# Integrate e^(-pi^2*x^2)

1. Mar 8, 2008

### Gear300

How would I integrate e^(-pi^2*x^2) from -infinite to infinite? I know that the integral of e^(-x^2) from -infinite to infinite is sqr(pi), in which sqr is square root.

2. Mar 8, 2008

### CompuChip

In general once can prove that
$$\int_{-\infty}^{\infty} e^{-\alpha x^2} \, \mathrm{d}x = \sqrt{\frac{\pi}{\alpha}},$$
for $\alpha > 0$ (and even $\alpha \in \mathbb{C}, \operatorname{Re}(\alpha) > 0$).

Then if you take $\alpha = 1$ or $\alpha = \pi^2$ you will get either of the integrals in your post.

3. Mar 8, 2008

### Gear300

I see...thank you.

4. Mar 8, 2008

### Schrodinger's Dog

Actually, I can see in this case that a=b so b-a=0 and therefore

$$\int_a^b e^{-\pi x^2}\,dx \rightarrow \int_{-\infty}^{\infty} e^{-\pi x^2}\,dx=1$$

But that's probably beside the point.