Principal value of Fourier Integral I have tried to find the principal value of
$$\int_{-\infty}^\infty {\sin(2x)\over x^3}\,dx.$$
As $ {\sin(2x)\over x^3}$ is an even function, its integral may not be zero in the given limits. I cannot calculate its principal value as I met with a third order pole at $z=0$. The expected value is $-2\pi$. How can I compute the principal value?  

Unfortunately $$PV\int_{-\infty}^\infty {\sin 2x\over x^3}\,dx$$ diverges.
The integrals $\int_\frac{\pi}{4}^{\infty} {\sin 2x\over x^3}\,dx=\int_{-\infty}^{-\frac{\pi}{4}}{\sin 2x\over x^3}\,dx$ exist.
Since $$\sin 2x\ge \frac{4}{\pi}x$$ for $0<x<\frac{\pi}{4}$, we have $$\frac{\sin 2x}{x^3}\ge \frac{4}{\pi}\frac{1}{x^2} $$ for $0<x<\frac{\pi}{4}$. Thus $$\int_{-\frac{\pi}{4}}^{-\epsilon} \frac{\sin 2x}{x^3} dx=\int_\epsilon^{\frac{\pi}{4}} \frac{\sin 2x}{x^3} dx\ge \int_\epsilon^{\frac{\pi}{4}} \frac{4}{\pi}\frac{1}{x^2} dx=\frac{4}{\pi}\left(\frac{1}{\epsilon}-\frac{4}{\pi}\right).$$ 
Therefore $$PV\int_{-\infty}^{\infty} \frac{\sin 2x}{x^3} dx =\lim_{ \epsilon\to 0} \left( \int_{-\infty}^{-\frac{\pi}{4}}
 +\int_{-\frac{\pi}{4}}^{-\epsilon}  +\int_\epsilon^{\frac{\pi}{4}}+\int_\frac{\pi}{4}^{\infty}\right) {\sin 2x\over x^3}\,dx=+\infty.$$
