# Misunderstanding with Fourier sine transform...

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...

For the transform to be invertible, the original function must be absolutely integrable, and $x \to \frac{1}{x}$ is not absolutely integrable.
• To confirm, does this mean we cannot invert $F_s(\xi)$ when $$\int_0^\infty \left|f(x)\sin(\xi x)\right|\text{d}x$$ diverges? Commented Jan 9, 2014 at 16:56
• You can't assume that the Fourier inversion theorem holds, unless you have some conditions on $f$. One sufficient condition is the absolute integrability and continuity at the integration point ($x$), but other conditions are sufficient too, like being Schwartz (that's stronger than the integrability) or others. See the Wikipedia article for more info. Commented Jan 9, 2014 at 17:17