I am trying to show that the following interchange of summation and integration is valid: $$\int_0^{+\infty} \sum_{n=1}^{+\infty} \frac{e^{-nx}}{n}\,dx=\sum_{n=1}^{+\infty} \int_0^{+\infty} \frac{e^{-nx}}{n}\,dx$$
I first tried using the Weierstrass $M$-test in order to show that the series summing the function $f_n(x)=\frac{e^{-nx}}{n}$ converges uniformly on $[0,+\infty)$, but I was unsuccessful. Indeed, I could not find a sequence $M_n$ such that $$\left|\frac{e^{-nx}}{n}\right|\le M_n \qquad x\in[0,+\infty)$$ and $$\sum_{n=1}^{+\infty}M_n\le +\infty$$ I tried $M_n=1/n$ but its series does not converge.
The series does not converge uniformly on $(0,\infty)$ since for $x_n = 1/n$,
$$\sup_{x \in (0,\infty)}\sum_{k = n+1}^{\infty} \frac{e^{-kx}}{k}\geqslant \sum_{k = n+1}^{2n} \frac{e^{-kx_n}}{k} > n \frac{e^{-2nx_n}}{2n} = \frac{e^{-2}}{2}.$$
So you will not have any luck with the Weierstrass M-test.
However, note that as $n \to \infty$ we have monotone convergence:
$$\sum_{k=1}^n \frac{e^{-kx}}{k} \uparrow \begin{cases}+\infty, \quad x = 0\\ -\log(1 - e^{-x}), \quad x > 0 \end{cases} $$
We can apply the monotone convergence theorem, and since the limit function is Lebesgue integrable on $(0,\infty),$ we have
$$-\int_{(0,\infty)} \log(1 - e^{-x}) = \frac{\pi^2}{6} = \sum_{n=1}^\infty\frac{1}{n^2} = \sum_{n=1}^\infty \int_{(0,\infty)} \frac{e^{-nx}}{n}$$