Integral of polynomial times exp(-x^2)

[Mr Davis 97]

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.

 

[Mr Davis 97]

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]

for $t+t^\ast>0\\ \int_{-\infty}^\infty \mathrm{f}(x^2)e ^{-t x^2} dx=\mathrm{f}\left(-\dfrac{d}{dt}\right)\int_{-\infty}^\infty e ^{-t x^2} dx%=\mathrm{f}\left(-\dfrac{d}{dt}\right)\sqrt{\frac{\pi}{t}}$
your case is f(x)=x
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

 

[Fred Wright]

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