I am trying to show that
\begin{equation} \int_0^{\infty} \frac{x}{e^x-1} dx = \sum_{n=0}^{\infty} \frac{1}{n^2} \end{equation}
Please note that this is the Lebesgue integral.
My current strategy is to note that \begin{equation} \frac{x}{e^x-1} = \frac{x}{e^x} \frac{1}{1-e^{-x}} \end{equation} And to write the RHS as a geometric series, which I've written as: $\sum_{n=0}^{\infty}xe^{-x(n+1)}$.
So my integral is thus: \begin{equation}\int_0^{\infty}\sum_{n=0}^{\infty}xe^{-x(n+1)} \end{equation}
I know that I can flip the integral and the summation (by monotone convergence theorem), i.e.: Work with $\sum \int$, but is this worth doing?
Also, I'm fairly certain I will take the limit of the improper integral, am I right in thinking that the upper variable of integration will be $n$so I will be taking the limit as $n$ approaches infinity of the integral from $0$ to $n$ of the geometric series from infinity to $n$?
Any input on this problem will be much appreciated!
Yes, after using the MCT, you have $$ \sum_{n=1}^{\infty} \int_0^{\infty} x e^{-nx} \, dx, $$ and the integral evaluates to $$ \int_0^{\infty} x e^{-nx} \, dx = \frac{1}{n^2} \int_0^{\infty} ye^{-y} \, dy = \frac{1}{n^2}, $$ by integrating by parts or using the definition of the Gamma-function.