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!

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 $$2I = \frac{4}{3}\zeta(3) - \frac{2\pi^2}{9} \approx -0.59050288491817620143.$$
