Complex integration lemma: shorter proof?












4
















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?






complex-analysis complex-integration







share|cite|improve this question



























    4
















    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?






    complex-analysis complex-integration







    share|cite|improve this question

























      4












      4








      4


      2







      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?






      complex-analysis complex-integration







      share|cite|improve this question















      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?






      complex-analysis complex-integration




      complex-analysis complex-integration






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked Nov 19 '18 at 0:27









      Szeto

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          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.






          share|cite|improve this answer



















          • 1




            You may accept your own answer(?)
            – Tianlalu
            Nov 30 '18 at 7:36











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          1 Answer
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          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.






          share|cite|improve this answer



















          • 1




            You may accept your own answer(?)
            – Tianlalu
            Nov 30 '18 at 7:36
















          1














          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.






          share|cite|improve this answer



















          • 1




            You may accept your own answer(?)
            – Tianlalu
            Nov 30 '18 at 7:36














          1












          1








          1






          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.






          share|cite|improve this answer














          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 22 '18 at 5:30

























          answered Nov 20 '18 at 10:17









          Szeto

          6,4362926




          6,4362926








          • 1




            You may accept your own answer(?)
            – Tianlalu
            Nov 30 '18 at 7:36














          • 1




            You may accept your own answer(?)
            – Tianlalu
            Nov 30 '18 at 7:36








          1




          1




          You may accept your own answer(?)
          – Tianlalu
          Nov 30 '18 at 7:36




          You may accept your own answer(?)
          – Tianlalu
          Nov 30 '18 at 7:36


















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