Collaboration between Integration and Summation

When it is permissible, it is often convenient when evaluating certain integrals to interchange the order of a summation and an integration. Let's use the following definite integral as an example to illustrate this technique:$$\int_0^{\infty}\frac{x}{e^x-1} \mathrm{d}x$$The graph is as shown in Figure 1.

Figure 1: created in GeoGebra

Let's begin the integration:$$\begin{align} \int_0^{1}\frac{x}{e^x-1} \mathrm{d}x&=\int_0^{1}\frac{x}{\dfrac{1}{e^{-x}}-1} \mathrm{d}x\\
&=\int_0^{1}\frac{x \, e^{-x}}{1-e^{-x}}\mathrm{d}x\\
&=\int_0^{1} x \, e^{-x} \left ( \dfrac{1}{1-e^{-x}} \right ) \mathrm{d}x\\
&=\int_0^{1} x \, e^{-x} \sum_{n=0}^{\infty} e^{-nx} \, \mathrm{d}x\\
&=\sum_{n=0}^{\infty} \int_0^{1} x \, e^{-x(n+1)} \, \mathrm{d}x\\
&=\sum_{n=1}^{\infty} \int_0^{1} x \, e^{-nx} \, \mathrm{d}x \\
\text{Now } \int x \, e^{-nx} \, \mathrm{d}x &= \int x \frac{\mathrm{d}}{\mathrm{d}x} \left (\frac{e^{-nx}}{-n} \right ) \, \mathrm{d}x\\
&=\frac{-x \, e^{-nx}}{n}+\frac{1}{n} \int e^{-nx} \, \mathrm{d}x\\
&=\frac{-x \, e^{-nx}}{n}- \frac{e^{-nx}}{n^2} +C\\
\text{So }\int_0^{1}\frac{x}{e^x-1} \mathrm{d}x&=\sum_{n=1}^{\infty} \left [ \frac{-x \, e^{-nx}}{n}- \frac{e^{-nx}}{n^2} \right ]_0^1\\
&=\sum_{n=1}^{\infty} \left ( \frac{-e^{-n}}{n}-\frac{e^{-n}}{n^2}+\frac{1}{n^2} \right )\\
&=\sum_{n=1}^{\infty} \frac{1}{n^2} -\sum_{n=1}^{\infty}\frac{e^{-n}}{n} -\sum_{n=1}^{\infty} \frac{e^{-n}}{n^2}\\
\text{Now }\sum_{n=1}^{\infty} \frac{1}{n^2} &=\zeta(2)\\
&=\frac{\pi^2}{6}\\
\sum_{n=1}^{\infty}\frac{e^{-n}}{n} &= 1-\ln(e-1) \\
\sum_{n=1}^{\infty}\frac{e^{-n}}{n^2} &= \text{Li}_2 \left (\dfrac{1}{e} \right )\\
\text{Hence } \int_0^{1}\frac{x}{e^x-1} \mathrm{d}x&=\frac{\pi^2}{6}+\ln(e-1) -\text{Li}_2 \left (\dfrac{1}{e} \right )-1\\
&\approx 0.777505

\end{align}$$Looking at Figure 1, it can be seen that this result looks just about right. I'm thankful to Mathematics M1's YouTube video for detailing the steps required to reach this result. I'm reproducing the steps here as a means of practising my LaTeX skills but it's also to remind me how interesting the integration technique is, specifically the way in which the integral and the summation are swapped around. Li is the Eulerian Logarithmic Integral and deserves a blog post of its own.$$ \mathrm{Li}(x)=\int_2^x \dfrac{\mathrm{d}t}{\ln t }= \mathrm{li}(x)-\mathrm{li}(2) \text{ where } \mathrm{li}(x)=\int_0^x \dfrac{\mathrm{d}t}{\ln t }$$