# Gaussian Integral

## Homework Statement

Solve:

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

## 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 $$I_o = \frac{1}{2} \sqrt{\frac{\pi}{\lambda}}$$

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

I2p = $$\int_0^\infty x^{2p} e^{- \lambda x^2} dx$$

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

## Answers and Replies

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Is this what you are asking?

$$\frac{\partial}{\partial \lambda} I_0 = \frac{\partial}{\partial \lambda} \int_{0}^{\infty} e^{-\lambda x^2} dx = \int_{0}^{\infty} -x^2 e^{-\lambda x^2} dx = -I_2$$

You can apply this recursively p times to get $$(-1)^p I_{2p}$$

Is this what you are asking?

$$\frac{\partial}{\partial \lambda} I_0 = \frac{\partial}{\partial \lambda} \int_{0}^{\infty} e^{-\lambda x^2} dx = \int_{0}^{\infty} -x^2 e^{-\lambda x^2} dx = -I_2$$

You can apply this recursively p times to get $$(-1)^p I_{2p}$$
Oh yeah I see the pattern if you take the derivative of Io in integral form instead of what it actually is.

I also did it for the odds and got nice cases for both :D

thanks.