How to inverse Fourier transform the Fourier transform of sin(x)? In a paper, I found $u(x, t) = \mathcal{F}^{-1}(\mathcal{F}(\sin(x))e^{-i\beta k t})$ where $\mathcal{F}$ is the Fourier transform and $k$ denotes frequency in the Fourier domain. Is it possible to solve for $u(x, t)$?
I am not familiar with the Fourier transform.
 A: The Fourier and inverse Fourier transforms can generally be defined as
$$F(\omega)=\mathcal{F}_x[f(x)](\omega)=\sqrt{\frac{| b|}{(2 \pi)^{1-a}}} \int\limits_{-\infty}^\infty f(t)\ e^{i b \omega t}\,dt\tag{1}$$
$$f(t)=\mathcal{F}_{\omega}^{-1}[F(\omega)](x)=\sqrt{\frac{| b|}{(2 \pi)^{a+1}}} \int\limits_{-\infty}^\infty F(\omega)\ e^{-i b \omega t}\,d\omega\tag{2}$$
where the Fourier parameters $\{a,b\}$ are arbitrary constants (see formulas (15) and (16) here).

Assuming $f(x)=\sin(x)$ and setting $\{a,b\}=\left\{-1,-\frac{1}{b}\right\}$ leads to the following evaluations
$$F(k)=\mathcal{F}_x[\sin(x)](k)=\frac{i\ \delta\left(\frac{k}{b}+1\right)}{2 \sqrt{|b|}}-\frac{i\ \delta\left(\frac{k}{b}-1\right)}{2 \sqrt{|b|}}\tag{3}$$
$$u(x,t)=\mathcal{F}_k^{-1}\left[F(k) e^{-i \beta k t}\right](x)=\sin(x-b \beta  t)\tag{4}$$
where $\delta(y)$ is the Dirac delta function.

These results are related to the modulation/frequency shifting property of the Fourier transform, but the linked Wikipedia article is based on the Fourier parameters $\{a,b\}=\{0,-2 \pi\}$.
