I Got this function: $$F(x)=\frac{1}{\sqrt{x^2-a}}\int_{-1}^1d\omega \frac{\sqrt{\omega^2-a}}{(x-\omega)}\phi(\omega)$$
$\phi$ is a well-behaved function and $a$ is a positive real number.
I want to calculate the discontinuity across the cut: $$G(x)=F(x-i\epsilon)-F(x+i\epsilon)$$
So: $$F(x+ i\epsilon)=\frac{1}{\sqrt{(x+ i\epsilon)^2-a}}\int_{-1}^1d\omega \frac{\sqrt{\omega^2-a}}{(x-\omega+ i\epsilon)}\phi(\omega) $$
I can use Plemelj formula: $$\int_{-1}^1d\omega \frac{\sqrt{\omega^2-a}}{(x-\omega+ i\epsilon)}\phi(\omega)=\int_{-1}^1d\omega \phi(\omega)\sqrt{\omega^2-a}\left[\mathcal{P}\left(\frac{1}{x-\omega}\right) - i\pi\delta(x-\omega)\right]$$ but for the factor $\frac{1}{\sqrt{(x+ i\epsilon)^2-a}}$ I'm not sure what to do. Can I assure that $$\frac{1}{\sqrt{(x+ i\epsilon)^2-a}}= \frac{1}{\sqrt{x^2-a}} \ ?$$ If so, then: $$F(x+ i\epsilon)=\frac{1}{\sqrt{x^2-a}}\int_{-1}^1d\omega \mathcal{P}\left(\frac{\phi(\omega)\sqrt{\omega^2-a}}{x-\omega}\right) - i\pi\phi(x)$$
Claiming that $$\frac{1}{\sqrt{(x- i\epsilon)^2-a}}= \frac{1}{\sqrt{x^2-a}}$$ we obtain
$$G(x)=2i\pi\phi(x)$$
Is my procedure incorrect?