Sam Longoria

2022-01-03

${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{1}{2}{x}^{2}}dx$ and ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^{2}{e}^{-\frac{1}{2}{x}^{2}}dx$
how i compute these integrals via Gauss Integral?

Jonathan Burroughs

Make the substitution $x=\sqrt{2}t$ in both, and use integration by parts in the second, by writing
$2{t}^{2}{e}^{-{t}^{2}}=-t×\frac{d{e}^{-{t}^{2}}}{dt}$

Stuart Rountree

An interesting alternate way to evaluate the second integral ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^{2}{e}^{-\frac{1}{2}{x}^{2}}dx$ using the results form the first one is to utilize a technique called ''differentiation under the integral sign''.
From a substitution and our first integral's value in hand, we can conclude that $\mathrm{\forall }a>0\in \mathbb{R}$
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-a{x}^{2}}dx=\frac{\sqrt{\pi }}{\sqrt{a}}$
Now we differentiate both sides with respect to a (under the integral sign on the left), and we obtain
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}-{x}^{2}{e}^{-a{x}^{2}}dx=-\frac{\sqrt{\pi }}{2a\sqrt{a}}$
Factoring out the $-1$ from both sides and plugging in $a=\frac{1}{2}$ gives us our answer:
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^{2}{e}^{-\frac{1}{2}{x}^{2}}dx=\frac{\sqrt{\pi }}{\sqrt{\frac{1}{2}}}=\sqrt{2\pi }$

karton

The one on the left is what I usually think of when I hear the term ''Gaussian integral''. But some people, maybe especially physicists, omit the fraction 1/2. Including the 1/2 makes sense because then the variance of the probability distribution that you get when you normalize the function to be a probability density is 1.
If you know that
$\int {e}^{-{w}^{2}}dw=\sqrt{\pi }$
then you can write $x=\sqrt{2}w$ and $dx=\sqrt{2}dw$. As w goes from $-\mathrm{\infty }$ to $\mathrm{\infty }$, so does x. So
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{1}{2}{x}^{2}}dx={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-{w}^{2}}\sqrt{2}dw=\sqrt{2\pi }$
For ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{1}{2}{x}^{2}}dx$ begin by observing that it's an even function integrated over an interval that is symmetric about 0, so it's equal to
$2{\int }_{0}^{\mathrm{\infty }}{x}^{2}{e}^{-\frac{1}{2}{x}^{2}}dx$
Then substitute , so $\frac{du}{\sqrt{2u}}=dx$. The integral becomes
$2{\int }_{0}^{\mathrm{\infty }}2u{e}^{-u}\frac{du}{\sqrt{2u}}=\frac{4}{\sqrt{2}}{\int }_{0}^{\mathrm{\infty }}{u}^{\frac{3}{2}-1}{e}^{-u}du=2\sqrt{2}T\left(\frac{3}{2}\right)=2\sqrt{2}\cdot \frac{1}{2}\left(\frac{1}{2}\right)=\sqrt{2\pi }$

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