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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...

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For the transform to be invertible, the original function must be absolutely integrable, and $x \to \frac{1}{x}$ is not absolutely integrable.

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  • $\begingroup$ 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? $\endgroup$ – Pixel Jan 9 '14 at 16:56
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    $\begingroup$ 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. $\endgroup$ – rewritten Jan 9 '14 at 17:17

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