Solve Gaussian Integral: Even/Odd Cases

[csnsc14320]

EDIT: meant to post this is the math forums, if you can remove this I'm going to switch it over

Homework Statement

Solve:

I_n = $$\int_{0}^{\infty} x^n e^{-\lambda x^2} dx$$

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 I_o.

For the I_o case, I evaluated it and obtained $$I_o = \frac{1}{2} \sqrt{\frac{\pi}{\lambda}}$$

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

$$\frac{d^p}{d \lambda^2} I_o = \frac{\prod_{p=1}^p (1 - 2p)}{2^{p+1}} \sqrt{\pi} \lambda^{-\frac{(2p + 1)}{2}}$$

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

I_2p = $$\int_0^\infty x^{2p} e^{- \lambda x^2} dx$$

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

 

[Tian WJ]

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}}