# How to solve this integral (right below)?

Show that: $$p.v.\int_{-\infty}^\infty\frac{1}{(\omega' - \omega_0)^2 + a^2}\frac{1}{\omega' - \omega}d\omega' = \frac{\pi}{a}\frac{\omega - \omega_0}{(\omega - \omega_0)^2 + a^2}$$

Where $p.v.$ is the principal value of the integral.

I suspect it has something to do with the "integral analog" of cauchy-riemann equations (below) which came from Cauchy Integral formula. $$f(x_0) = \frac{1}{\pi i}p.v.\int_{-\infty}^\infty\frac{u(x) + iv(x)}{x-x_0}dx$$

One can separate the values of $u(x_0)$ and $v(x_0)$ separating real and imaginary parts from above formula.

$$u(x_0) = \frac{1}{\pi}p.v\int_{-\infty}^\infty\frac{v(x)}{x-x_0}dx \\ v(x_0) = -\frac{1}{\pi}p.v\int_{-\infty}^\infty\frac{u(x)}{x-x_0}dx$$

However.. I am really stuck from here.... Indeed, the answer seems really close to these relations... But, I have no idea how this helps to solve integral, because, integration under $v(x)$ will give $u(x_0)$ and so forth.

Any help? How to solve?
