How can we prove this identify: $\int_0^{2\pi} \exp(i a\cos(x))\, dx = 2 \pi I_0(a)$

Question

How can we argue that $$ \int_0^{2\pi} \exp(i a\cos(x)) \,dx = 2 \pi I_0(a) $$ where $I_0(a)$ is a modified Bessel function.

I tried simplifying it as below: \begin{align} \int_0^{2\pi} \exp(i a\cos(x))\, dx & = \int_0^\pi \exp(i a\cos(x))\, dx + \int_\pi^{2\pi} \exp(i a\cos(x))\, dx\\ & = \pi I_0(a) + \int_0^\pi \exp(i a\cos(\theta + \pi))\, d\theta\\ & = \pi I_0(a) + \int_0^\pi \exp(-i a\cos(\theta)) \, d\theta \end{align}

How can I show that $$ \int_0^\pi \exp(-i a\cos(\theta)) \, d\theta = \pi I_0(a) \text{ ?} $$

Answer 1

Enforcing the substitution $x\mapsto 2\pi-x$, we see that

$$\int_\pi^{2\pi}e^{ia\cos(x)}\,dx=\int_0^\pi e^{ia\cos(x)}\,dx$$

Hence, we assert that

$$\begin{align} \int_0^{2\pi} e^{ia\cos(x)}\,dx&=2\int_0^\pi e^{ia\cos(x)}\,dx\\\\ &=2\pi I_0(a) \end{align}$$

as was to be shown!