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I wanted to evaluate $\int_0^\infty \frac{xe^{-x}}{1+e^{-x}} \, dx$ and found this answer where it is shown

$$\int_0^\infty \frac{xe^{-x}}{1+e^{-x}} \, dx = \int_0^\infty \sum_{k=0}^\infty(-1)^kxe^{-(k+1)x}\, dx = \sum_{k=0}^\infty(-1)^k\int_0^\infty xe^{-(k+1)x}\, dx $$

In the last step, switching the sum and integral is justified by answerer "because the sum and integral converge." There are many examples where changing the order of a convergent infinite series and convergent infinite integral is not allowed.

How is this justified here? Monotone convergence does not help.

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    $\begingroup$ Dominated convergence. Take the absolute value of each term, then you still get a convergent sum. $\endgroup$ Commented Jul 27, 2020 at 18:41

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In this case we can be brutal, taking the absolute value of each term still leads to a convergent sum and integral because the factor $x$ cancels the pole of $\frac{1}{e^x-1}$ at $0$ if we do so. Thus changing the order is justified by the dominated convergence theorem.

We can be more conservative too, if we wish (and if we considered the integral of $\frac{1}{e^x+1}$ instead of $\frac{x}{e^x+1}$ we would have to be). The series is alternating, with the terms decreasing in absolute value (except at $0$, where all terms have absolute value $1$), thus $$0 \leqslant \sum_{k = 0}^{N} (-1)^ke^{-(k+1)x} \leqslant e^{-x}$$ on $[0, +\infty)$ for every $N \in \mathbb{N}$, and we can appeal to the dominated convergence theorem with dominating function $xe^{-x}$. This works for $$\int_0^{+\infty} \frac{f(x)}{e^x+1}\,dx$$ whenever $f(x)e^{-x}$ is Lebesgue-integrable on $[0,+\infty)$. (Whether it helps determining the value of the integral is a different question.)

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