7

What is the integral of

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

My working is here:

= $-e^(-1/2x^2)/x$ from negative infinity to infinity.

What is the value of this? Not sure how to carry on from here. Thank you.

amWhy
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lakshmen
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  • Do you mean $\int e^{-\frac{x^2}2},dx$? – mrf Sep 23 '13 at 11:17
  • yes I meant that... – lakshmen Sep 23 '13 at 11:18
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    See http://en.wikipedia.org/wiki/Gaussian_integral – mrf Sep 23 '13 at 11:23
  • multiply by $\frac{\sqrt{2\pi}}{\sqrt{2\pi}}$, then compare the integral with the pdf of a standard normal distribution N(0,1). If you integration from $-\infty$ to $\infty$ over the standard normal pdf, you get 1. $\int \limits_{-\infty}^{\infty} f_{X}(x)~dx = 1$ , where $f_{X}(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}$. Also, note that standard normal distribution is even symmetry, so if you integrate from 0 to $\infty$ you get 1/2. – Bill Moore Apr 10 '20 at 21:20

2 Answers2

35

$$\left(\int\limits_{-\infty}^\infty e^{-\frac12x^2}dx\right)^2=\int\limits_{-\infty}^\infty e^{-\frac12x^2}dx\int\limits_{-\infty}^\infty e^{-\frac12y^2}dy=\int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty e^{-\frac12(x^2+y^2)}dxdy=$$

Change now to polar coordinates:

$$=\int\limits_0^{2\pi}\int\limits_0^\infty re^{-\frac12r^2}drd\theta=\left.-2\pi e^{-\frac12r^2}\right|_0^\infty=2\pi$$

So your integral equals $\;\sqrt{2\pi}\;$

DonAntonio
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9

Here is a solution using the gamma function $$ \int_{-\infty}^{\infty}e^{-x^2/2}dx = 2\int_{0}^{\infty}e^{-x^2/2}dx = \sqrt{2}\int_{0}^{\infty} y^{-1/2}\,e^{-y}dy = \sqrt{2}\Gamma(1/2), $$

Where $\Gamma(x)$ is the gamma function. Note that, the change of variables $y=\frac{x^2}{2}$.

Alex Ortiz
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Mhenni Benghorbal
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