# Fourier and Mellin transforms of Hilbert Transform

I am reading Hilbert transform recently and meet two questions. The book I am reading is Debnath and Bhatta "Integral Transforms and Their Applications".

If we define the Hilbert transform on the real line is, for $$x \in \mathbb{R}$$, $$H_x\{f(t)\}=\frac{1}{\pi}PV\int_{-\infty}^\infty \frac{f(t)}{t-x}dt=\lim_{\epsilon \rightarrow 0}\left(\int_{-\infty}^{t-\epsilon} +\int_{t+\epsilon}^\infty\right)\frac{f(t)}{t-x}dt,$$ where PV stands for the Cauchy principal value. Then the Fourier transform of the Hilbert transform, which can be considered as a convolution, is $$F_k\{H_x\{f(t)\}\}=i sgn(k) F_k\{f(t)\},$$ where $$i^2=-1$$, $$sgn(x)=\begin{cases} 1, x>0,\\ -1,x<0 \end{cases}$$ and $$F_k\{f(t)\}=\frac{1}{\sqrt{2 \pi}}\int_0^\infty e^{-ikt}f(t)dt$$, since $$F_k\{\sqrt{\frac{2}{\pi}}\left(-\frac{1}{x}\right)\}=i sgn(k)$$.

The first question, Eq.(9.4.3), if we let a complex variable $$z=x+i y$$, then why $$F_k\{H_z\{f(t)\}\}=2 i e^{-ky} H(k) F_k\{f(t)\},$$ where $$H(k)=\frac{1}{2}(1+sgn(k))$$. Which variable are we transforming here, $$x\rightarrow k$$ or $$z\rightarrow k$$?

The second question appears in the Mellin transform of the Hilbert transform. There is an integral I do not know how to solve it: $$PV\int_0^\infty\frac{x^{p-1}}{t-x}dx=\pi Cot(\pi p).$$ I understand we need to separate the integral or use the residue theorem to get rid of the singularity. But I cannot go further.

• For $Im(z) > 0$ let $h(z) = \frac{1}{\pi} \int_{-\infty}^\infty \frac{f(t)}{t-z}dt$. If $f \in L^1$ then for every $y$, $h(.+iy) \in L^1$ and $h$ is complex analytic. Then $H(y,\xi)=\int_{-\infty}^\infty h(x+iy) e^{-2i \pi \xi (x+iy)} dx$ doesn't depend on $y > 0$. Proof : $H(y,\xi)$ is complex analytic in $y$ and it doesn't depend on $Im(y)$, thus it is constant and it doesn't depend on $y$. Do the same with $h$ defined for $Im(z) < 0$ and conclude with $pv. \frac{1}{\pi} \int_{-\infty}^\infty \frac{f(t)}{t-x}dt = \frac12 \lim_{y \to 0} h(x+iy)+h(x-iy)$. – reuns Dec 4 '18 at 2:38
• That's what you asked, the Fourier transform of $h(z+.)$ – reuns Dec 4 '18 at 2:49