# Splitting up exponential terms when integrating

1. Oct 9, 2011

### leviathanX777

1. Relevant problem

integrate from 0 to infinity of r^2exp^(-r/a0)dr

2. Relevant equations

I'm also given; integral from 0 to infinity of x^nexp^-x dx = n!

3. The attempt at a solution

I'm just wondering if I can split up the exponential to make it look like this form. Eg;

integrate from 0 to infinity of (r^2e^(-r/a0)dr becomes; integrate from 0 to infinity of (r^2e^(-r)dr times integrate from 0 to infinity of (e^(1/a0)dr however I'm pretty sure when I split up the integral, the second term isn't correct. Can anyone help? I just don't want to integration by parts alot of times. As there's two other terms with higher powers of r to go through.

2. Oct 9, 2011

### Staff: Mentor

It looks like you are thinking that e-r/a = e-r * e1/a, which is not true. Review the properties of exponents. This wikipedia page has a summary.

The fastest approach to your integral, I believe, is by integration by parts. One application should get you to a form similar to the one you show in your relevant equations.

3. Oct 9, 2011

### leviathanX777

I did it already using integration by parts. First time I did it didn't yield something similar to the hint I was given. Had to do integration by parts twice and the exponential was still divided by ao all the way through.