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This question already has an answer here:

How to compute integral $\int_{-\infty}^\infty e^{-\frac{1}{2}x^2}dx$? I try to change it to polar coordinates but I have only one variable.

Thanks in advance!

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marked as duplicate by Semiclassical, zahbaz, user223391, Davide Giraudo, Namaste calculus May 9 '17 at 17:30

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    $\begingroup$ Square and change to polar coordinates. Very common integral. $\endgroup$ – user223391 May 9 '17 at 15:32
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    $\begingroup$ Have a look here ... $\endgroup$ – Paul Aljabar May 9 '17 at 15:34
  • $\begingroup$ I thought the univariate normal pdf had no closed form antiderivative in terms of elementary functions. $\endgroup$ – Michael McGovern May 9 '17 at 15:35
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    $\begingroup$ @MichaelMcGovern no antiderivative needed, when you alter to polar coordinates, you can integrate over "nice" regions like $\mathbb{R}$ or $\mathbb{R}^+$... $\endgroup$ – gt6989b May 9 '17 at 15:37
  • $\begingroup$ $\int_{R}\frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}dx=1$ can you get if from here? $\endgroup$ – Mesmerized student May 9 '17 at 15:37
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Let $I$ be the integral in question and note that $$ I^2=\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\frac{1}{2}(x^2+y^2)}\,dx\,dy =\int_{0}^{2\pi}\int_{0}^\infty e^{-\frac{1}{2}r^2}r\,dr\,d\theta $$ by changing to polar coordinates. The last integral is easy to compute.

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