# Thread: Is this function integrable?

1. ## Is this function integrable?

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

If so, how can you integrate it?

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

2. 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.

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

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

4. 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?

5. 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.

6. Originally Posted by pickslides
$\int x^2e^{-x^2}$
$u=x^2$
$dv=e^{-x^2}dx$
$du=2xdx$
$v=xe^{-x^2}+2\int x^2e^{-x^2}dx$

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

7. 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

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

9. 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)).

10. 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?

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

12. 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 :]

13. 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

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

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

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