How to Integrate e^(-pi^2*x^2) from -infinite to infinite?

[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.

 

[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.

 

[Gear300]

I see...thank you.

 

[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.