Integrating $\frac {sgn(x)} {x^3}$ When I try $$\int\limits_{-\infty}^{\infty} \frac {\mathrm{sgn}(x)} {x^3} dx$$ in Wolfram-Alpha, it tells me that it doesn't converge. Here $\mathrm{sgn}(x)$ is the sign function.
But when I try $$ \text{FourierTransform} \left (\frac {\mathrm{sgn}(x)} {x^3} \right) $$, it gives me $$\omega^2 \frac {2 \log|\omega| + 2 \gamma -3} {2\sqrt{2 \pi}}$$, edit: whose value at $\omega = 0$ is $0$ (as a limit)
I was under the impression that area under a curve equals FT evaluated at $\omega = 0$. I think that probably Dirichlet conditions for FT are not satisfied and therefore I cannot expect to see meaningful FT. I am not sure though. Can someone please point out the error in my reasoning and why I am seeing this discrepancy?
 A: I guess this question is not likely to get an answer, given the fact that this question is now buried several pages below. So I am going to answer it myself to the best of my understanding.
Ans: Wolfram-Alpha is not calculating the FT correctly.
Reasoning: To calculate the FT at $\omega = 0$, we can use the convolution property. $$\mathrm{FT} \left(\frac{1}{x^3}\right) = \frac{-i}{2} \sqrt{\frac{\pi}{2}}\omega^2 \mathrm{sgn}(\omega)$$
$$\mathrm{FT}(\mathrm{sgn}(x)) = i \sqrt{\frac{2}{\pi}} \frac{1}{\omega}$$
By convolution Property of FT,
$$\mathrm{FT}\left(\frac{\mathrm{sgn}(x)}{x^3}\right) = \mathrm{FT} \left(\frac{1}{x^3}\right) \star \mathrm{FT}(\mathrm{sgn}(x)) = \frac{1}{2} \int\limits_{\lambda= -\infty}^{\infty}\lambda^2\mathrm{sgn(\lambda)} \frac{1}{\omega - \lambda} d\lambda$$
At $\omega=0$, the integral reduces to $\int\limits_{\lambda=-\infty}^{\infty} -|\lambda| d\lambda$
This doesn't converge as with the original integral $\int\limits_{x=-\infty}^{\infty} \frac{\mathrm{sgn}(x)}{x^3}dx$ in the question (though I am not able to account for the negative sign).
This approach reconciles the seeming discrepancy. Therefore the FT expression given by WA must be wrong.
