I've stumbled upon a problem I can not solve in the book Mathematical Methods for Electrical Engineers by Thomas B.A. Senior(Page 171). The book gives the following instruction:
Using Cauchy's general integral formula, show that
$$
\left ( \frac{a^{n}}{n!} \right )^{2} = \frac{1}{2\Pi i }\oint_{C}^{ }\frac{a^{n}e^{az}}{n! z^{n+1}}dz
$$
where C is any closed contour surrounding the origin. Since
$$
\sum_{n=0}^{\infty }\frac{t^{n}}{n!}=e^{t}
$$
show that
$$
\sum_{n=0}^{\infty }\left ( \frac{a^{n}}{n!} \right )^2 = \frac{1}{2\pi}\int_{0}^{2\pi}e^{2acos\Theta }d\Theta
$$
I showed the first part, $$ g^{(n)}(0)=\frac{n!}{2\pi i}\oint_{C}^{ }\frac{a^{n}e^{az}}{n!z^{n +1}}dz = (\frac{a^{n}e^{az}}{n!})^{(n)}|_{z=0} = \frac{a^{2n}}{n!} => \frac{1}{2\pi i}\oint_{C}^{ }\frac{a^{n}e^{az}}{n!z^{n +1}}dz = (\frac{a^{n}}{n!})^{2} $$ But for the second part: $$ \sum_{n=0}^{\infty}(\frac{a^{n}}{n!})^{2}= \sum_{n=0}^{\infty}\frac{1}{2\pi i}\oint_{C}^{ }\frac{a^{n}e^{az}}{n!z^{n+1}}dz = \frac{1}{2\pi i}\oint_{C}^{ }e^{az}\sum_{n=0}^{\infty}\frac{a^{n}}{n! z^{n+1}}dz... $$ I do not know how to find this sum.
Write the integrand as
\begin{align*} \frac 1 {2\pi i} e^{az} \sum_{n = 0}^{\infty} \frac{a^n}{n! z^{n + 1}} &= \frac{e^{az}}{2 \pi i z} \sum_{n = 0}^{\infty} \frac{(a/z)^n}{n!} \\ &= \frac{e^{az}}{2 \pi i z} e^{a/z} \\ &= \frac{e^{a(z + 1/z)}}{2 \pi i z} \end{align*}
Now take $C$ to be the contour which is the unit circle, oriented counterclockwise around the origin (exercise: Understand why we can choose the contour to be what we want). In this case, using the usual $z = e^{i\theta}$, $$z + \frac 1 z = 2 \cos \theta$$ and $dz = i z d\theta$, so the integral reduces to
$$\int_0^{2\pi} \frac{1}{2\pi i z} e^{a(2 \cos \theta)} i z d\theta$$
and a line of algebra later, you're done.