In my copy of Table of Integrals, Series, and Products (Gradshteyn & Ryzhik) on p.1121, it says that the Fourier sine transform is defined $$F_s(\xi) = \sqrt{\frac{2}{\pi}}\int_0^\infty f(x)\sin(\xi x)\text{d}x.$$ It goes on to say that "... knowledge of ... $F_s(\xi)$ ... enables $f(x)$ to be recovered," and gives the following inversion integral $$f(x)=\sqrt{\frac{2}{\pi}}\int_0^\infty F_s(\xi)\sin(\xi x)\text{d}\xi.$$ On the next page it gives some examples, so I chose $f(x)=1/x$. Mathematica gives me $$F_s(\xi) = \sqrt{\frac{2}{\pi}}\int_0^\infty \frac{\sin(\xi x)}{x}\text{d}x = \sqrt{\frac{\pi}{2}}.$$ This is great as it matches what the book says. Next I tried to reverse this using the inversion formula to convince myself that it works. I attempted the following, $$f(x) = \sqrt{\frac{2}{\pi}}\int_0^\infty\sqrt{\frac{\pi}{2}}\sin(\xi x)\text{d}\xi=\int_0^\infty\sin(\xi x)\text{d}\xi,$$ which clearly does not converge (which Mathematica confirms)
My question is simply: what is it that I can't see? One thought is that the table of transforms is just a formal table, but if that's the case, then what is the point of the specific example $f(x)=1/x$ ? Or maybe these transforms are simply not always invertible, but they are listed because the "forwards" transform could turn out to be useful...
