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Gaussian Integral

  • Thread starter csnsc14320
  • Start date
  • #1
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1
EDIT: meant to post this is the math forums, if you can remove this I'm going to switch it over

Homework Statement


Solve:

In = [tex] \int_{0}^{\infty} x^n e^{-\lambda x^2} dx [/tex]


Homework Equations





The Attempt at a Solution


So my teacher gave a few hints regarding this. She first said to evaluate when n = 0, then consider the cases when n = even and n = odd, comparing the even cases to the p-th derivative of Io.

For the Io case, I evaluated it and obtained [tex] I_o = \frac{1}{2} \sqrt{\frac{\pi}{\lambda}} [/tex]

Now, for the "p-th" derivative of Io, i got

[tex] \frac{d^p}{d \lambda^2} I_o = \frac{\prod_{p=1}^p (1 - 2p)}{2^{p+1}} \sqrt{\pi} \lambda^{-\frac{(2p + 1)}{2}} [/tex]

I don't see how this related to n = 2p (even case) where

I2p = [tex] \int_0^\infty x^{2p} e^{- \lambda x^2} dx [/tex]

And even when I do figure this out, does this all combine into one answer, or is it kind of like a piecewise answer?

Any help with what to do with the even/odd cases would be greatly appreciated

Thanks
 

Answers and Replies

  • #2
26
0
Hello, I will give some hints here, and probably I will rearrange this quick reply in 24 hours.

\int_0^\infty e^{-ax^2}dx = \frac{1}{2} \sqrt{\frac{\pi}{a}}

\int_0^\infty x e^{-ax^2}dx = \frac{1}{2a}

\int_0^\infty x^n e^{-ax^2}dx = \frac{(n-1)!}{2(2a)^{n/2}} \sqrt{\frac{\pi}{a}}

\int_0^\infty x^{2n+1} e^{-ax^2}dx = \frac{n!}{2 a^{n+1}}

\int_0^\infty x^{2n} e^{-ax^2}dx = \frac{(2n-1)!}{2^{2n+1} a^n} \sqrt{\frac{\pi}{a}}
 

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