# Integral $\int_{-\infty}^\infty e^{-\frac{1}{2}x^2}dx$ [duplicate]

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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!

## marked as duplicate by Semiclassical, zahbaz, user223391, Davide Giraudo, Namaste calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 9 '17 at 17:30

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• Square and change to polar coordinates. Very common integral. – user223391 May 9 '17 at 15:32
• Have a look here ... – Paul Aljabar May 9 '17 at 15:34
• I thought the univariate normal pdf had no closed form antiderivative in terms of elementary functions. – Michael McGovern May 9 '17 at 15:35
• @MichaelMcGovern no antiderivative needed, when you alter to polar coordinates, you can integrate over "nice" regions like $\mathbb{R}$ or $\mathbb{R}^+$... – gt6989b May 9 '17 at 15:37
• $\int_{R}\frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}dx=1$ can you get if from here? – Mesmerized student May 9 '17 at 15:37

## 1 Answer

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.