$\int_{0}^{2\pi} \frac{1}{\pi} e^{-x e^{-j\pi\sin(\theta)}} d\theta$

Question

Math StackExchange community,

I've been working on analyzing the integral:

$$ \int_{0}^{2\pi} \frac{1}{\pi} e^{-x e^{-j\pi\sin(\theta)}} d\theta, $$

where $x$ is a complex number ($x = a + jb$), and $j = \sqrt{-1}$. My goal is to understand this integral's behavior, particularly in terms of its real and imaginary parts, and to find a theoretical solution or express it in terms of well-known functions or integrals.

Previous Attempts:

1. Imaginary Part Analysis: Through symmetry arguments and considering the periodic and symmetric properties of the sine and cosine functions, we established that the imaginary part of the integral is zero across the domain of (x).

2. Real Part Behavior: By analyzing the first and second derivatives of the function inside the integral with respect to (x), we proved that the real part of the integral is increasing and monotonic, as well as convex, with respect to (x).

3. Simplification Attempt: We considered simplifying the real part of the integral, focusing on:

$$ \int_{0}^{2\pi} \frac{1}{\pi} e^{-x \cos(\pi\sin(\theta))} d\theta. $$

This led to a discussion on whether this form has similarities with the modified Bessel function of the first kind, ($I_0(x)$), given its resemblance to integrals related to Bessel functions. However, the unique composition involving ($\cos(\pi\sin(\theta))$) does not straightforwardly match the standard form or known variations of ($I_0(x)$).

Questions:

1. Is there a closed-form expression for this integral, or can it be expressed in terms of well-known functions or integrals?

2. Given the properties we've established (imaginary part being zero, monotonicity, and convexity of the real part), are there any theoretical tools or approaches that could simplify the analysis or provide a direct solution?

I appreciate any insights or references to similar integrals or techniques that could aid in solving or approximating this integral in a more theoretical framework.

Thank you for your time and assistance.

Answer 1

I was able to make a bit of a mess thusly;

$$\frac{1}{\pi}\int_{0}^{2\pi}e^{-xe^{-i\pi\sin\theta}}d\theta = \frac{1}{\pi}\sum_{n=0}^{\infty}\frac{(-x)^n}{n!}\int_{0}^{2\pi}e^{-in\pi\sin\theta}d\theta$$

$$=\frac{1}{\pi}\sum_{n=0}^{\infty}\frac{(-x)^n}{n!}\sum_{k=0}^{\infty}\frac{(-in\pi)^k}{k!}\int_{0}^{2\pi}\sin^k\theta d\theta$$ The integral $\int_{0}^{2\pi}\sin^k\theta d\theta =0$ for all odd k. Therefore, we can leverage some symmetry and massage it into a Beta Function...

$$=\frac{4}{\pi}\sum_{n=0}^{\infty}\frac{(-x)^n}{n!}\sum_{k=0}^{\infty}\frac{(-n^2\pi^2)^{k}}{(2k)!}\int_{0}^{\frac{\pi}{2}}\sin^{2k}\theta d\theta$$

$$=\frac{2}{\pi}\sum_{n=0}^{\infty}\frac{(-x)^n}{n!}\sum_{k=0}^{\infty}\frac{(-n^2\pi^2)^{k}}{(2k)!}B\big((k+\frac{1}{2}),\frac{1}{2}\big)$$

$$=\frac{2}{\sqrt{\pi}}\sum_{n=0}^{\infty}\frac{(-x)^n}{n!}\sum_{k=0}^{\infty}\frac{(-n^2\pi^2)^{k}}{(2k)!}\frac{\Gamma(k+\frac{1}{2})}{k!}$$

$$=2\sum_{n=0}^{\infty}\frac{(-x)^n}{n!}\sum_{k=0}^{\infty}\frac{(-n^2\pi^2)^{k}}{4^k(k!)^2}$$

An application of the definition of the Bessel Function, $J_0$, reveals our final result;

$$2\sum_{n=0}^{\infty}\frac{(-x)^n}{n!} J_{0}(n\pi)$$