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
[tex]\int_{-\infty}^{\infty} e^{-\alpha x^2} \, \mathrm{d}x = \sqrt{\frac{\pi}{\alpha}}, [/tex]
for [itex]\alpha > 0[/itex] (and even [itex]\alpha \in \mathbb{C}, \operatorname{Re}(\alpha) > 0[/itex]).

Then if you take [itex]\alpha = 1[/itex] or [itex]\alpha = \pi^2[/itex] 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

[tex]\int_a^b e^{-\pi x^2}\,dx \rightarrow \int_{-\infty}^{\infty} e^{-\pi x^2}\,dx=1[/tex]

But that's probably beside the point.