# Integration Trickiness

1. May 6, 2007

### bartieshaw

Hi,

This integral seems to be coming up a fair bit in questions involving normalisation of wave functions

$$\int x^2\exp{(-ax^2)}dx$$

and my tutor and lecturer both say to just use the fact that

$$x^2\exp{(-ax^2)}=-\frac{d}{da}\exp{(-ax^2)}$$

my question simply is how do we use this. it may be obvious can you say

$$\int x^2\exp{(-ax^2)}dx=\int-\frac{d}{da}\exp{(-ax^2)}dx=\int-\frac{dx}{da}\exp{(-ax^2)}d=-\frac{dx}{da}\exp{(-ax^2)}$$

...?

cheers for any help,

Bart

Last edited: May 6, 2007
2. May 6, 2007

### quasar987

It's like this:

$$\int_{-\infty}^{+\infty}x^2\exp{(-ax^2)}dx=\int_{-\infty}^{+\infty}-\frac{d}{da}\exp{(-ax^2)}dx=-\frac{d}{da}\int_{-\infty}^{+\infty}\exp{(-ax^2)}dx$$

And he assumes that you know what $\int_{-\infty}^{+\infty}\exp{(-ax^2)}dx$ equals to.

3. May 6, 2007

### bartieshaw

well i do know that

$$\int_{-\infty}^{+\infty}\exp{(-ax^2)}dx=\sqrt{\frac{\pi}{a}}$$

so then my next question is how do we work that out (rather than me just knowing it). i started off with integration by parts and then kept on going in circles gettin back to an integral similar to

$$\int_{-\infty}^{+\infty}x^2\exp{(-ax^2)}dx$$

this isnt an essential question now to what im doing, just out of interest really....

cheers

bart

4. May 6, 2007

5. May 6, 2007

### quasar987

Talking about the identity $\int_{-\infty}^{+\infty}\exp{(-ax^2)}dx=\sqrt{\frac{\pi}{a}}$, Lord Kelvin once said "A mathematician is one to whom that is as obvious as two twice two makes four to you. Liouville was a mathematician." (quote taken from Spivak's calc. on manifolds)

Welcome to the club of the non-mathematicians.

6. May 6, 2007

### bartieshaw

haha....

looks like i was maybe taking a little too simple approach trying to use integratoin by parts.

cheers for the help

bart

7. May 7, 2007

### HallsofIvy

Staff Emeritus
Lord Kelvin may have said "twice two is four" but he surely didn't say "two twice two"!