Solve Integral: x^2[e^(-x^2)] from -∞ to ∞

[newton1]

how to do this integration??
\int x^2 [e^{-x^2}] dx
from - \infty \rightarrow \infty

 

[graphic7]

Break it into two integrals:

\int_{-\infty}^0 x^2 e^{-x^2}dx
\int_{0}^{\infty} x^2 e^{-x^2}dx

You must do this because you will not be able to do two limits in your definite integration.

From there, it's just simple integration-by-parts, taking the limit of what you get by doing the integrations, and adding them.

You are familiar with improper integrals?

Edit: Before you begin implementing those integration skills of your's, make sure the integral is convergent, otherwise, you'll just be wasting time.

 

[HallsofIvy]

Cute! My first thought was that integration by parts wouldn't work because you can't integrate e^{-x^2}[/tex] but if you let u= x, dv= xe^{-x^2} ,it works nicely!

 

[Zurtex]

However you tackle this you'll soon realize the answer involves the error function.

However even so it's still not that difficult to work out what you need.

 

[Galileo]

I learned a nice trick by using:
I(a)=\int \limits_{-\infty}^{+\infty}\exp(-ax^2)dx=\sqrt{\frac{\pi}{a}}
Differentiate both sides with respect to a:
\frac{dI}{da}=-\int \limits_{-\infty}^{+\infty}x^2\exp(-ax^2)dx=-\frac{1}{2}\sqrt{\frac{\pi}{a^3}}
So
\int \limits_{-\infty}^{+\infty}x^2\exp(-ax^2)dx=\frac{1}{2a}\sqrt{\frac{\pi}{a}}
Might be faster than partial integration. It's definitely more fun.

 

[Hurkyl]

You have to prove something about the integral before you can do that, though... I can't remember what off hand.

 

[Galileo]

You have to prove something about the integral before you can do that, though... I can't remember what off hand.

EDIT: The integrand should be differentiable en it's derivative continuous.

It's legal.

 

[Hurkyl]

That's it? I thought I recalled that there had to be some sort of uniform convergence or something going on. (Do those conditions prove the uniform convergence necessary?)

 

[Galileo]

Yeah, I guess that's true.
There are probably two theorems you need. One for bringing the derivative under the integral (that's Leibniz rule), which I know is true if the integration limits do not extend to infinity. For this the integrand must be differentiable and it's derivative continuous. (not the value I(a), my mistake).

The exponential function has a power series expansion and is uniformly convergent and eeh..

Aaack, whatever. The exponent is a very nice function, so all the tricks probably work.
I learned this trick in PHYSICS class okay? They don't care about that.

 

[Daniel90]

You can prove that integral(-inf-->+inf) exp(-x^2) dx = sqrt(pi) by converting (squaring it) to a double integral or you can use the residue theorem after converting it to a complex integral. The first way is easier.