Compute the Fourier transform of $f(t) = \sin t$?

Question

Compute the Fourier transform of $f(t) = \sin t$. Does that converge?

Answer 1

If you consider $f:=\sin$ as a function on $\mathbb R$ then it has a Fourier transform only in the distribution sense, as remarked in H.L.'s comment. If, however, you consider $\sin$ as a function on $\mathbb R/(2\pi)$ then its Fourier transform $\hat f$ is a function on $\mathbb Z$, namely the doubly infinite sequence $(c_k)_{k\in\mathbb Z}$ obtained by integration of $\sin$ against the functions $t\mapsto e^{-ikt}$: $$c_k={1\over 2\pi}\int_{-\pi}^\pi \sin(t)e^{-ikt}dt\qquad (k\in\mathbb Z).\qquad(*)$$ As $\sin$ is a nice function Fourier theory guarantees that we can get $\sin (t)$ back using the $c_k$ as follows: $$\sin (t)=\sum_{k=-\infty}^\infty c_k e^{ikt}.$$ Now, by Euler's formula, $\sin(t) =(e^{it}-e^{-it})/(2i)$, so we can immediately read off $\hat f(1)=c_1=-i/2$, $\hat f(-1)=c_{-1}= i/2$ and $\hat f(k)=c_k=0$ otherwise, without doing the integration (*).

Answer 2

$\sin(t)=\frac{e^{it}-e^{-it}}{2i}$. Since, $e^{it}$ and $e^{-it}$ are basis functions of the Fourier transform. From this easily can be computed using duality property and there will be no convergence issue.