Finding contour integral of $e^{t(z+z^{-1})}z^{-2}$

Question

I am really struggling with a contour integration question, which I am revising for an exam. I want to show that the contour integral of

$$\int_\Gamma e^{t(z+z^{-1})}z^{-2} dz=\sum_{m=0}^\infty b_mt^{2m+1}$$ where the $b_m$ are constants to be found, and $\Gamma$ is the positively oriented circle with centre $0$ and radius $R$.

I was wondering, as I instinctivey noticed that $z+z^{-1}=2cos(\theta)$ where, $z=Re^{i\theta}$ if we should rewrite this integral, so that it becomes $$\int _0^{2\pi}\frac{e^{2tRcos(\theta)}}{(Re^{i\theta})^2(iRe^{i\theta})}d\theta$$

But then, I don't see that this has any poles at all? Any help appreciated, thank you.

Answer 1

Hint:

$$e^{t\left(z+z^{-1}\right)}z^{-2}=z^{-2}\sum_{n=0}^{\infty}\dfrac{t^{n}}{n!}\left(z+z^{-1}\right)^{n}=\\\\=z^{-2}\sum_{n=0}^{\infty}\dfrac{t^{n}}{n!}\sum_{k=0}^{n}\binom{n}{k}z^{-k}z^{n-k}=\sum_{n=0}^{\infty}\dfrac{t^{n}}{n!}\sum_{k=0}^{n}\binom{n}{k}z^{n-2\left(k+1\right)} $$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \oint_{\Gamma}{\expo{t\pars{z + z^{-1}}} \over z^{2}}\,\dd z & = \oint_{\Gamma}{\expo{tz}\expo{t/z} \over z^{2}}\,\dd z \\[5mm] & = \oint_{\Gamma}{\bracks{\sum_{m = 0}^{\infty}\pars{tz}^{m}/m!} \bracks{\sum_{n = 0}^{\infty}\pars{t/z}^{n}/n!} \over z^{2}}\,\dd z \\[5mm] & = \sum_{m = 0}^{\infty}\sum_{n = 0}^{\infty} {t^{m + n} \over m!\, n!}\ \overbrace{\oint_{\Gamma}{\dd z \over z^{2 - m + n}}} ^{\ds{2\pi\ic\,\delta_{2 - m + n,1}}} \\[5mm] & = \sum_{m = 0}^{\infty} {2\pi\ic \over m!\pars{m - 1}!}\,t^{m + \pars{m - 1}}\bracks{m - 1 \geq 0} \\[5mm] & = \bbx{\sum_{m = 0}^{\infty}\ \underbrace{2\pi\ic \over \pars{m + 1}!\,m!}_{\ds{ b_{m}}}\ t^{2m + 1}} \end{align}

Answer 3

$$e^{tz} = \sum_{n=0}^{+\infty}t^n\frac{z^n}{n!}\,,$$ $$e^{t/z} = \sum_{m=0}^{+\infty}t^m\frac{z^{-m}}{m!}\,.$$

Therefore, the integrand is:

$$e^{tz}e^{t/z}z^{-2}=\sum_{n=0}^{+\infty}\sum_{m=0}^{+\infty} t^{n+m}\frac{z^{n-m-2}}{n!m!}\,.$$

For each value of $m$, we pick a residue from the singularity at the origin only when $n=m+1$, (with residue $t^{2m+1}/m!(m+1)!$). Therefore the value of the integral is:

$$ \sum_{m=0}^{+\infty}\frac{2\pi i}{m!(m+1)!}t^{2m+1}\,.$$

The value of $b_m$ is:

$$b_m = \frac{2\pi i}{m!(m+1)!}$$