so no attempt here!
 [/INST] Your conjecture, unfortunately, is not correct, but it is instructive to find the exact value.  Let
$$\begin{align*} I &= \int_{x=1}^\infty \frac{\lfloor x \rfloor - \lfloor x \rfloor^2}{x^4} \, dx \\
&= \sum_{k=1}^\infty \int_{x=0}^1 \frac{k - k^2}{(k+x)^4} \, dx \\
&= \sum_{k=1}^\infty k(1-k) \left[-\frac{1}{3(x+k)^3}\right]_{x=0}^1 \\
&= \sum_{k=1}^\infty \frac{1 + 2k - 3k^3}{3k^2(1+k)^3} \\
&= \frac{1}{3} \sum_{k=1}^\infty \left(\frac{1}{k+1} - \frac{1}{k} + \frac{1}{k^2} - \frac{3}{(k+1)^2} + \frac{2}{(1+k)^3} \right) \\
&= \frac{1}{3} \left(-1 + \zeta(2) - 3(\zeta(2) - 1) + 2 (\zeta(3) - 1) \right) \\
&= \frac{2}{3}(\zeta(3) - \zeta(2)), 
\end{align*}$$
where $\zeta(s) = \sum_{k=1}^\infty k^{-s}$ is the Riemann zeta function, and in particular, $\zeta(2) = \pi^2/6$ and $\zeta(3)$ is Apéry's constant.  Thus $$2I = \frac{4}{3}\zeta(3) - \frac{2\pi^2}{9} \approx -0.59050288491817620143.$$
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