Principal value of Fourier Integral

Question

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?

Answer 1

For this you want to use the function

$$f(z)={e^{2iz}\over z^3}$$

then use the fact that $\sin(2z)=\Im( e^{2iz})$ on the real axis. The exponential has the right decay for a semicircular contour with a bump around the origin, that's why we choose it.

Now, since we only go halfway around, we get only half the residue

$${1\over 2}\operatorname{Res}\left({e^{2iz}\over z^3},0\right)={1\over 2\pi i}\int_C {e^{2iz}\over z^3}\,dz$$

and taking imaginary parts of this we get

$$PV\left(\int_{-\infty}^\infty {\sin(2x)\over x^3}\,dx\right)$$

The last equality with our original integral comes from the fact that cosine part goes to $0$ since ${\cos(2z)\over z^3}$ is an odd function, so the PV is $0$.

The contour is made up of pieces in the form

$$C=[-R,-\epsilon]\cup C_\epsilon\cup [\epsilon, R]\cup C_R$$

where the $C_r$ are semicircles of radius $r$ centered at the origin and in the upper half plane, the $\epsilon$ circle is clockwise to agree with the direction of the contour traveling from $-R\to\epsilon$ and then the next segment $\epsilon\to R$, the $R$ circle is counter-clockwise aimed to connect the two segments who's ends are distance $R$ from the origin. The circular integrals go to $0$ by the standard estimates, so all we will be left with in the limit will be

$$PV\left(\int_{-\infty}^\infty {\sin(2x)\over x^3}\,dx\right).$$

Computing the residue we see this is ${1\over 2!}{d^2\over dz^2}\bigg|_{z=0}\left(e^{2iz}\right)=-2$, hence we get that the value of the integral is

$$-{2\pi i\over 2}\cdot 2$$

and since the sine portion is the imaginary part, we see that our original integral is just $-2\pi$, as desired.

Answer 2

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 $$PV\int_{-\infty}^{\infty} \frac{\sin 2x}{x^3} dx =\lim_{ \epsilon\to 0} \left( \int_{-\infty}^{-\frac{\pi}{4}} +\int_{-\frac{\pi}{4}}^{-\epsilon} +\int_\epsilon^{\frac{\pi}{4}}+\int_\frac{\pi}{4}^{\infty}\right) {\sin 2x\over x^3}\,dx=+\infty.$$