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

At lunch with a math friend years ago, he showed me an integral whose solution was, he said, so beautiful that it compelled him to become a professional mathematician. I scribbled the integral down on a napkin, but for the life of me I cannot remember the trick he found so compelling.

And these things are hard to Google...

$$ \int_{-\infty}^{\infty} e^{-x^2} dx $$

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marked as duplicate by Ross Millikan, t.b., Arturo Magidin, Zev Chonoles, yunone Apr 26 '11 at 3:47

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
You can search maths in Latex here –  Emre Apr 26 '11 at 3:41
    
en.wikipedia.org/wiki/Gaussian_integral ; standard method is to compute the integral over the entire plane of $e^{-(x^2+y^2)}$ using polar coordinates, and then show that it equals the square of the integral you want. –  Arturo Magidin Apr 26 '11 at 3:41
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@Arturo & Ross: both links give the same proof... which is indeed quite slick. Cheers. (I would delete my question now, but it already has an answer, so this is disallowed.) –  Fixee Apr 26 '11 at 3:47

1 Answer 1

The integral is equal to $\sqrt{\pi}$. The proof that I know (which is quite beautiful, though it may not be the same one your math friend is referring to) uses the extremely clever trick of squaring the integral we want to find, and switching to polar coordinates. See here for the details.

There is a quote, attributed to Lord Kelvin, about the fact that $\int_{-\infty}^\infty e^{-x^2}=\sqrt{\pi}$:

"A mathematician is one to whom that is as obvious as that twice two makes four is to you. Liouville was a mathematician."

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