# Fourier transform of sin(3pix/L)

1. May 11, 2016

### spacetimedude

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

2. Relevant equations

3. The attempt at a solution
So we want sine in terms of the exponentials when we take the fourier transform $$F(k)=\int_{-\infty}^{\infty}f(x)e^{-ikx}dx$$ where $f(x)=\sin(3\pi x/L)$. Let a=3pi/L. Then $$\sin(ax)=\frac{e^{iax}-e^{-iax}}{2i}$$.
(Is this correct?)
Then we can take the fourier transform:
$$F(k)=\int_{-\infty}^{\infty}\frac{e^{iax}-e^{-iax}}{2i}e^{-ikx}dx$$. Rearranging gives $$\frac{1}{2i}[\delta(K+a)-\delta(K-a)]$$. But my notes says there is $\sqrt{2\pi}$ in front and I'm not sure where it came from?
Any help will be appreciated.

2. May 11, 2016

### Ray Vickson

3. May 12, 2016

### spacetimedude

I don't understand how to apply the dirac delta function here? I just used the integral representation of delta to get to the last line.

4. May 12, 2016

### spacetimedude

Oh I am missing 2pi when I integrate. But where is the square root coming from?

5. May 12, 2016

### vela

Staff Emeritus
You're probably using a different convention for the Fourier transform compared to what was done in your notes (or your notes are wrong). I think you also made a sign error.