Question: Show that, for $a \in (0,1)$, $$h(x) = \frac{e^{-ax} - e^{(a-1)x}}{1-e^{-x}}$$ is an integrable function on $\mathbb{R}$.
I can't see how to begin with this problem.
Attempt 1: On $(0, \infty)$ we have that $e^{-x} \leq 1$ and hence we can expand binomially to get that $$ h(x) = \sum_{k=0}^{\infty} e^{-(a+k)x} - e^{ax-a-kx}.$$ But this expansion isn't valid on the whole of $\mathbb{R}$ and we don't have symmetry of $h$ to utilise, so I can't really see how this approach would go anywhere.
We also have that $h$ is a continuous function away from $0$, and thus is a measurable function in $\mathbb{R}$, but I cannot also see how I might introduce a bound of an integrable function on $h$ to apply the comparison test either.
Any help would be appreciated.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\ds{\int_{0}^{\infty}{\expo{-ax} - \expo{-\pars{1 - a}x} \over 1 - \expo{-x}}\,\dd x}} \,\,\,\stackrel{x\ =\ -\ln\pars{t}}{=}\,\,\, \int_{1}^{0}{t^{a} - t^{1 - a} \over 1 - t}\,\pars{-\,{\dd t \over t}} \\[5mm] & = \int_{0}^{1}{1 - t^{-a} \over 1 - t}\,\dd t - \int_{0}^{1}{1 - t^{a - 1} \over 1 - t}\,\dd t = H_{-a} - H_{a - 1} = \bbx{\pi\cot\pars{\pi a}}\,,\quad \Re\pars{a} \in \pars{0,1}. \end{align}