Complex integration lemma: shorter proof?











up vote
4
down vote

favorite
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?










share|cite|improve this question


























    up vote
    4
    down vote

    favorite
    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?










    share|cite|improve this question
























      up vote
      4
      down vote

      favorite
      2









      up vote
      4
      down vote

      favorite
      2






      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?










      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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 19 at 0:27









      Szeto

      6,1442726




      6,1442726






















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          0
          down vote













          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























            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














             

            draft saved


            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004332%2fcomplex-integration-lemma-shorter-proof%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            0
            down vote













            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



























              up vote
              0
              down vote













              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

























                up vote
                0
                down vote










                up vote
                0
                down vote









                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 yesterday

























                answered Nov 20 at 10:17









                Szeto

                6,1442726




                6,1442726






























                     

                    draft saved


                    draft discarded



















































                     


                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004332%2fcomplex-integration-lemma-shorter-proof%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    Wiesbaden

                    Marschland

                    Dieringhausen