Understanding $s \int_1^\infty \frac{[x]}{x^{s+1}} \, \mathrm{d}x = s\sum_{n=1}^\infty n \int_{n}^{n+1} \frac{1}{x^{s+1}} \, \mathrm{d}x$ 
$$s \int_1^\infty \frac{[x]}{x^{s+1}} \, \mathrm{d}x = s\sum_{n=1}^\infty n \int_{n}^{n+1} \frac{1}{x^{s+1}} \, \mathrm{d}x$$

Can anyone explain how this identity works? Thanks!!
 A: So, the fundamental identity at play here is (assuming that $[x]$ is the "floor" function $\lfloor x \rfloor$, i.e. that $[x]$ is the largest integer not exceeding $x$),
$$\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor}
\newcommand{\dd}{\mathrm{d}}
\int_1^\infty \frac{\floor{x}}{x^{s+1}} \, \dd x = \sum_{n=1}^\infty n \int_n^{n+1} \frac{\dd x}{x^{s+1}}$$
Well, consider the left-hand side, split up similarly to the right-hand side:
$$\begin{align*}
\int_1^\infty \frac{\floor{x}}{x^{s+1}} \, \dd x
&= \int_1^2 \frac{\floor{x}}{x^{s+1}} \, \dd x + \int_2^3 \frac{\floor{x}}{x^{s+1}} \, \dd x + \int_3^4 \frac{\floor{x}}{x^{s+1}} \, \dd x + \cdots\\
&= \sum_{n=1}^\infty \int_n^{n+1} \frac{\floor{x}}{x^{s+1}} \, \dd x
\end{align*}$$
Now, look at this remaining integral. Since we're integrating from $n$ to $n+1$, this means that $x \in (n,n+1)$.
What is $\floor{x}$, then? Clearly, $\floor{x} = n$.
This $n$ is independent of $x$, and so can be pulled out of the integral, giving the identity you want.
