Cosine Convolution Theorem

Question

I have found a number of sources that suggest:

$$\int_0^\infty f(t)g(t)\sin(xt) dt = \frac{\pi}{2}\int_0^\infty \left(\int_0^\infty f(s) \sin(ts) ds\right) \left(\int_0^\infty g(s)\sin(xs)\sin(ts)ds\right)dt$$

For example, here http://dsp-book.narod.ru/TAH/ch03.pdf, formula 3.3.22, where $$2\sin(xs)\sin(ts)=\cos((x-t)s)-\cos((x+t)s)$$

But this doesn't appear to work at all. Let $f(t)=g(t)=e^{-t}$, then according to mathematica: $$LHS = \sqrt{\frac{2}{\pi}}\frac{x}{4+x^2}$$ $$ RHS = \frac{i\pi}{2x(x-2i)}\ \ \ Im(x)<-1$$

How are these the same??? They don't appear similar at all. What am I misunderstanding?

Answer 1

I didn't realize that a Fourier Transforms could have multiple answers. I guess that makes sense when crossing an abscissa, just wasn't paying attention; why would Mathematica default to Im<-1, instead of Reals. It appears I need to include the assumption that x is real. Also it appears I had worked out the wrong coefficient in the front.

Sqrt[2/ Pi]*
 Integrate[
  FourierSinTransform[Exp[-t], t, s]*
   FourierSinTransform[Exp[-t]*Sin[x*t], t, s], {s, 0, Infinity}, 
  Assumptions -> x \[Element] Reals]

and

FourierSinTransform[Exp[-2 t], t, x, 
 Assumptions -> x \[Element] Reals]