# Definite integral resolution [duplicate]

Possible Duplicate:
Proving $\int_{0}^{+\infty} e^{-x^2} dx = \frac{\sqrt \pi}{2}$

My calculus is a bit rusty, how should I solve this in order to calculate the solution?

$\int^{\infty}_{-\infty}e^{-x^{2}}dx=\sqrt{\pi}$

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## marked as duplicate by Pedro Tamaroff, Alex Becker♦, David Mitra, William, DonAntonioSep 14 '12 at 3:18

if $I=\int_{-\infty}^{\infty}e^{-x^2}$ then $$I^2=\int\int e^{-(x^2+y^2)}dxdy=\int\int re^{-r^2}drd\theta=\pi$$