# Fourier transform of sin(3pix/L)

## 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
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## 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.

BvU