$\displaystyle \int_{0}^{\infty}x^2e^{-x^2}$

If so, how can you integrate it?

Also, what about $\displaystyle \int_0^\infty e^{-x^2}$

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- Feb 7th 2010, 11:51 AMpaupsersIs this function integrable?
$\displaystyle \int_{0}^{\infty}x^2e^{-x^2}$

If so, how can you integrate it?

Also, what about $\displaystyle \int_0^\infty e^{-x^2}$ - Feb 7th 2010, 11:54 AMpickslides
- Feb 7th 2010, 11:58 AMfelper
A function is integrable $\displaystyle \neq$ is possible to find it's primitve o it's exact numerical value (using elemental functions).

- Feb 7th 2010, 01:24 PMpaupsers
- Feb 7th 2010, 01:27 PMpickslides
- Feb 7th 2010, 01:37 PMpaupsers
- Feb 7th 2010, 01:43 PMpickslides
Parts won't work this time, my apologies

Wolfram Mathematica Online Integrator - Feb 7th 2010, 01:48 PMfelper
You can do this integral using doble integrals, or some know values of the gamma function.

- Feb 7th 2010, 02:08 PMvince
this can be done.

Choose

$\displaystyle u=x$

$\displaystyle dv=xe^{-x^2}dx$

Notice that then v=$\displaystyle \frac{-e^{-x^2}}{2}$

You will then notice that the**uv**term of the integration by parts is 0 (remember that for definite integrals, this term gets evaluated at the main integral's bounds, i.e., $\displaystyle \infty$ and 0.) What you're left with is

$\displaystyle \frac{1}{2}\int_0^\infty e^{-x^2}$

A rather cool method for that one will use polar coordinates. here's a link:

Gaussian integral - Wikipedia, the free encyclopedia

(note the limits of integration for the Gaussian integral: You will need to divide by 2 (justified as the integrand is an even function)). - Feb 7th 2010, 02:30 PMpaupsers
- Feb 7th 2010, 02:31 PMpaupsers
Oops, I just saw that you edited your post to explain that part. Thanks!!

- Feb 7th 2010, 02:33 PMvince
- Feb 7th 2010, 03:02 PMRandom Variable
Use the fact that $\displaystyle \int_{0}^{\infty} e^{-x^{2}} \ dx = \frac{\sqrt{\pi}}{2} $

Make the substitution $\displaystyle x^{2}= au^{2} $

which implies that $\displaystyle 2xdx=2audu $

so $\displaystyle dx=\frac{au}{x} dx = \frac{ax}{\sqrt{a}x}dx = \sqrt{a} dx $

then $\displaystyle \int^{\infty}_{0} e^{-x^{2}} \ dx = \sqrt{a} \int^{\infty}_{0} e^{-au^{2}} du = \frac{\sqrt{\pi}}{2} $

and $\displaystyle \int^{\infty}_{0} e^{-au^{2}} \ du =$ $\displaystyle \frac{1}{2} \sqrt{\frac{\pi}{a}}$

notice that $\displaystyle - \int^{\infty}_{0} \frac{\partial}{\partial a} \ e^{-au^{2}} \ du =$$\displaystyle \int^{\infty}_{0} u^{2} e^{-au^{2}} \ du $

now swtich the order of integration and differentiation (which in this case is allowed)

$\displaystyle - \frac{\partial}{\partial a} \int^{\infty}_{0} e^{-au^{2}} \ du = - \frac{\partial}{\partial a} \ \frac{1}{2} \sqrt{\frac{\pi}{a}} = $$\displaystyle \ \frac{\sqrt{\pi}}{4}a^{-3/2}$

so $\displaystyle \int^{\infty}_{0} u^{2} e^{-au^{2}} \ du = \frac{\sqrt{\pi}}{4}a^{-3/2} $

letting a=1

$\displaystyle \int^{\infty}_{0} u^{2} e^{-u^{2}} \ du = \frac{\sqrt{\pi}}{4} $ - Feb 7th 2010, 03:44 PMfelper
$\displaystyle I=\int_0^{\infty}x^2e^{-x^2}dx$

Set $\displaystyle x^2=t$ then $\displaystyle I=\dfrac{1}{2}\int_0^{\infty}\sqrt{t}e^{-t}=\dfrac{1}{2}\Gamma(\dfrac{1}{2})=\dfrac{\sqrt{\ pi}}{4}$