How does one find the Fourier transform of $f(x):=\mathbf 1_{[0,2\pi]}(x)\sin(x)$?
I have tried to use the definition from my text:
\begin{align*} \hat f(\xi) & = \frac{1}{\sqrt{2\pi}} \int_0^{2\pi}\sin(x)e^{-ix\xi} \ dx \\ & = \frac{1}{\sqrt{2\pi}} \int_0^{2\pi}\frac{1}{2i}\left(e^{ix}-e^{-ix}\right)e^{-ix\xi} \ dx \end{align*}
It doesn't seem to lead anywhere. (Anywhere in the sense that, since we are decomposing sine, I was expecting the transform to be something simple). Is this the right way to go?
Wolfram alpha returns:
$$i \sqrt\frac{\pi}{2}\delta(\omega-1)-i \sqrt\frac{\pi}{2}\delta(\omega+1)$$
If this is relevant to the problem: how does Dirac delta function come into play?
$$ \begin{align*} \hat f(\xi) & = \frac{1}{\sqrt{2\pi}} \int_0^{2\pi}\frac{1}{2i}\left(e^{ix}-e^{-ix}\right)e^{-ix\xi} \ dx \\ &=\frac{1}{\sqrt{2\pi}} \frac{1}{2i}\int_0^{2\pi}e^{ix-ix\xi}-e^{-ix-ix\xi} \ dx \\ &=\frac{1}{\sqrt{2\pi}} \frac{1}{2i}\int_0^{2\pi}e^{(i-i\xi)x}-e^{(-i-i\xi)x} \ dx \\ \end{align*} $$
Now simply use
$$ \int_0^{2\pi}e^{\alpha x}= \frac{e^{\alpha x}}{\alpha}|_0^{2 \pi}=\frac{e^{2 \pi \alpha}-1}{\alpha}$$
The dirac function comes from the fact that you asked WA to calculate the FT of $\sin(x)$ over the reals not restricted to $[0, 2 \pi]$.