Fourier Transform of $|t|^{-\frac{4}{3}}$

Question

I need to compute

$I(\omega)=\int_{-\infty}^{+\infty}dte^{-i\omega t}|t|^{-\frac{4}{3}}\tag{1}$

for $\omega\geq0$. It is clear that the integral diverges around $t=0$, yet I was hoping to find out if it is possible to understand it in terms of some distribution (similarly to the case $\delta(\omega)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}dte^{-i\omega t}$). I found some results for the Fourier transform of $|t|^\alpha$ when $\alpha\geq-1$, but this is not the case here.

I have also a question related. I tried to compute $I(\omega)$ by doing: $I(\omega)=\int_{-\infty}^{+\infty}dte^{-i\omega t}|t|^{-\frac{4}{3}}=\int_{0}^{\infty}dte^{i\omega t}|-t|^{-\frac{4}{3}}+\int_{0}^{\infty}dte^{-i\omega t}|t|^{-\frac{4}{3}}=2\int_{0}^{\infty}dt\cos(\omega t)t^{-\frac{4}{3}}\tag{2}$

Then using Mathematica I computed

$\frac{dI(\omega)}{d\omega}=-2\int_{0}^{\infty}dt\sin(\omega t)t^{-\frac{1}{3}}=-2\frac{\pi\,\textrm{sign}(\omega)}{\Gamma(\frac{1}{3})|\omega|^{2/3}}\tag{3}$

Then for $\omega\geq0$ one has:

$\frac{dI(\omega)}{d\omega}=-2\frac{\pi}{\Gamma(\frac{1}{3})\omega^{2/3}}\tag{4}$

which implies

$I(\omega)=-\frac{6\pi}{\Gamma(\frac{1}{3})}\omega^{1/3}\tag{5}$.

Last equation does not seems to make sense, since implies $I(\omega=0)=0$, while for $\omega=0$ the integral $I(\omega)$ should diverge. What is the mistake in the steps above?

Thanks in advance for the help!

Answer 1

Let us define the distribution $\psi(t)=\left(|t|^{-4/3}\right)$ as

$$\langle \psi,\phi\rangle=\int_{-\infty}^\infty \frac{\phi(t)-\phi(0)}{|t|^{4/3}}\,dt$$

for $\phi\in \mathbb{S}$. Then, we see that

$$\begin{align} \langle \mathscr{F}\{\psi\},\phi\rangle &=\langle \psi,\mathscr{F}\{\phi\}\rangle\\\\ &=\int_{-\infty}^\infty \frac1{|t|^{4/3}}\int_{-\infty}^\infty \phi(\omega)(e^{i\omega t}-1)\,d\omega\,dt\\\\ &=2\int_{-\infty}^\infty \phi(\omega)\int_0^\infty \frac{\cos(\omega t)-1}{t^{4/3}}\,dt\,d\omega\\\\ &=2\int_{-\infty}^\infty \phi(\omega) |\omega|^{1/3}\underbrace{\int_0^\infty \frac{\cos(t)-1}{t^{4/3}}\,dt}_{=\sin(2\pi/3)\Gamma(-1/3)}\,d\omega \end{align}$$

Therefore, we find that the Fourier transform of the distribution $\psi$ is

$$\mathscr{F}\{\psi\}(\omega)=2\sin(2\pi/3)\Gamma(-1/3)|\omega|^{1/3}$$

where $f(\omega)=|\omega|^{1/3}$ is locally integrable and has inverse Fourier transform.

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Here, we show that the inverse Fourier Transform of $|\omega|^{1/3}$ is given by the distribution $\frac{1}{2\sin(2\pi/3)\Gamma(-1/3)}\psi(t)=\frac{1}{2\sin(2\pi/3)}\left(|t|^{-4/3}\right)$. To that end, we now proceed.

Let $f(\omega)=|\omega|^{1/3}$. Then, for any $\phi\in \mathbb{S}$ we have

$$\begin{align} \langle \mathscr{F}^{-1}\{f\},\phi \rangle &=\langle f,\mathscr{F}^{-1}\{\phi\} \rangle\\\\ &=\frac1\pi\text{Re}\int_0^\infty \omega^{1/3} \int_{-\infty}^\infty \phi(t) e^{-i\omega t}\,dt\,d\omega\tag1 \end{align}$$

Integrating by parts the inner integral in $(1)$ reveals

$$$$\begin{align} \langle \mathscr{F}\{f\},\phi \rangle &=\frac1\pi\text{Re}\left(-i \int_0^\infty \omega^{-2/3}\int_{-\infty}^\infty \phi'(t) e^{-i\omega t}\,dt\,d\omega\right)\\\\ &=\frac1\pi\text{Im}\left(\int_{-\infty}^\infty \phi'(t) \underbrace{\int_0^\infty \omega^{-2/3}e^{-i\omega t}\,d\omega}_{=e^{-i\text{sgn}(t)\pi/6}\frac{\Gamma(1/3)}{|t|^{1/3}}}\,dt\right)\\\\ &=-\frac{\Gamma(1/3)}{2\pi}\int_{-\infty}^\infty \frac{\phi'(t)}{|t|^{1/3}}\text{sgn}(t)\,dt\\\\ &=-\frac{\Gamma(4/3)}{2\pi}\int_{-\infty}^\infty \frac{\phi(t)-\phi(0)}{|t|^{4/3}}\,dt\\\\ &=\frac{1}{2\sin(2\pi/3)\Gamma(-1/3)}\int_{-\infty}^\infty \frac{\phi(t)-\phi(0)}{|t|^{4/3}}\,dt \end{align}$$ $$

as expected!