Integral of x²e^-x²

1. Feb 17, 2008

arsmath

I am working on an integral I am finding tricky, and I think I'm missing something.
I need to integrate on the interval 0 to infinity, x²e^-x².
We have proved that on the interval of -∞ to ∞, e^-x²=√∏ so from o to ∞, it equals √∏/2. I can use this in my proof, but I don't see how. When I try integrating by parts I have trouble getting a finite answer. I would love some help,

2. Feb 17, 2008

John Creighto

The functions symmetric so I don't see your issue. Out of curiosity how did you prove the first integral as if I recall correctly the improper integral is done using residues. The proper integral can also be solved if you don't mind solutions in terms of error functions.

Last edited: Feb 17, 2008
3. Feb 17, 2008

Hurkyl

Staff Emeritus
Well, then show what you did!

4. Feb 17, 2008

arsmath

its the euler poisson gaus integral and evaluate it as a double integral, to be honest the proof is a bit over my head.

5. Feb 17, 2008

John Creighto

I've heard of that method, I believe it uses polar coordinates. That said the better way to solve it is via residues:
http://en.wikipedia.org/wiki/Methods_of_contour_integration

6. Feb 17, 2008

arsmath

ok here goes:
from 0-∞∫ x²e^-x²dx
u= x² dv = e^-x²dx
du=2xdx v=∫e^-x²dx

(x²∫e^-x²dx) – (2∫xdx)(∫e^-x²dx)
since I have that from 0-∞∫ e^-x²dx = √∏/2
I have x²(√∏/2) -√∏ which when evaluated on the interval 0-∞ goes to ∞. . .
This makes me think I’ve done it incorrectly

7. Feb 17, 2008

sutupidmath

cant one just go like this

integ(x^2e^-x^2)dx=integ(x*x*e^-x^2)dx, and then take u=x, du=dx, and v=integ(x*e^-x^2)dx, this last one can be done using the sub -x^2=t=> -2xdx=dt, xdx=-dt/2.

THis is easier i guess!

8. Feb 17, 2008

arsmath

I think that may help.

9. Feb 17, 2008

arsmath

that is my all time favorite xkcd comic.

10. Feb 17, 2008

John Creighto

I'm not sure what you have done is valid as this:
(2∫xdx)(∫e^-x²dx)
seems illegal.

11. Feb 17, 2008

arsmath

i think you're right, and I appreciate the feedback. . .so I'll try to hit it from another angle. thanks

12. Feb 17, 2008

George Jones

Staff Emeritus
Did you try to work through post #7?

13. Feb 17, 2008

John Creighto

The first substitution you did does not change the integral. You made a mistake.

14. Feb 17, 2008

John Creighto

Why not write out how you solved it for the limits from -oo to oo and we can see if we can extend the method. If you're interested I can show you how to solve the equation for the proper integral but I don't think that is what the teacher is looking for as you haven't learned about the error function yet.

15. Feb 17, 2008

sutupidmath

Yes you are right that, indeed, it does not change the integral at all, it ends up with
integ e^(-x^2)dx. Besides this what mistake are you talking about?

16. Feb 17, 2008

John Creighto

no the integrand remains unchanged. That is you now have

u^2e^(-u^2)du

and I wasn't talking about any other mistake.

17. Feb 17, 2008

sutupidmath

integ(x^2e^-x^2)dx=integ(x*x*e^-x^2)dx, and then take u=x, du=dx, and v=integ(x*e^-x^2)dx, this last one can be done using the sub -x^2=t=> -2xdx=dt, xdx=-dt/2.
now v=-1/2e^(-x^2), now going back to integ by parts we have

-1/2 xe^-x^2 +1/2 integ (e^-x^2) dx

i do not see how you will end up with the same integrand.!!

18. Feb 17, 2008

John Creighto

Oh sorry, my mistake.

19. Feb 17, 2008

sutupidmath

now i guess he can express the last part in terms of the error function,

20. Feb 18, 2008

Mute

If you've proved

$$\int_0^\infty dx~e^{-x^2} = \sqrt{\pi}$$

you should be able to extend that result to the more arbitrary case

$$\int_{-\infty}^\infty dx~e^{-\alpha x^2} = \sqrt{\frac{\pi}{\alpha}}$$
by substitution.

Now, the Liebniz rule for integrals says that (in the case of constant limits)

$$\frac{d}{d\alpha}\int_a^b dx f(x,\alpha) = \int_a^b dx\frac{\partial f(x,\alpha)}{\partial \alpha}$$

With this in mind, can you figure out how to evaluate your integral?

(I don't know if you're allowed to do it this way if this is homework, but it's a clever trick...)

Last edited: Feb 18, 2008