Hilbert transform in soliton paper

Question

I asked this question over at the Mathematics SE, see here, but have not gotten any responses, so I figured I might as well try here as well. While the question is mathematical, it does appear in a physics context, and I think there is a high likelihood that users here will have helpful insights.

Equation (5) in this paper by H. H. Chen, Y. C. Lee, and N. R. Pereira says that

$$H\left(\frac{1}{x - a}\right) = \frac{i}{x - a},$$

where $a$ is a complex constant with $\mathfrak{Im}(a) < 0$. $H$ is the Hilbert transform,

$$Hf(x) =\frac{1}{\pi}\text{p.v.} \int_{-\infty}^\infty \frac{f(z)}{z - x} dz,$$

where $\text{p.v.}$ denotes the Cauchy principal value.

I am having trouble computing $H\frac{1}{x-a}$ to verify the above. Apparently it should be "easy to see", so I'm probably missing some computational tools. It seems like complex (contour) integration should not be necessary, since the integrand is a function of a real variable, but maybe that would still make it easier? I have unfortunately not had the opportunity to study complex analysis yet, so I'm not sure. This is also my first time encountering the Hilbert transform.

I welcome both answers with explicit computation (preferably elementary) and answers that point me towards the necessary tools/concepts.

Answer 1

The first step is to split up the integrand using partial fractions, $$\int dz\left(\frac{1}{z-a}\frac{1}{z-x}\right)=\int dz\left(\frac{1}{x-a}\frac{1}{z-x}-\frac{1}{x-a}\frac{1}{z-a}\right).$$ Then the $1/(x-a)$ can be pulled out of the $z$-integration, $$\int dz\left(\frac{1}{z-a}\frac{1}{z-x}\right)=\frac{1}{x-a}\int dz\left(\frac{1}{z-x}-\frac{1}{z-a}\right).$$ Taking the principal value, the first term in the remaining integrand gives zero, since it has just a simple pole on the real line contour of integration, $$\mathcal{P}\int_{-\infty}^{+\infty}\frac{dz}{z-x}=\lim_{\epsilon\rightarrow0}\left( \int_{-\infty}^{x-\epsilon}\frac{dz}{z-x}+\int_{x+\epsilon}^{+\infty}\frac{dz}{z-x}\right)\\=\lim_{\epsilon\rightarrow0}\left[\lim_{t\rightarrow+\infty}\left( \int_{-t+x}^{x-\epsilon}\frac{dz}{z-x}+\int_{x+\epsilon}^{t+x}\frac{dz}{z-x}\right)\right]=0.$$ (This is really just a shifted version of $\mathcal{P}\int dz/z=0$, the basic identity for principal value integrals.) With the first term in the integrand taken care of, that just leaves the second term, which has no singularity on the real axis, so we can evaluate it directly as an improper integral (no principal value required), $$-\int_{-\infty}^{+\infty}\frac{dz}{z-a}=-\lim_{t\rightarrow+\infty}\int_{-t}^{t}\frac{dz}{z-a}=\lim_{t\rightarrow+\infty}\left[-\ln\left(\frac{t-a}{-t-a}\right)\right].$$ As $t\rightarrow+\infty$, the complex number $t-a$, written in polar form, approaches $te^{i0}$, while $-t-a\,(\approx-|t|)$ approaches $te^{i\pi}$. (Because the contour of integration passes above the pole at $a$, the argument of the complex number is decreasing from the lower limit at $-\infty$ to the upper limit at $+\infty$.) Therefore the logarithm reduces to $$\lim_{t\rightarrow+\infty}\left[-\ln\left(\frac{t-a}{-t-a}\right)\right]=-\ln\frac{e^{i0}}{e^{i\pi}}=i\pi.$$ Combining that with the overall factor of $1/(x-a)$ that was pulled out of the integrations, we have the Hilbert transform $$H\frac{1}{x-a}=\frac{1}{\pi}\mathcal{P}\int_{-\infty}^{+\infty} dz\left(\frac{1}{z-a}\frac{1}{z-x}\right)=\frac{i}{x-a}.$$

Answer 2

This can be evaluated several ways. The most straightforward one is through the complex contour integral. You need to specify that $x\in \mathbf R$ though.

Because $f(z)$ is holomorphic in the closed upper half complex plane, by Cauchy's formula $$\bigg(\text{p.v.}\int_{-\infty}^\infty+\int_{z=\epsilon e^{i\theta},\,\theta\in[\pi,0]}+\int_{z=Re^{i\theta},\,\theta\in[0,\pi]}\bigg) \frac{f(z)}{z-x}=0,\,\forall 0<\epsilon<R.$$ Denote the above integrations by $I_1,\,I_2,\,I_3$ respectively. $$\lim_{\epsilon\to 0^+}I_2=-i\pi f(x),$$ $$\lim_{R\to \infty}I_3=0.$$ Then $$I_1 = i\pi f(x).$$ You just need to substitute your particular $f$ to get the particular formula.

Answer 3

Here's how you solve these kind of questions if you don't really know what you're doing but quickly need a result:

1. Solve the integral with some calculator (i. e. https://www.integral-calculator.com/):

$\int\frac{\frac{1}{x-a}}{z-x}dz = \dfrac{\ln\left(z-x\right)}{x-a}$

2. Scroll through Wikipedia until you find something helpful:

$\ln{-z} = \ln{z} +i\pi$

3. Approximate the solution by ignoring small things:

$\dfrac{\ln\left(z-x\right)}{x-a}\big \vert_{-\infty}^{+\infty} = \frac{1}{x-a}\left(\ln{z}-\ln{z}+i\pi\right) = \frac{i\pi}{x-a}$