I want to evaluate the following,
$$I(z)=\int_0^{\pi}\cos(z\sin x)e^{\cos (x)}\ dx,\quad z\in\mathbb{N}$$
using Bessel functions. My attempt where I ended up with a divergent series is below. I think the error comes from the switching of integration and summation, but I hope there is a way around this.
Using the known expansion in terms of the Bessel function $J_\alpha(z)$,
$$\cos(z \sin x)=J_0(z)+2\sum_{k=1}^\infty J_{2k}(z)\cos(2kx)$$
$$\begin{align*}I(z)=\int_0^{\pi}\cos(z\sin x)e^{\cos (x)}\ dx&=\int_0^\pi J_0(z)e^{\cos(x)}\ dx+2\int_0^\pi \sum_{k=1}^\infty J_{2k}(z)\cos(2kx)e^{\cos(x)}\ dx\\&=\underbrace{J_0(z)\int_0^\pi e^{\cos(x)}\ dx}_{T_1}+\underbrace{2\sum_{k=1}^\infty J_{2k}(z)\int_0^\pi\cos(2kx)e^{\cos(x)}\ dx}_{T_2}.\end{align*}$$
We evaluate $T_1$ by subbing $z=e^{ix}$ and integrating over a circle contour $\gamma$ surrounding the origin in a counter clockwise sense,
$$\begin{align*}J_0(z)\int_0^\pi e^{\cos(x)}\ dx&=\frac{J_0(z)}{2}\int_{-\pi}^\pi e^{\cos(x)}\ dx
\\&=\frac{J_0(z)}{2}\int_{-\pi}^\pi \exp\left(\frac{e^{ix}+e^{-ix}}{2}\right)\frac{ie^{ix}}{ie^{ix}}\ dx \\&=\frac{J_0(z)}{2i}\oint_\gamma\frac{1}{z}\exp\left(\frac{z+z^{-1}}{2}\right)\ dz\end{align*}$$
the only singularity is at $z=0$, and we find the residue at the point by extracting the coefficient of the $z^{-1}$ term in the laurent expansion of the integrand,
$$[z^{-1}]\left(\frac{1}{z}\exp\left(\frac{z+z^{-1}}{2}\right)\right)=[z^{-1}]\left(\frac{1}{z}\sum_{n=0}^\infty\frac{z^n}{n!2^n}\sum_{n=0}^\infty\frac{z^{-n}}{n!2^n}\right)=\sum_{n=0}^\infty\frac{1}{(n!2^{n})^2}=I_0(1)$$
where $I_0(z)$ is the Modified Bessel function and $[z^n]f(z)$ is the coefficient of $z^n$ in $f$. Thus by Cauchy's formula,
$$T_1=\pi J_0(z)I_0(1).$$
For $T_2$ we can evaluate the integral in terms of the Modified Bessel function again, given the representation,
$$I_\alpha(z)=\frac{1}{\pi}\int_0^\pi e^{z\cos(x)}\cos(\alpha x)\ dx-\frac{\sin(\alpha \pi)}{\pi}\int_0^{+\infty} e^{-z\cosh(x)-\alpha x}\ dx$$
we have,
$$2\sum_{k=1}^\infty J_{2k}(z)\int_0^\pi\cos(2kx)e^{\cos(x)}\ dx=2\sum_{k=1}^\infty J_{2k}(z)\left(\pi I_{2k}(1)+\sin(2\pi k)\int_0^{+\infty} e^{-\cosh(x)-2kx}\ dx\right)$$ the integral on the RHS vanishes due to the $\sin(2\pi k)$ factor,
$$T_2=2\pi\sum_{k=1}^\infty J_{2k}(z)I_{2k}(1)$$
hence,
$$I(z)=\pi J_0(z)I_0(1)+2\pi\sum_{k=1}^\infty J_{2k}(z)I_{2k}(1)$$
Edit: The convergence of the series above is as follows.
For fixed $z$ and large $k$,
$$J_{2k}(z)\sim\frac{1}{\sqrt{4\pi k}}\left(\frac{ez}{4k}\right)^{2k},\quad I_{2k}(1)\sim\frac{1}{\sqrt{4\pi k}}\left(\frac{e}{4k}\right)^{2k}\\
J_{2k}(z)I_{2k}(1)\sim\frac{1}{4\pi k}\left(\frac{e}{4k}\right)^{4k}z^{2k}\sim\frac{1}{k^k}$$
which is convergent. Here we have made use of asymptotic expansions $(10.19.1)$ and $(10.14.1)$ as suggested in the comments.
Best Answer
Let the parameter $a\in\mathbb R$, and write the integral as
$$I(a)=\int_0^{\pi}\cos(a\sin x)e^{\cos (x)}\ dx$$
Note that $I(a)=I(-a)$, and we get
$$I(a)=\frac12\int_{-\pi}^{\pi}\cos(a\sin x)e^{\cos (x)}\ dx=\frac12\Re\int_{-\pi}^{\pi}e^{ia\sin x}e^{\cos (x)}\ dx$$
Let $z=e^{ix}$, and choose the unit circle as the contour,
$$I(a)=\frac12\Re\oint e^{ia\frac1{2i}(z-\frac1z)}e^{\frac12(z+\frac1z)}\frac{1}{iz}\ dz=\frac12\Re\oint e^{\frac{a+1}{2}z}e^{\frac{1-a}2\cdot \frac1z}\frac{1}{iz}\ dz$$
where $z=0$ is the only sigularity inside the contour
$$I(a)=\frac12\cdot\Re\left[2\pi i~\text{Res}(z=0)\right]=\pi\cdot \Re\left[\text{Res}\left( e^{\frac{a+1}{2}z}e^{\frac{1-a}2\cdot \frac1z}\frac{1}{ z},~ z=0\right)\right]$$
We take the Laurent series,
$$e^{\frac{a+1}{2}z}e^{\frac{1-a}2\cdot \frac1z}\frac{1}{ z}=\frac{1}{ z}\sum_{k=0}\frac{1}{k!}\left( \frac{a+1}{2}\right)^kz^k\sum_{n=0}\frac{1}{n!}\left( \frac{1-a}{2}\right)^nz^n$$
the coefficient of the $z^{-1}$ term is
$$\sum_{k=0}\frac{1}{k!}\left( \frac{a+1}{2}\right)^k\frac{1}{k!}\left( \frac{1-a}{2}\right)^k=\sum_{k=0}\frac{1}{(k!)^2}\left( \frac{1-a^2}{4}\right)^k$$
therefore,
$$\boxed{\begin{align} \int_0^{\pi}\cos(a\sin x)e^{\cos (x)}\ dx&=\pi I_0(\sqrt{1-a^2}),~~~a\in (-1,1)\\ \\ \int_0^{\pi}\cos(a\sin x)e^{\cos (x)}\ dx&=\pi,~~~a=\pm1\\ \\ \int_0^{\pi}\cos(a\sin x)e^{\cos (x)}\ dx&=\pi J_0(\sqrt{a^2-1}),~~~a\in (-\infty, -1)\cup(1, \infty) \end{align}}$$