Finding the values of $a$ and $\alpha$ for which the integral exists In my exam I was asked to find the values for $a>0$ and $\alpha \in \mathbb{C}$ for which the following integral exists and is finite:
$$I(\alpha,a) = \int_0^\infty \frac{x^{\alpha-1}dx}{e^{a x}-1}$$
This gave me a lot of trouble and I was not able to give a solution. I know that the integrand is defined at the endpoints due to the comparison with asymptotic functions that converge at those points, but I have no idea on how to move from there. Any hint or help will be immensely appreciated. Thank you!
 A: $\require{amsmath}$
$\newcommand\norm[1]{\left\lVert#1\right\rVert}$
At $+\infty$ the integrand is $\mathcal{O}(x^{\alpha-1}e^{-\alpha x})$ and at $x\to0^+$ the integrand is  $\mathcal{O}(x^{\alpha-2}/\alpha)$, therefore the integral converges only for $\Re(\alpha)>1$. Set $ax=u$, obtaining
$$I(\alpha,a)=\frac{1}{a^\alpha}{\displaystyle\int_0^{+\infty}\frac{u^{\alpha-1}}{e^u-1}\,du}$$
Now,
$$
\begin{aligned}{\displaystyle\int_0^{+\infty}\frac{u^{\alpha-1}}{e^u-1}\,du}&={\displaystyle\int_0^{+\infty}\frac{u^{\alpha-1}e^{-u}}{1-e^{-u}}\,du}={\displaystyle\int_0^{+\infty}u^{\alpha-1}e^{-u}\sum_\limits{n=0}^\limits{+\infty}e^{-nu}\,du=}\\&={\displaystyle\int_0^{+\infty}\sum_\limits{n=0}^\limits{+\infty}u^{\alpha-1}e^{-(n+1)u}\,du}.
\end{aligned}$$
Since $\vert u^{\alpha-1}\vert=u^{\Re(\alpha)-1},\,$the series has only positive terms if $\alpha$ is real. Let's assume that $\alpha>0$, we can interchange the sum with the integral sign $(*)$, and, let's set $s=(n+1)u$, giving
$${\displaystyle\int_0^{+\infty}u^{\alpha-1}e^{-(n+1)u}\,du}={\displaystyle\int_0^{+\infty}\frac{s^{\alpha-1}}{(n+1)^{\alpha}}e^{-s}\,ds}=\frac{1}{(n+1)^{\alpha}}{\displaystyle\int_0^{+\infty}s^{\alpha-1}e^{-s}\,ds}=\frac{\Gamma(\alpha)}{(n+1)^{\alpha}}\,.$$
It directly follows that
$$I(\alpha,a)=\frac{\Gamma(\alpha)}{a^{\alpha}}\zeta(\alpha), \quad \alpha>1$$
if $\alpha>0$, but also for $\Re(\alpha)>1$, since the series of the norms
$$\sum_\limits{n=0}^\limits{+\infty}\norm{u^{\alpha-1}e^{-(n+1)u}}_1$$
converges.

$(*)$ $\textbf{Theorem:}$ If $f_j$ is a sequence of positive measurable functions and $f(x)=\sum_{j=0}^{\infty}f_j(x)$, then 
$$\int_Xf=\sum_{j=0}^\infty\int_Xf_j.$$
Proof: Follows immediately from the Sigma additivity and the Monotone Convergence Theorem applied to the sequence $f_0+\cdots+f_m.$
