Finding $f(x)$ in a Fourier Integral

Question

I'm having trouble trying to find that function $f(x)$ or trying to find that it doesn't exist.

$$\int_0^\infty f(x)\sin(\alpha x)dx= \left\{ \begin{array}{lr} \alpha & \alpha \in (0, 2\pi) \\ e^{-(\alpha - 2\pi)} & \alpha \in (2\pi, \infty) \end{array} \right.$$

I know that I am supposed to analyse that comparing to a Fourier Integral but I am stuck trying to analyse $\alpha $ and its values when <0,=0,>0.. Still, I don't have much idea of what to do in this question..

edit: we can assume that the function is odd

Sorry for the latex and thank you in advance!

Answer 1

The given equation is the Sine Transform of $f(x)$.

$$F_s(\omega)=2\int_{0}^{\infty}f(x)\sin(2\pi\omega x) dx$$

So with change of variable $\alpha=2\pi\omega,$ $$F_s(\omega)=2\int_0^\infty f(x)\sin(2\pi\omega x)dx= \left\{ \begin{array}{lr} 4\pi\omega & \omega \in (0, 1) \\ 2e^{-2\pi(\omega - 1)} & \omega \in (1, \infty) \end{array} \right.$$

The inverse transform is

$$f(x)=\int_{0}^{\infty}F_s(\omega)\sin(2\pi\omega x) d\omega$$

Thus

$$\begin{align} f(x)&=\int_{0}^{1}4\pi\omega\sin(2\pi\omega x) d\omega+\int_{1}^{\infty}2e^{-2\pi(\omega - 1)}\sin(2\pi\omega x) d\omega\\ &=\frac{\sin(2\pi x)-2\pi x\cos(2\pi x)}{\pi x^2}+\frac{x\cos(2\pi x)+\sin(2\pi x)}{\pi(1+x^2)} \end{align}$$