I need to evaluate this integral and holy Wolfram did it for me:
${\displaystyle\int_{-\pi}^{\pi} \exp\left(-\frac{1}{2}\left(\frac{2r\sin(x/2)}{\sigma}\right)^2\right) \textrm{d}x} = 2 \pi \exp(-\frac{r^2}{\sigma^2}) I_0(\frac{r^2}{\sigma^2})$
Where $I_0(x)$ is the modified Bessel function of the first kind.
Nevertheless, since this is part of my master thesis, I'd also like to have a step-by-step solution, but I have absolutely no idea how to solve this integral since Wolfram kindly asks me to go Pro. I must confess that this is the first time I encounter the Bessel function and, although I red something, I'm not very familiar with it. What would be the approach for the solution of this integral? Any advice?
Thanks!
Place $k \equiv \frac{r^2}{\sigma^2}$, I would do like this:
$$ \begin{aligned} \int_{-\pi}^{\pi} \exp\left(-2\,k\,\sin^2\left(\frac{x}{2}\right)\right)\text{d}x & = 2\int_0^{\pi} \exp\left(-2\,k\,\sin^2\left(\frac{x}{2}\right)\right)\text{d}x \\ & = 2\int_0^{\pi} \exp\left(-k\left(1 - \cos x\right)\right)\text{d}x \\ & = 2\exp(-k)\int_0^{\pi} \exp\left(k\cos x\right)\text{d}x \\ & = 2\pi\,\exp(-k)\,\frac{1}{\pi}\int_0^{\pi}\exp\left(k\cos x\right)\text{d}x \\ & = 2\pi\,\exp(-k)\,I_0(k) \end{aligned} $$
as reported in Abramowitz and Stegun 1972, p. 376.