Integration Problem: Solving \oint e-ax^2dx = 1/4\sqrt{}\pi/a^3

[henryc09]

Homework Statement

Given that:

\oint e^{-ax^2}dx = \sqrt{}\pi/a (between infinity and minus infinity)

show that

\ointx²e^{-ax^2}dx = 1/4\sqrt{}\pi/a^3 (between 0 and infinity)

Homework Equations

The Attempt at a Solution

I assume integration by parts may be necessary, but not really sure how to go about it, ie. what is first expression really telling you as the limits are different to the second. I think it might also be the infinities confusing me so if someone could point me in the right direction that'd be appreciated.

 

[CompuChip]

I assume that you meant
\int_{-\infty}^\infty
(click to see the LaTeX code) rather than \oint.

As for the problem, try differentiating with respect to a ;)

 

[henryc09]

Hmmm ok so I have it starting to look right but I really can't quite see how they all relate to each other. If you differentiate \sqrt{\pi/a} with respect to a you get -1/2\sqrt{\pi/a^3} and if you differentiate e^{-ax^2} with respect to a you get -x²e^{-ax^2}

which is looking good however maybe I'm just being stupid but I can't see what to do from there.

 

[henryc09]

ok actually I think I have it now, thanks for the help!

 

[CompuChip]

Yep, so the trick is to take the first (standard) integral, derive it, and then pull the derivative inside the integral.

In fact, you can use that to derive a formula for
\int x^{2n} e^{-a x^2} \, dx
for any n = 0, 1, 2, ...