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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 ? $

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    The related question searcher seems to be spotty. I am surprised both by what it finds and by what it misses. No problem.2011-05-13

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Such an integral is called a Gaussian Integral

This link should help you out.