# Inegral evaluation

1. Jan 24, 2010

### dionysian

This is not a homework question but I am trying to follow the proof on wolfram that $$\int_{-\infty }^{ \infty }{e}^{{x}^{-2}} dx = \sqrt{\pi}$$ and I am haveing trouble at one point where they state $$\int_{0 }^{ \infty }r{e}^{{r}^{-2}} dr = \left[- \frac{ 1}{ 2} {e }^{ {-r }^{2 } } \right ] \infty \rightarrow 0$$.

How the hell do they make this jump? The only way i would know to evaluate this integral is to use integration by parts and this would eventually leave you with another integral that is itself the gaussian. I am sure there is an easy were than parts. Does anyone have any insight here?

btw the$$\infty \rightarrow 0$$ is suppose to mean the integral is evaluated from 0 to infinity but i dont know how to do this proper in latex.

2. Jan 24, 2010

### tiny-tim

Hi dionysian!

Use a substitution: u = r2.

3. Jan 24, 2010

### dionysian

Thanks... I am a little rusty...