Showing $\frac{e^{-ax} - e^{(a-1)x}}{1-e^{-x}}$ is integrable on $\mathbb{R}$ 
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.
 A: $\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}}}
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\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}

where $\ds{H_{z}}$ is a Harmonic Number. The final result is found with the Euler Reflection Formula.

