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

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

Help please because i can't find the solution.

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marked as duplicate by t.b., Srivatsan, Dylan Moreland, Sivaram Ambikasaran, Jonas Teuwen Dec 8 '11 at 22:46

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This is a famous integral known as the Gaussian integral. Call $\displaystyle I=\int_{-\infty}^{\infty}e^{-x^2}dx$. Clearly then $\displaystyle I=\int_{-\infty}^{\infty}e^{-y^2}dy$. Thus, $\displaystyle I^2=\left(\int_{-\infty}^{\infty}e^{-x^2}dx\right)\left(\int_{-\infty}^{\infty}e^{-y^2}dy\right)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dx dy$. Now, try switching to polar coordinates and see what happens.

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