# [SOLVED] Tricks to evaluate this integral?

• Mar 6th 2009, 01:04 PM
Amanda1990
[SOLVED] Tricks to evaluate this integral?
Let $J_0 (x) = 2/\pi \int_{0}^{\pi/2} cos (x cos \theta) d\theta$. Show that $\int_{0}^{\infty} J_0(x) \exp(-ax) dx = 1/\sqrt{(1+a^2)}$ if $a>0$.

I'm guessing that you have to reverse the order of integration in order to make the integrals easier to evaluate, but it still seems quite tricky. Any ideas?
• Mar 6th 2009, 02:14 PM
Krizalid
Yes, just reverse integration order.

We need to deal with $\int_{0}^{\infty }{e^{-ax}\cos (x\cos \theta )\,dx}.$ The best way on solving it, it's via complex numbers, but to do that, first put $\beta=\cos\theta.$ (It's just a constant, it's just to make computations easier.)

Thus the integral equals $\int_{0}^{\infty }{e^{-ax}\cos (\beta x)\,dx}=\text{Re}\int_{0}^{\infty }{e^{-(a+\beta i)x}\,dx}=\text\,\frac{1}{a+\beta i}=\frac{a^{2}}{a^{2}+\beta ^{2}},$ thus the remaining challenge is to compute $\frac{2}{\pi }\int_{0}^{\frac{\pi }{2}}{\frac{a}{a^{2}+\cos ^{2}\theta }\,d\theta }.$