I'm trying to solve the following problem:
Let $t \in \mathbb{R}$ be fixed and let
$$ e^{[\frac{t(z - 1/z)}{2}]} = \sum_{n \in \mathbb{Z}} J_n(t)z^n$$ be the Laurent expansion in $\mathbb{C}$*. Show that
(a) $J_n(t) = \frac{1}{\pi} \int_0^\pi \cos(t\sin(\theta) - n\theta)\,d\theta.$
(b) $J_{-n}(t) = (-1)^n J_n(t)$.
I've done a few basic calculations so far but haven't really gotten anywhere. For part (a), using the theorem about the representation of holomorphic functions as Laurent series, I know that $$J_n(t) = \frac{1}{2\pi{i}} \int_{\gamma} \frac{(e^{\frac{t(z - \frac{1}{z})}{2}})}{z^{n+1}}\, dz,$$ where $\gamma(s) = e^{2\pi{i}s}$ for any $s \in [0, 1]$. But I'm not quite sure of where to start in order to put this in the desired form stated in (a).
Any help would be appreciated!
Answer
For $a)$, consider the path $\gamma(t) = e^{i 2 \pi t}$ for $0 \leq t \leq 1$. This is a unit disk, it doesn't get much better than this. If we express the function as a Laurent series centered at 0 as $\sum \limits_{n \in \mathbb{Z}}c_nz^n$, we know that the coefficients $c_n$ is given by $\frac{1}{2 \pi i}\int_\gamma \frac{f(z)}{z^{n + 1}}$. It is routine to verify that this integral cannot have an imaginary part by using the power series representation of the exponential function. That is, it is an honest to goodness real-valued number. The result then follows from the definition of the integral. Then Bob's your uncle!
For $b)$, consider the power series representation of $J_n$.
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