Approximating an Integral by Expanding it's Integrand I am struggling with the following integral:
$$
\int_{\Gamma}^{\infty} \left[ \frac{1 + \frac{1}{z}}{e^{z} - 1} + \frac{1}{(e^{z} - 1)^{2}} \right] dz
$$
where $0 < \Gamma \ll 1$. 
There is no closed form expression for this integral. I have however, noticed the following facts:
$$
\frac{1 + \frac{1}{z}}{e^{z} - 1} + \frac{1}{(e^{z} - 1)^{2}} \ = \ \frac{2}{z^{2}} - \frac{1}{2z} + \frac{z^{2}}{360} + \mathcal{O}(z^{4})
$$
$$
\lim_{z \to \infty} \left\{ \frac{1 + \frac{1}{z}}{e^{z} - 1} + \frac{1}{(e^{z} - 1)^{2}} \right\} = 0
$$
So to me at first glance, it would seem that my integral will vanish at the limit $z \to \infty$, and $maybe$ I can say that:
$$
\int_{\Gamma}^{\infty} \left[ \frac{1 + \frac{1}{z}}{e^{z} - 1} + \frac{1}{(e^{z} - 1)^{2}} \right] dz \approx \left[ 0 \right] - \left[ \int \left( \frac{2}{z^{2}} - \frac{1}{2z} \right) dz \right] \bigg|_{z = \Gamma} = \frac{1}{2}\log(\Gamma) + \frac{2}{\Gamma}
$$
I realize this is really informal and there is probably something wrong with this...but is there anything that can be said along these lines? I am really interested in a power series approximation of my integral!
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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Lets $\ds{\mrm{f}\pars{z} \equiv
{1 + 1/z \over \expo{z} - 1} + {1 \over \pars{\expo{z} - 1}^{2}}.\qquad
0 < \Gamma \ll 1}$.

\begin{align}
\int_{\Gamma}^{\infty}\mrm{f}\pars{z}\,\dd z & =
\int_{\Gamma}^{1}\mrm{f}\pars{z}\,\dd z +
\int_{1}^{\infty}\mrm{f}\pars{z}\,\dd z
\\[5mm] & =
\braces{\int_{\Gamma}^{1}\pars{{2 \over z^{2}} - {1 \over 2z}}\,\dd z +
\int_{\Gamma}^{1}\bracks{\mrm{f}\pars{z} - {2 \over z^{2}} + {1 \over 2z}}
\,\dd z} +
\int_{1}^{\infty}\mrm{f}\pars{z}\,\dd z
\\[5mm] & =
-2 + {2 \over \Gamma} + {1 \over 2}\,\ln\pars{\Gamma} -
\int_{0}^{\Gamma}\bracks{\mrm{f}\pars{z} - {2 \over z^{2}} + {1 \over 2z}}
\,\dd z\label{1}\tag{1}
\\ & +\
\underbrace{\int_{0}^{1}\bracks{\mrm{f}\pars{z} - {2 \over z^{2}} +
{1 \over 2z}} + \int_{1}^{\infty}\mrm{f}\pars{z}\,\dd z}
_{\ds{\equiv\ \alpha + 2}: \mbox{a constant}}
\end{align}

The integral in \eqref{1} can be estimated by expanding
  $\ds{\mrm{f}\pars{z} - {2 \over z^{2}} + {1 \over 2z}}$ in power of $\ds{z}$.

For instance,
$$
\int_{0}^{\Gamma}\bracks{\mrm{f}\pars{z} - {2 \over z^{2}} + {1 \over 2z}}
\,\dd z \sim
\int_{0}^{\Gamma}{z^{2} \over 360}\,\dd z = {\Gamma^{3} \over 1080}
\qquad\mbox{as}\quad \Gamma \to 0^{+}
$$
such that
$$
\bbx{\int_{\Gamma}^{\infty}\mrm{f}\pars{z}\,\dd z \sim
{1 \over 2}\,\ln\pars{\Gamma} + {2 \over \Gamma} + \alpha + 
{\Gamma^{3} \over 1080}\qquad\mbox{as}\quad \Gamma \to 0^{+}}
$$

Numerically, $\ds{\alpha \approx -1.1304}$. 

