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

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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 $$ \int_{-\infty}^{\infty} \exp(-x^2) \,\mathrm{d}x = \sqrt{\pi} \space ? $$

@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

score 6 · Accepted Answer · answered May 13 '11 at 13:45

6

Such an integral is called a Gaussian Integral

This link should help you out.

answered May 13 '11 at 13:45

kodyv

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

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