# 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