What is the integral of $e^{-x^2/2}$ over $\mathbb{R}$ [closed]

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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.

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2 Answers

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

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 $\;\sqrt{2\pi}\;$

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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 $y=\frac{x^2}{2}$.

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