# Fourier transform

1. Jun 12, 2012

### robertjford80

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

3. The attempt at a solution

I don't understand this step. It's got to be some sort of identity that I missed. I also don't understand why the limits of integration change.

2. Jun 12, 2012

### clamtrox

Here's an easy way of seeing it:

Remember that the integral over an even interval of an odd function is zero
$$\int_{-L}^L f(x) dx = 0$$
if $f(-x) = -f(x)$.

You can see fairly easily that $\frac{\sin(\alpha)}{\alpha}$ is an even function and $\sin(\alpha x)$ is an odd function; therefore $\frac{\sin(\alpha) \sin(\alpha x)}{\alpha}$ is odd and it's integral vanishes over an even support interval.

3. Jun 12, 2012

### robertjford80

ok, I understand what you mean, although it took me about 30 minutes to get it. I still understand why the limits of integration change. I also don't understand why 1/pi changes to 2/pi though I think it has something to with the change in the limits of integration.

4. Jun 12, 2012

### clamtrox

For an even function f(-x) = f(x), you can show that $\int_{-L}^L f(x) dx = 2 \int_0^L f(x) dx$

5. Jun 12, 2012

cool