# 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.

• Yes, it will just be $u(x,t) = \sin(x-\beta t)$ or similar (might depend on exact choice of Fourier transform). Jul 28, 2022 at 21:13
• @md2perpe I am not sure on the exact choice of Fourier transform. Can the solution be something like $u(x, t) = \sin\left(a x-b\beta t\right)$? Jul 28, 2022 at 21:23
• I believe the solution can be $u(x,t)=\sin (x-b \beta t)$, but not $u(x,t)=\sin (a x-b \beta t)$. Jul 29, 2022 at 3:35
• I agree with @StevenClark. The factor $b$ is probably something like $2\pi$ or $1/(2\pi).$ Jul 29, 2022 at 7:54

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 $$\{a,b\}=\{0,-2 \pi\}$$.