Value of the integral $\int_1^\infty \frac{\lfloor x\rfloor}{x^3} \, dx$ Evaluate
$ \displaystyle \int_1^\infty \frac{\lfloor x\rfloor}{x^3} \, dx$.
(I am relatively confident of my answer below but I would still like to make sure that it is correct.)

\begin{align}
\int_1^\infty \frac{\lfloor x\rfloor}{x^3} \, dx &= \sum_{n=1}^\infty \int_n^{n+1} \frac n{x^3} \, dx \\
&= -\frac 12 \sum_{n=1}^\infty \frac n{x^2} \bigg\vert_n^{n+1} \\ 
&= \frac 12 \sum_{n=1}^\infty \left( \frac 1n - \frac n{(n+1)^2} \right) \\ 
&= \frac 12 \sum_{n=1}^\infty \frac{2n+1}{n(n+1)^2} \\ 
&= \sum_{n=1}^\infty \frac 1{(n+1)^2}+ \frac 12\sum_{n=1}^\infty \frac 1{n(n+1)^2} \\ 
&= \sum_{n=1}^\infty \frac 1{(n+1)^2}+ \frac 12\sum_{n=1}^\infty \left(\frac 1n - \frac 1{n+1} \right) - \frac 12\sum_{n=1}^\infty \frac 1{(n+1)^2} \\ 
&= \frac 12\sum_{n=1}^\infty \frac 1{(n+1)^2} + \frac 12 \\ 
&= \frac 12\sum_{k=2}^\infty \frac 1{k^2}+\frac 12 \\ 
&= \frac 12\left(\frac{\pi^2}6-1 \right)+\frac 12 \\ 
&= \boxed{\frac{\pi^2}{12}}
\end{align}
