Is this function integrable?

**paupsers** · February 7th 2010, 11:51 AM

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

If so, how can you integrate it?

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

**pickslides** · February 7th 2010, 11:54 AM

Originally Posted by paupsers

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

If so, how can you integrate it?

For this I would try integration by parts twice.

$\int x^2e^{-x^2}~dx$

**felper** · February 7th 2010, 11:58 AM

A function is integrable $\neq$ is possible to find it's primitve o it's exact numerical value (using elemental functions).

**paupsers** · February 7th 2010, 01:24 PM

Originally Posted by pickslides

For this I would try integration by parts twice.

$\int x^2e^{-x^2}~dx$

I've tried that and I still haven't been able to solve it, I just end up going in circles. Any more advice?

**pickslides** · February 7th 2010, 01:27 PM

Originally Posted by paupsers

I've tried that and I still haven't been able to solve it, I just end up going in circles.

Can you post your workings.

**paupsers** · February 7th 2010, 01:37 PM

Originally Posted by pickslides

Can you post your workings.

$\int x^2e^{-x^2}$
$u=x^2$
$dv=e^{-x^2}dx$
$du=2xdx$
$v=xe^{-x^2}+2\int x^2e^{-x^2}dx$

I've also tried it with letting $u=e^{-x^2}$
and I end up getting stuck as well.

**pickslides** · February 7th 2010, 01:43 PM

Originally Posted by paupsers

$\int x^2e^{-x^2}$
$u=x^2$
$dv=e^{-x^2}dx$
$du=2xdx$
$v=xe^{-x^2}+2\int x^2e^{-x^2}dx$

I've also tried it with letting $u=e^{-x^2}$
and I end up getting stuck as well.

Parts won't work this time, my apologies

Wolfram Mathematica Online Integrator

**felper** · February 7th 2010, 01:48 PM

You can do this integral using doble integrals, or some know values of the gamma function.

**vince** · February 7th 2010, 02:08 PM

this can be done.

Choose
$u=x$
$dv=xe^{-x^2}dx$

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

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., $\infty$ and 0.) What you're left with is
$\frac{1}{2}\int_0^\infty e^{-x^2}$

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)).

**paupsers** · February 7th 2010, 02:30 PM

Originally Posted by vince

this can be done.

Choose
$u=x$
$dv=xe^{-x^2}dx$

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

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., $\infty$ and 0.) What you're left with is
$\frac{1}{2}\int_0^\infty e^{-x^2}$

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)).

Hmmm, I don't understand why the uv term is 0.
If u = x and v = -(1/2)e^(-x^2), why does multiplying them make it 0?

**paupsers** · February 7th 2010, 02:31 PM

Oops, I just saw that you edited your post to explain that part. Thanks!!

**vince** · February 7th 2010, 02:33 PM

Originally Posted by paupsers

Oops, I just saw that you edited your post to explain that part. Thanks!!

you may be compelled to click 'thanks'...im a thanks hog :]

**Random Variable** · February 7th 2010, 03:02 PM

Use the fact that $\int_{0}^{\infty} e^{-x^{2}} \ dx = \frac{\sqrt{\pi}}{2}$

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

which implies that $2xdx=2audu$

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

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

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

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

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

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

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

letting a=1

$\int^{\infty}_{0} u^{2} e^{-u^{2}} \ du = \frac{\sqrt{\pi}}{4}$

**felper** · February 7th 2010, 03:44 PM

$I=\int_0^{\infty}x^2e^{-x^2}dx$

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

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