Contour integration with sign function

Question

I am trying to deal with integral $$\int_{-\infty}^{\infty}\frac{dx}{w-x+i0\,\mathrm{sgn}\,w}.$$ I know that answer should be proportional to $\mathrm{sng}\,w$, but do not understand how to obtain it. Naively I represent integrand as $$\frac{1}{w-x+i0\,\mathrm{sgn}\,w}=\frac{1}{w-x}-i\pi\delta(w-x)\,\mathrm{sgn}\,w$$ but I do not understand how to perform further derivation.

Answer 1

I do not know about your initial integrand with the $i0$ notation, but if you want to find

$$\int_{-\infty}^\infty dx \left\{ \frac{1}{\omega-x}-i\pi\delta(\omega-x)\,\mathrm{sgn}\,\omega \right\}$$

for real $\omega$, then

$$ \int_{-\infty}^\infty dx ~ i\pi\delta(\omega-x)\,\mathrm{sgn}\,\omega = i\pi ~\mathrm{sgn}\,\omega$$ by the definition of $\delta(x)$. The integral $$\int_{-\infty}^\infty dx \frac{1}{\omega-x}$$ diverges, but if you take the Cauchy principal value, then $$P.V. \int_{-\infty}^\infty dx \frac{1}{\omega-x} = \lim_{R \to \infty} \int_{-R}^R dx \frac{1}{\omega-x} = \lim_{R \to \infty}-\ln\left(\omega-x \right)|_{-R}^R = \lim_{R \to \infty} \ln\left( \frac{\omega+R}{\omega-R} \right) = \ln(1)=0$$