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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2Square and change to polar coordinates. Very common integral. – May 09 '17 at 15:32
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1Have a look here ... – Paul Aljabar May 09 '17 at 15:34
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I thought the univariate normal pdf had no closed form antiderivative in terms of elementary functions. – Michael McGovern May 09 '17 at 15:35
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1@MichaelMcGovern no antiderivative needed, when you alter to polar coordinates, you can integrate over "nice" regions like $\mathbb{R}$ or $\mathbb{R}^+$... – gt6989b May 09 '17 at 15:37
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$\int_{R}\frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}dx=1$ can you get if from here? – Mesmerized student May 09 '17 at 15:37
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Surely this is a duplicate, but I can't find one. – Matthew Leingang May 09 '17 at 15:38
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@MichaelMcGovern with these bounds it does. – GuPe May 09 '17 at 15:38
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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.
Sri-Amirthan Theivendran
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