Conceptual passage in geometric series I have a question about a passage in solving an integral. The integral in question is rather known:
$$\int_0^{+\infty} \dfrac{x^{s-1}}{e^x-1} \text{d}x = \Gamma(s)\zeta(s)$$
Now, in solving the integral what one does is to collect $e^x$ and using the geometric series:
$$\int_0^{+\infty} \dfrac{x^{s-1}}{e^x(1-e^{-x}} \text{d}x \to \int_0^{+\infty} \dfrac{x^{s-1}}{e^x}\sum_{k = 0}^{+\infty} e^{-kx} \text{d}x$$
Where indeed we used the geometric series:
$$\dfrac{1}{1-e^{-x}} = \sum_{k = 0}^{+\infty} e^{-kx}$$
Now, the question which will surely be dumb but today I'm really messed up.
Why exactly can we use the geometric series without worries? I mean I know that we can becahse $|e^{-x}| < 1$ but the domain of the integral starts from $0$.
I would mind that this would give some problem, since $\dfrac{1}{1-e^{-x}}$ as $x = 0$ is problematic.
How can we rigirously justify that passage? Does it have something to do with the absolute convergence? I'm rust.
Thank you!

It can be justified using two very well known theorems of measure theory: the monotone convergence theorem and the dominated convergence theorem.
First assume that $s>1$, then $f(x):=\frac{x^{s-1}}{e^x-1}>0$ for all $x>0$, and define $g_n(x):=\sum_{k=0}^n e^{-kx}$ and $f_n(x):=\frac{x^{s-1}}{e^x}g_n(x)$. Now the monotone convergence theorem says to us that if we have a sequence $\{h_n\}_{n\in\mathbb{N}}$ of non-negative measurable functions such that $h_n$ increases pointwise to a function $h$ then
$$
\lim_{n\to \infty }\int_{\mathbb{R}}h_n(x)\,d x=\int_{\mathbb{R}}h(x)\,d x\tag1
$$
Above the integrals are Lebesgue integrals respect to the Lebesgue measure. In your case, as $f_n\uparrow f$ pointwise  this means that
$$
\sum_{k\geqslant 0}\int_{(0,\infty )}\frac{x^{s-1}}{e^x}e^{-kx}\,d x=\int_{(0,\infty )}\frac{x^{s-1}}{e^x-1}\,d x\tag2
$$
Now, if $s\in \mathbb{C}$ and $\operatorname{Re}(s)>1$ we have that
$$
\left| \frac{x^{s-1}}{e^x}g_n(x) \right|\leqslant \left| \frac{x^{s-1}}{e^x-1} \right|\leqslant \frac{x^{\operatorname{Re}(s)-1}}{e^x-1}\tag3
$$
and so, as the LHS of (2) is finite when $s$ is real and bigger than one, we can apply the dominated convergence theorem to reach again (2) but this time for $s\in \mathbb{C}$ such that $\operatorname{Re}(s)>1$.
Finally, it can be proved, using the monotone convergence theorem, that when $f$ is a non-negative improperly Riemann integrable function we have that
$$
\int_{0}^{\infty }f(x)\,d x=\int_{(0,\infty )}f(x)\,d x\tag4
$$
where the LHS is an improper integral of Riemann and the RHS is a Lebesgue integral.
