# Bessel's integral, how to actually evaluate?

I am just about to study Bessel functions and I have recently seen one of its integral representations given by: $$J_ \alpha (x) = \frac{1}{\pi} \int_0 ^ \pi \cos(\alpha \tau - x\sin\tau) d\tau - \frac{\sin(\alpha \pi)}{\pi}\int_0^ \infty e^{-x\sinh(t)-\alpha t}dt$$ as you can see from the wikipedia page.

I am mainly interested on how to evaluate the integral on the last term, $$\int_0^ \infty e^{-x\sinh(t)-\alpha t}dt$$

when x is a positive real number and $\alpha$ is a real parameter. I have tried the elementary methods like integration by parts and looking for differentials but with no luck. I'm actually on my way to study Bessel functions until I derive the integral representation above but I'd like to know how to actually evaluate these kinds of integrals so that I can use the representation for certain cases.

Can anyone give me an advice on how to solve the integral at the bottom analytically? I'm thinking that there may be some contour trick involved but nothing really comes to mind.

## 1 Answer

$\int_0^\infty e^{-x\sinh(t)-\alpha t}~dt=\pi\mathbf{A}_\alpha(x)$ (according to http://dlmf.nist.gov/11.10#E4)

When $\alpha$ is not an integer,

$J_\alpha(x)=\dfrac{1}{\pi}\int_0^\pi\cos(\alpha\tau-x\sin\tau)~d\tau-\dfrac{\sin(\alpha\pi)}{\pi}\int_0^\infty e^{-x\sinh(t)-\alpha t}~dt=\mathbf{J}_\alpha(x)-\dfrac{\sin(\alpha\pi)}{\pi}\int_0^\infty e^{-x\sinh(t)-\alpha t}~dt$ (according to http://en.wikipedia.org/wiki/Anger_function)

$\therefore\int_0^\infty e^{-x\sinh(t)-\alpha t}~dt=\dfrac{\pi(\mathbf{J}_\alpha(x)-J_\alpha(x))}{\sin(\alpha\pi)}$

• Well, that's cool. But I'm actually trying to claw my way out of evaluating this into a closed form containing Anger or Bessel functions. Sorry if I didn't include that in the question earlier. Apr 18, 2014 at 12:37