A textbook I am going through claims that for smooth $f: \mathbb{R} \rightarrow \mathbb{R}$, it holds that
$$\lim_{\epsilon \rightarrow 0^+} \int_{-1}^1 \frac{f(t)}{t-i\epsilon}dt = \text{PV} \int_{-1}^1 \frac{f(t)}{t} dt + i\pi f(0)$$
where PV denotes the Cauchy principal value. I am trying to prove this to myself but I am stuck. Could anyone provide a hint?
You can rewrite the left hand side as $$ \lim\limits_{\epsilon\to 0^+}\int\limits_{-1}^{1}\frac{t+i\epsilon}{t^2+\epsilon^2}f(t)\,dt=f(0)\lim\limits_{\epsilon\to 0^+}\int\limits_{-1}^{1}\frac{t+i\epsilon}{t^2+\epsilon^2}\,dt+\lim\limits_{\epsilon\to 0^+}\int\limits_{-1}^{1}\frac{t+i\epsilon}{t^2+\epsilon^2}[f(t)-f(0)]\,dt= $$
$$ =2if(0)\lim\limits_{\epsilon\to 0^+}\arctan\frac{1}{\epsilon}+\int\limits_{-1}^{1}\frac{f(t)-f(0)}{t}\,dt=i\pi f(0)+PV\, \int\limits_{-1}^1\frac{f(t)}{t}\,dt $$ The reason why $\int\limits_{-1}^{1}\frac{f(t)-f(0)}{t}\,dt$ is the principal value is explained in Mark Viola's comment below.
BTW, this is one of Sokhotski's formulas (1873), related to the existence of the limit $$ \lim\frac 1{t-i\epsilon} $$ as $\epsilon\to 0^+$, in the sense of distributions. Namely, the limit is $$ i\pi\delta+\mathcal{P}\frac 1{x} $$ where $\mathcal{P}$ is the principal value (as a distribution) and $\delta$ is Dirac's delta.