Show that

$$\mathcal{P} \int_{-\infty}^{\infty} \frac{t^{z-1}}{t-a} d t=\pi i a^{z-1},$$

where $a$ is real and positive, $0<\operatorname{Re}(z)<1$ and $\mathcal{P}$ denotes the Cauchy principal value; the principal branches of $t^{z}$ etc. are implied. Deduce that

$$\int_{0}^{\infty} \frac{t^{z-1}}{t+a} d t=\pi a^{z-1} \operatorname{cosec} \pi z$$

and that

$$\mathcal{P} \int_{0}^{\infty} \frac{t^{z-1}}{t-a} d t=-\pi a^{z-1} \cot \pi z$$

Use $(*)$ to show that, if $\operatorname{Im}(b)>0$, then

$$\int_{0}^{\infty} \frac{t^{z-1}}{t-b} d t=-\pi b^{z-1}(\cot \pi z-i)$$

What is the value of this integral if $\operatorname{Im}(b)<0$ ?