# A integral about expotential function

1. Jun 26, 2011

### athrun200

1. The problem statement, all variables and given/known data

2. Relevant equations
Unknown

3. The attempt at a solution

How to calculate this integral?
I have tried substitution and by parts. But they fail to get the answer.

#### Attached Files:

File size:
8.2 KB
Views:
113
• ###### 2.jpg
File size:
11.5 KB
Views:
106
2. Jun 26, 2011

### hunt_mat

There is a trick for this, square the integral:
$$\left(\int_{-\infty}^{\infty}e^{-x^{2}}dx\right)^{2}=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^{2}+y^{2})}dxdy$$
Now the idea is to make a change of co-ordinates from cartesian to polar...

3. Jun 26, 2011

### dextercioby

Are you taking quantum mechanics before knowing what a gaussian integral is ? Because you shouldn't...

4. Jun 26, 2011

### athrun200

Oh you are right.
In fact, I am self-learning quantum mechanics, and I don't know what maths skills I need to have.

5. Jun 27, 2011

### hunt_mat

So move on from my explanation:
$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^{2}+y^{2})}dxdy=\int_{0}^{2\pi}\int_{0}^{\infty}re^{-r^{2}}drd\theta =2\pi\int_{0}^{\infty}re^{-r^{2}}dr$$
I think I will leave the last bit to you as it's standard integration.