# Showing that $\int_{-\infty}^{\infty} \exp(-x^2) \,\mathrm{d}x = \sqrt{\pi}$ [duplicate]

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

The primitive of $f(x) = \exp(-x^2)$ has no analytical expression, even so, it is possible to evaluate $\int f(x)$ along the whole real line with a few tricks. How can one show that $$\int_{-\infty}^{\infty} \exp(-x^2) \,\mathrm{d}x = \sqrt{\pi} \space ?$$

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@Ross Millikan Damn, that one didn't show up in the related questions. It is quite a duplicate, the only difference is a predictable symmetry argument to double the answer. This question will probably be deleted. Oh, well. –  Luke May 13 '11 at 14:02
The related question searcher seems to be spotty. I am surprised both by what it finds and by what it misses. No problem. –  Ross Millikan May 13 '11 at 14:07