# (x^4)e^(-x^2/a^2) integral

1. May 20, 2010

### duckie

1. The problem statement, all variables and given/known data

$$\gamma\int\frac{1}{a\sqrt{\pi}}e^{\frac{-x^{2}}{a^{2}}}x^{4}dx$$
There should be a gamma coefficient. I've put it in using latex but it doesn't seem to be showing in the preview...

2. Relevant equations

?

3. The attempt at a solution

I know that the answer is (3/4)*gamma*a^4 (I can't seem to get Latex to work any more... sorry :/), my lecturer did all the steps leading up to the integral on the board then went straight to the answer so I'm guessing it's a definite integral.
I know that without the constant coefficient and with x^2 instead of x^4 the answer is (1/2)(pi*a^6)^0.5, it's given on a formula sheet. I have no idea what to do for x^4 though...
Both integrals are between + and - infinity.
Any help would be greatly appreciated.

Edit: I see the gamma is showing now. If you would like me to put the other couple of solutions in then let me know and I'll have another go

Last edited: May 20, 2010
2. May 20, 2010

### tiny-tim

Welcome to PF!

Hello duckie! Welcome to PF!

(have a pi: π and a gamma: γ and an integral: ∫ and try using the X2 tag just above the Reply box )
(there's been problems with previewing latex since the last server update … it doesn't affect the final result, and you can eliminate it by clicking "Refresh" on your browser with every preview )

Essentially, you're trying to integrate x4e-x2

if you keep integrating by parts, you should get down to ∫-∞ e-x2 dx, which you presumably know how to do.

3. May 20, 2010

### duckie

Re: Welcome to PF!

Thanks

I had a go at it this way earlier by setting u = x4 and dv = e-x2 but that ended pretty quickly as v = ∫-∞ e-x2 dx = π1/2... right?

4. May 20, 2010

### Cyosis

No that is wrong. You can't enter the boundary values when you go from dv to v. The product of uv in the partial integration needs to be evaluated as a whole.

Hint: $$x^4 e^{-x^2}=x^3 \left(xe^{-x^2}\right)$$

5. May 20, 2010

### tiny-tim

Hello duckie!

Think laterally …

try xe-x2

6. May 20, 2010

### duckie

Oh, do I use the substitution u = x2?

Edit: Just did it this way, I have an answer now! Which is more that I've had for a long while... though apparently it's wrong. On integrating x4e-x2 I got 3π1/2/4.
Though when I put the 1/a2 term in the exponential I ended up with a41/2/4. Which is logical as far as I can see. Though in the original question there is an additional coefficient of 1/a which will mean my result cancelling to an a3 term...
Is the answer I was given wrong or have I lost a factor of a somewhere?
Edit 2: Don't worry, I see where I lost it

Thanks for all the help!

Last edited: May 20, 2010