The black line is the branch cut.
Lemma
$$\lim_{\Delta\to0^+}\left(\int_{\gamma_1}+\int_{\gamma_2}\right)f(z)\ln(z-s)dz=-2\pi i\int_{pe^{i\theta}}^{qe^{i\theta}}f(t)dt$$ where $\arg(z-s)\in[\theta,\theta+2\pi)$, $f$ being holomorphic on the path of integration.
Many advanced users on this site use this lemma without stating, letting alone proving it. I wrote a proof here, but it is quite long.
Is there a shorter proof of this lemma?
I just found a short proof using integration by parts:
Let $\hat k=i\frac{s}{|s|}$.
Let $P=pe^{i\theta}, Q=qe^{i\theta}$.
Let $P^{\pm}=P\pm \Delta\hat k,Q^{\pm}=Q\pm \Delta\hat k$.
Let $F$ be the local antiderivative of $f$. (A local antiderivative exists due to local continuity.)
Then, $$ \begin{align} &~~~~~\lim_{\Delta\to0^+}\left(\int_{\gamma_1}+\int_{\gamma_2}\right)f(z)\ln(z-s)dz \\ &=\lim_{\Delta\to0^+}\left(\int_{P^+}^{Q^+}+\int_{Q^-}^{P^-}\right)f(z)\ln(z-s)dz \\ &=\lim_{\Delta\to0^+}\bigg[F(z)\ln(z-s)\bigg]_{P^+,Q^-}^{Q^+,P^-} -\lim_{\Delta\to0^+}\left(\int_{P^+}^{Q^+}+\int_{Q^-}^{P^-}\right)\frac{F(z)}{z-s}dz \\ &=\lim_{\Delta\to0^+}\bigg[F(z)\ln(z-s)\bigg]_{P^+,Q^-}^{Q^+,P^-}+0 \\ &=\lim_{\Delta\to0^+}\bigg[F(z)\ln(z-s)\bigg]_{P^+}^{P^-} +\lim_{\Delta\to0^+}\bigg[F(z)\ln(z-s)\bigg]_{Q^-}^{Q^+} \\ &=F(P)\lim_{\Delta\to0^+}\bigg[\ln(z-s)\bigg]_{P^+}^{P^-} +F(Q)\lim_{\Delta\to0^+}\bigg[\ln(z-s)\bigg]_{Q^-}^{Q^+} \\ &=F(P)(2\pi i)+F(Q)(-2\pi i) \\ &=-2\pi i\bigg(F(Q)-F(P)\bigg) \\ &=-2\pi i\int_{pe^{i\theta}}^{qe^{i\theta}}f(t)dt \end{align} $$
Q.E.D.
Essentially the proof is only 9 lines long.