Complex integration lemma: shorter proof? 
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?
 A: 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.
