Show that $\int_{0}^{2\pi}\frac{dk}{2\pi}\,e^{-s(e^{-ik}-1)+ikm}=\frac{s^{m}}{m!}e^{-s}$

Question

Book (Altland and Simons, Ch 10, Eq. 10.41) I am reading used the integral:

$$\int_{0}^{2\pi}\frac{dk}{2\pi}\,e^{-s(e^{-ik}-1)+ikm}=\frac{s^{m}}{m!}e^{-s},\quad m\in\mathbb{Z},\, s\in\mathbb{R}.$$

It says that it is easily shown using Taylor expansion of the exponent and and the identity:

$$\int_{0}^{2\pi}\frac{dk}{2\pi}\,e^{ink}=\delta_{n,0},\quad n\in\mathbb{N}.$$

I don't see how to solve this integral using the methods proposed in the book, can somebody fill in the details?

Answer 1

I'll assume that the $-s$ in the integrand's exponent means $s$ instead. $$\begin{align*}&\phantom{~=} \int_0^{2\pi}\frac 1 {2\pi}e^{s(e^{-ik}\;-1)+ikm}\,dk\\&= \int_0^{2\pi}\frac 1 {2\pi}e^{-s+ikm}\left(\sum_{n=0}^\infty \frac 1 {n!}\left(se^{-ik}\right)^n\right)dk\\&= \sum_{n=0}^\infty \frac {e^{-s}s^n} {n!}\left(\int_0^{2\pi}\frac 1 {2\pi}e^{ikm-ikn}dk\right)\\&= \sum_{n=0}^\infty \frac {e^{-s}s^n} {n!}\delta_{m,n}\\&= \frac{e^{-s}s^m}{m!} \end{align*}$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\int_{0}^{2\pi} \,\exp\pars{-s\bracks{\expo{-\ic k} - 1} + \ic km} {\dd k \over 2\pi} = \require{cancel}\xcancel{{s^{m} \over m!}\expo{-s}}:\ {\LARGE ?}}$.

\begin{align} &\bbox[10px,#ffd]{\int_{0}^{2\pi} \,\exp\pars{-s\bracks{\expo{-\ic k} - 1} + \ic km} {\dd k \over 2\pi}} = \expo{s}\int_{0}^{2\pi} {\exp\pars{-s/\expo{\ic k}} \over \pars{\expo{\ic k}}^{-m}}\, {\dd k \over 2\pi} \\[5mm] \stackrel{}{=}\,\,\,&\ \expo{s}\oint_{\verts{z} = 1}{\expo{-s/z} \over z^{-m}}\,{\dd k \over 2\pi} = \expo{s}\sum_{n = 0}^{\infty}\pars{-1}^{n}\,{s^{n} \over n!}\ \oint_{\verts{z} = 1}{1 \over z^{n - m}} \,{\dd z/\pars{\ic z} \over 2\pi} \\[5mm] = &\ \expo{s}\sum_{n = 0}^{\infty}\pars{-1}^{n}\,{s^{n} \over n!}\ \underbrace{\oint_{\verts{z} = 1}{1 \over z^{n - m + 1}} \,{\dd z \over 2\pi\ic}}_{\ds{\delta_{nm}}} \\[5mm] = &\ \expo{s}\bracks{\pars{-1}^{m}\,{s^{m} \over m!}}\bracks{m \in \mathbb{N}_{\geq 0}} = \bbx{\pars{-1}^{m}\,{s^{m} \over m!}\,\expo{s} \bracks{m \in \mathbb{N}_{\geq 0}}} \end{align}