# Integral of polynomial times exp(-x^2)

I have the integral ##\int_{-\infty}^{\infty} x^2 e^{-x^2} ~dx##. Is there any simple way to integrate this, given that that I already know that the value of the Gaussian integral is ##\sqrt{\pi}##?

Stephen Tashi
Try integration by parts.

Try integration by parts.
What I found was that ##\int_{-\infty}^{\infty}e^{-x^2} ~dx = 2\int_{-\infty}^{\infty}x^2 e^{-x^2} ~dx##, so by original integral is ##\sqrt{\pi} / 2##. To do this though I had to start from the original Gaussian integral and integrate by parts to get the integral that I have. Is that trick of integrating by parts the original Gaussian integral to get what I want what you had in mind?

Stephen Tashi
Is that trick of integrating by parts the original Gaussian integral to get what I want what you had in mind?

No, I was thinking of starting with the original problem and using ##f(x) = x, g'(x) = x e^{-x^2} ##.

lurflurf
Homework Helper
for ##t+t^\ast>0\\
\int_{-\infty}^\infty \mathrm{f}(x^2)e ^{-t x^2} dx=\mathrm{f}\left(-\dfrac{d}{dt}\right)\int_{-\infty}^\infty e ^{-t x^2} dx%=\mathrm{f}\left(-\dfrac{d}{dt}\right)\sqrt{\frac{\pi}{t}}##
the case f(x)=x^n imvolves (2n-1)!!
that comes up all the time in
chemistry http://www.colby.edu/chemistry/PChem/notes/Integral.pdf
psychology http://www.umich.edu/~chem461/Gaussian Integrals.pdf
http://www.math.tamu.edu/~manshel/papers/Hermite-integrals.pdf
http://mathworld.wolfram.com/GaussianIntegral.html

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Another approach would be to realize that the integrand is an even function and write:
$$I(x) =\int_{-\infty}^{\infty}x^2e^{-x^2}dx=2\int_{0}^{\infty}x^2e^{-x^2}dx$$
Let ##u=x^2##, ##du=2xdx## and ##x=\sqrt {u}##
$$I(u)=\int_{0}^{\infty}u^{\frac {1} {2}}e^{-u}du = \Gamma(\frac {3} {2}) = \frac {1} {2} \sqrt {\pi}$$

Peace,
Fred