contour integral with given contour












1












$begingroup$


enter image description here



I know that there are three simple poles



e^(pii/3),e^(pii),e^(5pi*i/3)



and Res(1/(x^3+1))=Res(1/3x^2)=Res(-x/3)



but i dont know how to deal with that kind of contour



I gussed e^(pi*i/3) is only pole that interior of contour



but using only residue of e^(pi*i/3) doesn’t work.



hint will be appreciated






complex-analysis contour-integration







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    enter image description here



    I know that there are three simple poles



    e^(pii/3),e^(pii),e^(5pi*i/3)



    and Res(1/(x^3+1))=Res(1/3x^2)=Res(-x/3)



    but i dont know how to deal with that kind of contour



    I gussed e^(pi*i/3) is only pole that interior of contour



    but using only residue of e^(pi*i/3) doesn’t work.



    hint will be appreciated






    complex-analysis contour-integration







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      0



      $begingroup$


      enter image description here



      I know that there are three simple poles



      e^(pii/3),e^(pii),e^(5pi*i/3)



      and Res(1/(x^3+1))=Res(1/3x^2)=Res(-x/3)



      but i dont know how to deal with that kind of contour



      I gussed e^(pi*i/3) is only pole that interior of contour



      but using only residue of e^(pi*i/3) doesn’t work.



      hint will be appreciated






      complex-analysis contour-integration







      share|cite|improve this question









      $endgroup$




      enter image description here



      I know that there are three simple poles



      e^(pii/3),e^(pii),e^(5pi*i/3)



      and Res(1/(x^3+1))=Res(1/3x^2)=Res(-x/3)



      but i dont know how to deal with that kind of contour



      I gussed e^(pi*i/3) is only pole that interior of contour



      but using only residue of e^(pi*i/3) doesn’t work.



      hint will be appreciated






      complex-analysis contour-integration




      complex-analysis contour-integration






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 11 '18 at 23:57









      simsimsimsim

      434




      434






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          For the integral along the real axis, we have : $x=te^{2pi i}=t, dx=dt$, so we have $int_{0}^{infty}frac{dt}{t^3+1}$ which is our original integral ($:=I$)



          For the integral along part of the circle, we have :$$x=R*e^{frac{2pi it}{3}}, 0 leq tleq 1, dx=frac{2pi i}{3}R*e^{frac{2 pi i t}{3}}$$ so we have $$frac{2pi i}{3}int frac{R*e^{frac{2 pi i t}{3}}}{R^3*e^{2pi it}+1}dt$$ which tends to zero as R $to infty$

          Now, for the last one : $x=te^{frac{2pi i}{3}}, dx=e^{frac{2pi i}{3}}dt$
          Our integral is $int_{infty}^{0}frac{e^{frac{2 pi i}{3}}dt}{t^3+1}=-e^{frac{2 pi i}{3}} int^{infty}_{0}frac{dt}{t^3+1}= -e^{frac{2 pi i}{3}}*I$



          So the residue theorem gives use $$(1-e^{frac{2 pi i}{3}})int_{0}^{infty}frac{dt}{t^3+1} + 0 = 2 pi i sum Res(f_i)$$



          Our only residue inside our domain is $x=e^{frac{pi i}{3}}$ If we compute our residue, we get $frac{1}{3e^{frac{2 pi i}{3}}}$



          Finally we have :$$int_{0}^{infty}frac{dt}{t^3+1}=frac{2 pi i}{(1-e^{frac{2 pi i}{3}})}*frac{1}{3e^{frac{2 pi i}{3}}}=frac{2 pi}{3 sqrt{3}}$$



          If I did any typo, please say it in the comments as soon as possible.



          Hope this helps !






          share|cite|improve this answer









          $endgroup$













            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',
            autoActivateHeartbeat: false,
            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%2f3036035%2fcontour-integral-with-given-contour%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









            1












            $begingroup$

            For the integral along the real axis, we have : $x=te^{2pi i}=t, dx=dt$, so we have $int_{0}^{infty}frac{dt}{t^3+1}$ which is our original integral ($:=I$)



            For the integral along part of the circle, we have :$$x=R*e^{frac{2pi it}{3}}, 0 leq tleq 1, dx=frac{2pi i}{3}R*e^{frac{2 pi i t}{3}}$$ so we have $$frac{2pi i}{3}int frac{R*e^{frac{2 pi i t}{3}}}{R^3*e^{2pi it}+1}dt$$ which tends to zero as R $to infty$

