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

  • Thread starter duckie
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  • #1
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Homework Statement



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

Homework Equations



?

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:

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hello duckie! Welcome to PF! :smile:

(have a pi: π and a gamma: γ and an integral: ∫ and try using the X2 tag just above the Reply box :wink:)
[tex]\gamma\int\frac{1}{a\sqrt{\pi}}e^{\frac{-x^{2}}{a^{2}}}x^{4}dx[/tex]
There should be a gamma coefficient. I've put it in using latex but it doesn't seem to be showing in the preview...

(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 :wink:)

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. :smile:
 
  • #3
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Hello duckie! Welcome to PF! :smile:
Thanks :smile:

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

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
Cyosis
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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: [tex]x^4 e^{-x^2}=x^3 \left(xe^{-x^2}\right)[/tex]
 
  • #5
tiny-tim
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Hello duckie! :smile:

Think laterally …

try xe-x2 :wink:
 
  • #6
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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 :smile:

Thanks for all the help!
 
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