Show that $\int_0^{2\pi} e^{\cos\theta} \ d\theta=\sum_{n=0}^{\infty} \frac{1}{(n!2^n)^2}$

Question

Question:

Show that $\int_0^{2\pi} e^{\cos\theta} \ d\theta=\sum_{n=0}^{\infty} \frac{1}{(n!2^n)^2}$

My attempt:

We are learning about laurent series and their integral representation. I think I have to make the substitution $z=e^{i\theta}$ and convert the integral into $$\frac{1}{2\pi i}\int_C\frac{e^{(z+\frac{1}{z})/2}}{z} \ dz,$$ where $C$ is the unit circle in the complex plane. I don't know what to do from here though.

Answer 1

$$\int_0^{2\pi} \exp(\cos\theta) \ d\theta = $$

$$ \int_{|z|=1}\frac{-i\exp(z+\frac{1}{z})}{z}dz = 2 \pi i \text{Res}(f,0) $$

Because $0$ is the only singularity of $f(z) := -iz^{-1}\exp(z+\frac{1}{z}) $

$\text{Res}(f,0) = $ coefficient of $z^{-1}$ in the Laurent expansion of $f$, let's find this expansion:

$f(z) = \sum_{n=0}^{\infty}\frac{-i(z+z^{-1})^n}{2^nz} = \sum_{n=0}^{\infty} \sum_{j=0}^{n} -i\frac{n!}{j!(n-j)!2^{2n}} z^{n-2j-1} $

So the $z^{-1}$ is determined by the equation $n-2j-1 = -1$, so $n = 2j$, let's plug this in:

$$ \int_0^{2\pi} \exp(\cos\theta) \ d\theta = 2\pi \sum_{n=0}^{\infty} \frac{(2j)!}{j!j!(2j)!2^{2j}} = 2\pi \sum_{n=0}^{\infty} \frac{1}{(j!)^22^{2j}} $$ as desired.

I have left the algebra for you to do. I think you left out a factor of $2\pi$ in the answer, see http://www.wolframalpha.com/input/?i=integrate+exp%28cos%28x%29%29+from+0+to+2pi and http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html