Using Laurent series to evaluate integral

Question

Use the Laurent series of $e^{1/z}$ on the punctured plane $\mathbb{C} - \{0\}$ to compute $1/\pi \int_0^\pi e^{\cos\theta}\cos(\sin\theta-n\theta) d\theta$ for $n=0,1,2,...$

I have been trying to attempt this question from my complex analysis book and I am not sure where to even begin. All I know is that the Laurent series of $e^{1/z}$ is $f(z)=\sum_{n=0}^{\infty} \frac{1}{n!}\frac{1}{z^n}$, for all $|z|>0$.

I have no idea how this relates the the integral though because there is no $e^{1/z}$ only a $e^{\cos\theta}$ in the integral. I thought maybe I have to put the $\cos\theta$ values into Taylor form but I'm not sure if that's correct.

Any help is greatly appreciated! Thanks :)

Answer 1

First note that the integrand is even, meaning we have that

$$\frac{1}{\pi}\int_0^\pi e^{\cos\theta}\cos(\sin\theta-n\theta)\:d\theta = \frac{1}{2\pi}\int_{-\pi}^\pi e^{\cos\theta}\cos(\sin\theta-n\theta)\:d\theta$$

This integral is the real part of

$$I = \frac{1}{2\pi}\int_{-\pi}^\pi e^{\cos\theta+i\sin\theta-in\theta}\:d\theta = \frac{1}{2\pi}\int_{-\pi}^\pi e^{e^{i\theta}}e^{-in\theta}\:d\theta$$

So letting $z = e^{-i\theta}$ gives us

$$I = \frac{1}{2\pi i} \oint_{|z|=1}e^{\frac{1}{z}}z^{n-1}\:dz$$

The minus sign goes away due to the integral needing to be positively oriented (counterclockwise). The residue theorem tells us that the only term that survives in the integral will be the $n$th power in the Laurent series since that is the coefficient of the $z^{-1}$ power, giving us

$$I = \frac{1}{n!}$$

which is already real, so no further simplification is necessary.

Answer 2

Let $\gamma$ be the unit circle, and consider the integral $$I_m=\int_\gamma z^m\exp(1/z)\,dz$$ where $m$ is an integer. Writing $z=e^{it}$ gives $$I_m=i\int_0^{2\pi}e^{im t}e^{\cos t-i\sin t}e^{it}\,dt =i\int_0^{2\pi}e^{\cos t}e^{i((m+1)t-\sin t)}\,dt.$$ The imaginary part of this integral is $$\int_0^{2\pi}e^{\sin t}\cos((m+1)t-\sin t)\,dt$$ which is more-or-less then integral you are after.

But looking at the Laurent series of $e^{1/z}$ and using calculus of residues gives $I_m=0$ for $m\le -2$ and $$I+m=\frac{2\pi i}{(m+1)!}$$ for $m\ge-1$.