Fourier transform of sin(3pix/L)

[spacetimedude]

Homework Statement

Homework Equations

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.

 

[Ray Vickson]

Homework Statement

Homework Equations

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.

Google is your friend; look up "Dirac delta function".

 

[spacetimedude]

Google is your friend; look up "Dirac delta function".

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.

 

[spacetimedude]

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

 

[vela]

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.