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

BucketOfFish

## Homework Statement

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:

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

## Homework Equations

Integration by parts:

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

## 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?

Homework Helper
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?

Homework Helper
Gold Member
Nice hint.

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?

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