# Homework Help: How do I evaluate <x^2> for a Gaussian function?

1. Sep 9, 2012

### BucketOfFish

1. The problem statement, all variables and given/known data

For some Gaussian distribution, let's say $e^{-x^2}$ times a constant, I want to find the expectation value of $x^2$. In other words, I want to evaluate:

$$\int^{\infty}_{-\infty} e^{-x^2}x^2dx$$

2. Relevant equations

Integration by parts:

$$\int udv = uv - \int vdu$$

3. The attempt at a solution

I tried a few things actually, but none of them got anywhere. One of the routes I attempted was u-substitution followed by integration by parts. First, since the function is even, you set it equal to:

$$2\int^{\infty}_{0} e^{-x^2}x^2dx$$

Letting $a = x^2, da = 2xdx$, we get:

$$\int^{\infty}_{0} e^{-a}\sqrt{a}da$$

Then, letting $u = \sqrt{a}, dv = e^{-a}da, du = \frac{1}{2\sqrt{a}}da, v = -e^{-a}$:

$$\int e^{-a}\sqrt{a}da = -\sqrt{a}e^{-a} + \int e^{-a}\frac{1}{2\sqrt{a}}da$$

And here I get stuck. Anybody have hints?

2. Sep 9, 2012

### Dick

Think about $\int^{\infty}_{-\infty} e^{- a x^2} dx$. You can work that out, right? Now what is d/da of that when a=1?

3. Sep 9, 2012

### LCKurtz

Nice hint.

4. Sep 10, 2012

### BucketOfFish

Ok, so I think I've got it. Basically, you take:

$$\int^{\infty}_{-\infty} -\frac{de^{-ax^2}}{da}dx$$

Then since x and a are independent of each other, you can bring the d/da outside the integral:

$$\frac{d}{da}\int^{-\infty}_{\infty} e^{-ax^2}dx$$

Then you do that integral to get $-\sqrt{\pi/a}$, and after taking the derivative with respect to a, you get $\frac{\sqrt{\pi}a^{-3/2}}{2}$. Plugging in a=1, you end up with $\frac{\sqrt{\pi}}{2}$ for your final answer.

I didn't offend any mathematicians with egregiously bad calculus, did I?

Last edited: Sep 10, 2012
5. Sep 10, 2012

### Dick

Maybe. But even physicists would be offended because the sign is obviously wrong. x^2*exp(-x^2) is a positive function. How can its integral be negative?

6. Sep 10, 2012

### BucketOfFish

Yep, I realized my terrible error and fixed it. Does it look alright now?

7. Sep 10, 2012

### Dick

The answer is right. I'm not sure you've put the sign correction in the right place everywhere. But I'm sure you can fix it.