Showing integral on contour tends to zero

Question

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

Answer 1

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\oint_{\verts{z}\ =\ R}{\exp\pars{t\bracks{z + 1/z}} \over z^2}\,\dd z:\ {\large ?}}$.

\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.