# Thread: [SOLVED] Tricks to evaluate this integral?

1. ## [SOLVED] Tricks to evaluate this integral?

Let $\displaystyle J_0 (x) = 2/\pi \int_{0}^{\pi/2} cos (x cos \theta) d\theta$. Show that $\displaystyle \int_{0}^{\infty} J_0(x) \exp(-ax) dx = 1/\sqrt{(1+a^2)}$ if $\displaystyle a>0$.

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?

2. Yes, just reverse integration order.

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

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