- #1

csnsc14320

- 57

- 1

## Homework Statement

Solve:

I

_{n}= [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 I

_{o}.

For the I

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

Now, for the "p-th" derivative of I

_{o}, 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

I

_{2p}= [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