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?

 

[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?

 

[LCKurtz]

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?

 

[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?

 

[BucketOfFish]

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

 

[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.