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!
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\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}
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 $$ \bbx{\int_{\Gamma}^{\infty}\mrm{f}\pars{z}\,\dd z \sim {1 \over 2}\,\ln\pars{\Gamma} + {2 \over \Gamma} + \alpha + {\Gamma^{3} \over 1080}\qquad\mbox{as}\quad \Gamma \to 0^{+}} $$