Laurent series of transformed exponential function

Question

This exercise is from Conways Complex Analysis (Chapter 5, exercise 12)

(a) Let $\lambda \in \mathbb{C}$ and show that $$ \exp \left\{\frac{1}{2} \lambda\left(z+\frac{1}{z}\right)\right\}=a_0+\sum_{n=1}^{\infty} a_n\left(z^n+\frac{1}{z^n}\right) $$ for $0<|z|<\infty$, where for $n \geq 0$ $$ a_n=\frac{1}{\pi} \int_0^\pi e^{\lambda \cos t} \cos n t d t $$ (b) Similarly, show $$ \exp \left\{\frac{1}{2} \lambda\left(z-\frac{1}{z}\right)\right\}=b_0+\sum_{n=1}^{\infty} b_n\left(z^n+\frac{(-1)^n}{z^n}\right) $$ for $0<|z|<\infty$, where $$ b_n=\frac{1}{\pi} \int_0^\pi \cos (n t-\lambda \sin t) d t $$

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I think the first one is already on StackExchange. I am rather interested in proving:

There are uniquely determined integer functions $J_{n}$ such that for all $(z, \zeta) \in \mathbb{C} \times(\mathbb{C} \backslash 0)$ the equation $\exp \left(\frac{z}{2}\left(\zeta-\frac{1}{\zeta}\right)\right)=\sum_{n=-\infty}^{\infty} J_{n}(z) \zeta^{n}$ holds.

This could mabe be done with Laurent development. The only question is how one is to know that the $J_n$ are holomorphic.

Answer 1

By https://proofwiki.org/wiki/Existence_of_Laurent_Series laurent series for $exp(\frac{z}{2}(\zeta-\frac{1}{\zeta}))$ exists and since laurents series exists we have by Laurent Series of the function : $J_n(z) = \frac{1}{2 \pi} \int_{-\pi}^{\pi} cos(nt-z sin(t)) dt$.

Answer 2

The functions we are looking for are called Bessel functions. With $$ \exp \left(\frac{z}{2}\left(\zeta-\frac{1}{\zeta}\right)\right)=\sum_{n=-\infty}^{\infty} J_{n}(z) \zeta^{n} $$ follows from Cauchy's integral formulas $$ J_n(z)=\frac{1}{2 \pi i} \int_{|\zeta|=1} \frac{1}{\zeta^{n+1}} \exp\left(\frac{z}{2}\left(\zeta-\frac{1}{\zeta}\right)\right) d\zeta $$ The integrand is an integer function in $z$, so $J_n(z)$ is also integer. With $\zeta(t)=\exp(it)$, it follows that $$ \begin{aligned} J_n(z) & =\frac{1}{2 \pi} \int_0^{2 \pi} \exp(-i(n+1)) \exp\left(\frac{z}{2}\left(\exp(it)-\exp(-it)\right)\right) \exp(it) d t\\\ &=\frac{1}{2 \pi} \int_0^{2 \pi} \exp(-i n t) \exp(i z \sin t) d t =\frac{1}{2 \pi} \underbrace{\int_{-\pi}^\pi \cos (z \sin t-n t) d t}_{=: I_1}+\frac{i}{2 \pi} \underbrace{\int{-\pi}^\pi \sin (z \sin t-n t) d t}_{=: I_2} \\ & =\frac{1}{\pi} \int_0^\pi \cos (n t-z \sin t) d t \end{aligned} $$ since the integrand of $I_1$ is even and that of $I_2$ is odd. The exponential series yields $$ \begin{aligned} \exp \left(\frac{z}{2}\left(\zeta-\frac{1}{\zeta}\right)\right) & =\sum_{k=0}^{\infty} \frac{1}{k !}\left(\frac{z}{2}\right)^k\left(\sum_{j=0}^k\binom{k}{j}(-1)^j \zeta^{k-2 j}\right) \\ & =\sum_{k=0}^{\infty} \frac{1}{k !}\left(\frac{z}{2}\right)^k\left(\sum_{j=0}^k\binom{k}{j}(-1)^{k-j} \zeta^{2 j-k}\right) \end{aligned} $$ We are looking for the coefficient $J_n(z)$ of $\zeta^n$ in the Laurent series expansion. Rearrangement is allowed because of absolute convergence. The first series representation yields for $n \geq 0$ (set $k=2 j+n$ ): $$ J_n(z)=\sum_{j=0}^{\infty} \frac{1}{(2 j+n) !}\binom{2j+n}{j}(-1)^j\left(\frac{z}{2}\right)^{2 j+n}=\sum_{k=0}^{\infty} \frac{(-1)^k}{k !(n+k) !}\left(\frac{z}{2}\right)^{n+2 k} . $$ The second row representation yields for $n<0$ (with $k=2 j-n)$ : $$ J_n(z) =\sum_{j=0}^{\infty} \frac{1}{(2 j-n) !}\binom{2j-n}{j}(-1)^{j-n}\left(\frac{z}{2}\right)^{2 j-n}=(-1)^n \sum_{j=0}^{\infty} \frac{(-1)^j}{(j-n) ! j !}\left(\frac{z}{2}\right)^{2 j-n}=(-1)^n J_{-n}(z) . $$

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Now I am not entirely sure about the holomorphy argument. Any ideas?