Integral $\int_{-\pi}^\pi d\theta\,e^{i \theta+a e^{i \theta} + b e^{-i \theta}}$ in terms of Bessel Function

Question

Is it possible to find

$$\int_{-\pi}^\pi d\theta\,e^{i \theta+a e^{i \theta} + b e^{-i \theta}}$$

in terms of Bessel functions?

Answer 1

I'll give the answer for the more general integral $$\frac{1}{2\pi}\int_{-\pi}^\pi\mathrm{d}\theta\,\mathrm{e}^{\mathrm{i} \theta n}\mathrm{e}^{a\mathrm{e}^{\mathrm{i} \theta}+b\mathrm{e}^{-\mathrm{i} \theta}}.$$ We begin by expanding the second exponential in the integral as a power series: $$\mathrm{e}^{a\mathrm{e}^{\mathrm{i} \theta}+b\mathrm{e}^{-\mathrm{i} \theta}}=\sum_{\mu,\nu\in\mathbb{Z}}\frac{1}{\mu!\nu!}a^\mu b^\nu\mathrm{e}^{\mathrm{i} \theta(\mu-\nu)},$$ where we are allowed to start at negative infinity for both indices, since $1/n!$ is zero for all natural numbers $n<0$. By use of the Cauchy product formula, we find for the double sum $$\sum_{\mu=-\infty}^\infty\left(\sum_{\nu=-\infty}^\mu\frac{a^\mu b^{\mu-\nu}}{\mu!(\mu-\nu)!}\mathrm{e}^{\mathrm{i} \theta\nu}\right).$$ Inserting this back into the original integral, we get from the orthogonality of the complex exponential: $$\sum_{\mu=-\infty}^\infty\sum_{\nu=-\infty}^\mu\frac{a^\mu b^{\mu-\nu}}{\mu!(\mu-\nu)!}\frac{1}{2\pi}\int_{-\pi}^\pi\mathrm{d}\theta\,\mathrm{e}^{\mathrm{i} \theta n}\mathrm{e}^{\mathrm{i} \theta\nu}=\sum_{\mu=-\infty}^\infty\sum_{\nu=-\infty}^\mu\frac{a^\mu b^{\mu-\nu}}{\mu!(\mu-\nu)!}\delta_{\nu,-n},$$ which allows us to simplify the sum as $$\sum_{\mu=0}^\infty\frac{a^\mu b^{\mu+n}}{\mu!(\mu+n)!}.$$ From there, its just a matter of identifying the modified Bessel function of the first kind: $$\left(\frac{b}{a}\right)^{\frac{n}{2}}\sum_{\mu=0}^\infty\frac{(\sqrt{ab})^{2\mu+n}}{\mu!(\mu+n)!}=\left(\frac{b}{a}\right)^{\frac{n}{2}}I_n(2\sqrt{ab}).$$ Therefore, we have for the original expression $$\boxed{\frac{1}{2\pi}\int_{-\pi}^\pi\mathrm{d}\theta\,\mathrm{e}^{\mathrm{i} \theta n}\mathrm{e}^{a\mathrm{e}^{\mathrm{i} \theta}+b\mathrm{e}^{-\mathrm{i} \theta}}=\left(\frac{b}{a}\right)^{\frac{n}{2}}I_n(2\sqrt{ab})}$$ Your desired integral is the case $n=1$ as provided by @metamorphy.