How to compute the Hilbert transform of $1/(x-a)$?

Question

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,

$$Hf(x) = \frac1\pi\text{p.v.} \int_{-\infty}^\infty \frac{f(z)}{z - x} dz,$$

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

I am having trouble computing $H(1/(x-a))$ 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 am a physics student that unfortunately haven't 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. Thank you!

Answer 1

You can also evaluate this particular integral by the partial fraction decomposition and use the integral $ \frac 1zdz=d\ln z +C.$

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. $$\lim_{\epsilon\to 0^+}I_2=-i\pi f(x),$$ $$\lim_{R\to \infty}I_3=0,\text{ since}\lim_{R\to\infty} f(e^{i\theta})=0,\, \forall \theta\in[0,\pi].$$ Then $$I_1 = i\pi f(x).$$ You just need to substitute your particular $f$ to get the particular formula.