Jtdylktuy

The black line is the branch cut.

Lemma

$$lim_{Deltato0^+}left(int_{gamma_1}+int_{gamma_2}right)f(z)ln(z-s)dz=-2pi iint_{pe^{itheta}}^{qe^{itheta}}f(t)dt$$ where $arg(z-s)in[theta,theta+2pi)$, $f$ 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 $hat k=ifrac{s}{|s|}$.

Let $P=pe^{itheta}, Q=qe^{itheta}$.

Let $P^{pm}=Ppm Deltahat k,Q^{pm}=Qpm Deltahat k$.

Let $F$ be the local antiderivative of $f$. (A local antiderivative exists due to local continuity.)

Then,
$$
begin{align}
&~~~~~lim_{Deltato0^+}left(int_{gamma_1}+int_{gamma_2}right)f(z)ln(z-s)dz \
&=lim_{Deltato0^+}left(int_{P^+}^{Q^+}+int_{Q^-}^{P^-}right)f(z)ln(z-s)dz \
&=lim_{Deltato0^+}bigg[F(z)ln(z-s)bigg]_{P^+,Q^-}^{Q^+,P^-} -lim_{Deltato0^+}left(int_{P^+}^{Q^+}+int_{Q^-}^{P^-}right)frac{F(z)}{z-s}dz \
&=lim_{Deltato0^+}bigg[F(z)ln(z-s)bigg]_{P^+,Q^-}^{Q^+,P^-}+0 \
&=lim_{Deltato0^+}bigg[F(z)ln(z-s)bigg]_{P^+}^{P^-}
+lim_{Deltato0^+}bigg[F(z)ln(z-s)bigg]_{Q^-}^{Q^+} \
&=F(P)lim_{Deltato0^+}bigg[ln(z-s)bigg]_{P^+}^{P^-}
+F(Q)lim_{Deltato0^+}bigg[ln(z-s)bigg]_{Q^-}^{Q^+} \
&=F(P)(2pi i)+F(Q)(-2pi i) \
&=-2pi ibigg(F(Q)-F(P)bigg) \
&=-2pi iint_{pe^{itheta}}^{qe^{itheta}}f(t)dt
end{align}
$$

Q.E.D.

Essentially the proof is only 9 lines long.