Showing integral on contour tends to zero I'm trying to prove:
$$\int \frac{e^{t(z+\frac{1}{z})}}{z^2} = \sum_0 ^{\infty} b_m t^{2m+1}$$
Where the integral is over a contour centre the origin, radius R, and the $b_m$ are some coefficients.
Now I can obtain these coefficients by considering the resiude at $z=0$ and multiplying the power series for $e^{tz}$ and $e^{t/z}$ however I am now having difficulty in showing that as $R \to \infty$ the integral around the contour tends to zero, any hints on this would be great.
Thanks
 A: $\newcommand{\+}{^{\dagger}}
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$\ds{\oint_{\verts{z}\ =\ R}{\exp\pars{t\bracks{z + 1/z}} \over z^2}\,\dd z:\
     {\large ?}}$.

\begin{align}
&\color{#66f}{\large\oint_{\verts{z}\ =\ R}
{\exp\pars{t\bracks{z + 1/z}} \over z^2}\,\dd z}
=\oint_{\verts{z}\ =\ R}{\exp\pars{tz}\exp\pars{t/z} \over z^2}\,\dd z
\\[3mm]&=\oint_{\verts{z}\ =\ R}{1 \over z^2}\,
\sum_{m = 0}^{\infty}{t^{m}z^{m} \over m!}
\sum_{n = 0}^{\infty}{t^{n}z^{-n} \over n!}\,\dd z
=\sum_{n = 0}^{\infty}\sum_{m = 0}^{\infty}{t^{m + n} \over m!\,n!}\
\overbrace{\oint_{\verts{z}\ =\ R}{\dd z \over z^{2 - m + n}}}
^{\ds{=\ 2\pi\ic\,\delta_{m,n + 1}}}
\\[3mm]&=2\pi\ic\sum_{n = 0}^{\infty}{t^{2n + 1} \over \pars{n + 1}!\,n!}
=\color{#66f}{\large 2\pi\ic\ {\rm I}_{1}\pars{2t}}
\end{align}

where $\ds{{\rm I}_{\alpha}\pars{x}}$ is the
Modified Bessel Function of the First Kind.