            Now, for the last one : $x=te^{frac{2pi i}{3}}, dx=e^{frac{2pi i}{3}}dt$
            Our integral is $int_{infty}^{0}frac{e^{frac{2 pi i}{3}}dt}{t^3+1}=-e^{frac{2 pi i}{3}} int^{infty}_{0}frac{dt}{t^3+1}= -e^{frac{2 pi i}{3}}*I$



            So the residue theorem gives use $$(1-e^{frac{2 pi i}{3}})int_{0}^{infty}frac{dt}{t^3+1} + 0 = 2 pi i sum Res(f_i)$$



            Our only residue inside our domain is $x=e^{frac{pi i}{3}}$ If we compute our residue, we get $frac{1}{3e^{frac{2 pi i}{3}}}$



            Finally we have :$$int_{0}^{infty}frac{dt}{t^3+1}=frac{2 pi i}{(1-e^{frac{2 pi i}{3}})}*frac{1}{3e^{frac{2 pi i}{3}}}=frac{2 pi}{3 sqrt{3}}$$



            If I did any typo, please say it in the comments as soon as possible.



            Hope this helps !






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              For the integral along the real axis, we have : $x=te^{2pi i}=t, dx=dt$, so we have $int_{0}^{infty}frac{dt}{t^3+1}$ which is our original integral ($:=I$)



              For the integral along part of the circle, we have :$$x=R*e^{frac{2pi it}{3}}, 0 leq tleq 1, dx=frac{2pi i}{3}R*e^{frac{2 pi i t}{3}}$$ so we have $$frac{2pi i}{3}int frac{R*e^{frac{2 pi i t}{3}}}{R^3*e^{2pi it}+1}dt$$ which tends to zero as R $to infty$

              Now, for the last one : $x=te^{frac{2pi i}{3}}, dx=e^{frac{2pi i}{3}}dt$
              Our integral is $int_{infty}^{0}frac{e^{frac{2 pi i}{3}}dt}{t^3+1}=-e^{frac{2 pi i}{3}} int^{infty}_{0}frac{dt}{t^3+1}= -e^{frac{2 pi i}{3}}*I$



              So the residue theorem gives use $$(1-e^{frac{2 pi i}{3}})int_{0}^{infty}frac{dt}{t^3+1} + 0 = 2 pi i sum Res(f_i)$$



              Our only residue inside our domain is $x=e^{frac{pi i}{3}}$ If we compute our residue, we get $frac{1}{3e^{frac{2 pi i}{3}}}$



              Finally we have :$$int_{0}^{infty}frac{dt}{t^3+1}=frac{2 pi i}{(1-e^{frac{2 pi i}{3}})}*frac{1}{3e^{frac{2 pi i}{3}}}=frac{2 pi}{3 sqrt{3}}$$



              If I did any typo, please say it in the comments as soon as possible.



              Hope this helps !






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                For the integral along the real axis, we have : $x=te^{2pi i}=t, dx=dt$, so we have $int_{0}^{infty}frac{dt}{t^3+1}$ which is our original integral ($:=I$)



                For the integral along part of the circle, we have :$$x=R*e^{frac{2pi it}{3}}, 0 leq tleq 1, dx=frac{2pi i}{3}R*e^{frac{2 pi i t}{3}}$$ so we have $$frac{2pi i}{3}int frac{R*e^{frac{2 pi i t}{3}}}{R^3*e^{2pi it}+1}dt$$ which tends to zero as R $to infty$

                Now, for the last one : $x=te^{frac{2pi i}{3}}, dx=e^{frac{2pi i}{3}}dt$
                Our integral is $int_{infty}^{0}frac{e^{frac{2 pi i}{3}}dt}{t^3+1}=-e^{frac{2 pi i}{3}} int^{infty}_{0}frac{dt}{t^3+1}= -e^{frac{2 pi i}{3}}*I$



                So the residue theorem gives use $$(1-e^{frac{2 pi i}{3}})int_{0}^{infty}frac{dt}{t^3+1} + 0 = 2 pi i sum Res(f_i)$$



                Our only residue inside our domain is $x=e^{frac{pi i}{3}}$ If we compute our residue, we get $frac{1}{3e^{frac{2 pi i}{3}}}$



                Finally we have :$$int_{0}^{infty}frac{dt}{t^3+1}=frac{2 pi i}{(1-e^{frac{2 pi i}{3}})}*frac{1}{3e^{frac{2 pi i}{3}}}=frac{2 pi}{3 sqrt{3}}$$



                If I did any typo, please say it in the comments as soon as possible.



