How to solve $\int_0^\infty f(x) \sin(wx) \, dx=\frac a {a^2+w^2}$ with using Fourier transform? $\int_0^\infty f(x) \sin(wx) \, dx=\frac{a}{a^2+w^2}$
I thought it is Fourier sine transform of $f(x)$
since  $\ F_s(f(x))=F_s(k)=\sqrt{\frac{2}{\pi}} \int_0^\infty f(x)\sin(wx) \, dx$
so $\sqrt{\frac{\pi}{2}} F_s(w)=\frac{a}{a^2+w^2}$. by taking invers sin transform of this
$$\sqrt{\frac{\pi}{2}}f(x)=\int_0^\infty \frac{a}{a^2+w^2} \sin(wx)\,dx$$
 A: This will be clear to those who have dealt with Laplace transforms. The Laplace transform of $\sin wt$ is given by ${w \over w^2 + s^2}$. In other words
$$\int_0^{\infty} e^{-st} \sin wt \,dt = {w \over w^2 + s^2}$$
Thus in your notation, the answer to your question is ${a \over w} e^{-ax}$.
A: Upon second thought, there seems to be nothing wrong with this question. In fact, if one restricts the domain of the FST to $\omega>0$, the transform has a unique inverse, which should exist as long as $f$ is absolutely integrable. Indeed it can be shown that the inverse defined by
$$F^{-1}_s\left[\int_{0}^{\infty}dxf(x)\sin{\omega x}\right](\omega)=\frac{2a}{\pi}\int_{0}^{\infty}\frac{\sin\omega t}{t^2+a^2}dt$$
converges absolutely and therefore it is equal to $f(\omega)$. This integral can only be done in terms of special functions. One can compute that
$$\int_{0}^{\infty}\frac{e^{-zx}}{x^2+a^2}dx=\int_{\infty}^{z}\frac{\sin a (z-z')}{z'}dz'=\frac{\sin az}{a}\text{Ci}(az)+\frac{\cos az }{a}\left(\frac{\pi}{2}-\text{Si}(az)\right), ~~\text{Re}~z\geq 0$$
where the last equality is given in terms of the cosine-sine integrals. We can set therefore $z=\pm i\omega $ in the expression above to get the final result
$$\frac{\pi}{2a}f(\omega)=\frac{\cosh a\omega}{a}\frac{\text{Si}(ia\omega)-\text{Si}(-ia\omega)}{2i}-\frac{\sinh a\omega}{a}\frac{\text{Ci}(ia\omega)+\text{Ci}(-ia\omega)}{2}$$
which converges for all $\omega>0$.