                Hope this helps !






                share|cite|improve this answer









                $endgroup$



                For the integral along the real axis, we have : $x=te^{2pi i}=t, dx=dt$, so we have $int_{0}^{infty}frac{dt}{t^3+1}$ which is our original integral ($:=I$)



                For the integral along part of the circle, we have :$$x=R*e^{frac{2pi it}{3}}, 0 leq tleq 1, dx=frac{2pi i}{3}R*e^{frac{2 pi i t}{3}}$$ so we have $$frac{2pi i}{3}int frac{R*e^{frac{2 pi i t}{3}}}{R^3*e^{2pi it}+1}dt$$ which tends to zero as R $to infty$

                Now, for the last one : $x=te^{frac{2pi i}{3}}, dx=e^{frac{2pi i}{3}}dt$
                Our integral is $int_{infty}^{0}frac{e^{frac{2 pi i}{3}}dt}{t^3+1}=-e^{frac{2 pi i}{3}} int^{infty}_{0}frac{dt}{t^3+1}= -e^{frac{2 pi i}{3}}*I$



                So the residue theorem gives use $$(1-e^{frac{2 pi i}{3}})int_{0}^{infty}frac{dt}{t^3+1} + 0 = 2 pi i sum Res(f_i)$$



                Our only residue inside our domain is $x=e^{frac{pi i}{3}}$ If we compute our residue, we get $frac{1}{3e^{frac{2 pi i}{3}}}$



                Finally we have :$$int_{0}^{infty}frac{dt}{t^3+1}=frac{2 pi i}{(1-e^{frac{2 pi i}{3}})}*frac{1}{3e^{frac{2 pi i}{3}}}=frac{2 pi}{3 sqrt{3}}$$



                If I did any typo, please say it in the comments as soon as possible.



                Hope this helps !







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 12 '18 at 22:22









                PoujhPoujh

                614516




                614516






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3036035%2fcontour-integral-with-given-contour%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

                    Tonle Sap (See)

                    Image
                    Tonle Sap Satellitenfoto (April 1994) Geographische Lage Kambodscha Zuflüsse Tonle-Sap-Fluss, Siem Reap River Abfluss Tonle-Sap-Fluss Orte in der Nähe Siem Reap, Battambang Daten Koordinaten 12° 52′ 34″  N , 104° 4′ 23″  O 12.876 104.073 Koordinaten: 12° 52′ 34″  N , 104° 4′ 23″  O Höhe über Meeresspiegel f1 0,5 Meter Fläche 2.700–25.000 km² dep1 f5 Länge 250 km (maximal) dep1 f6 Breite 100 km (maximal) dep1 f7 Volumen 80 km³ (maximal) dep1 f8 Maximale Tiefe 14 m f10 Mittlere Tiefe 2–3 m (minimal) f11 Besonderheiten größter See Südostasiens Der Tonle-Sap-See (Khmer: បឹង​ទន្លេសាប , Aussprache: [ boeung tunleː saːp ]) in Kambodscha ist der größte See Südostasiens und eines der fischreichsten Binnengewässer der Erde. Inhaltsverzeichnis 1 Geographie 2 Wirtschaftliche Bedeutung 2.1 Verkehr 2.2 Reis 2.3 Fisch 3 Ökologie und Gefährdung 3.1 Überfischung...
                    Read more

                    I get strange results when I access the Sqlitedatabase with Unity C# via XAMPP

                    Image
                    0 <?php $servername = "localhost"; $username="root"; $password=""; $dbName="escape_room"; $user_username = $_POST['Input_user']; $user_password = $_POST['Input_pass']; $conn = new mysqli($servername, $username, $password, $dbName); if(!$conn){ die("Cound not Connect: " . mysqli_connect_error()); } $sql = "SELECT pass FROM escape_room WHERE user = '".$user_username."' "; $result = mysqli_query($conn, $sql); //Get the result and confirm login if(mysqli_num_rows($result)>0){ while($row = mysqli_fetch_assoc($result)){ if($row['pass'] == $user_password){ echo "login success"; ...
                    Read more

                    Guatemaltekische Davis-Cup-Mannschaft

                    Image
                    Guatemala Kapitän Anthony Vásquez Aktuelles ITF-Ranking 73 Statistik Erste Teilnahme 1990 Davis-Cup-Teilnahmen 22 Bestes Ergebnis Amerika-Gruppenzone II HF (1995, 2000, 2001) Ewige Bilanz 42:30 Erfolgreichste Spieler Meiste Siege gesamt Daniel Chávez Morales (29) Meiste Einzelsiege Cristian Paiz (21) Meiste Doppelsiege Daniel Chávez Morales (17) Bestes Doppel Jacobo Chávez / Daniel Chávez Morales (9) Meiste Teilnahmen Cristian Paiz (32) Meiste Jahre Daniel Chávez Morales (10) Letzte Aktualisierung der Infobox: 3. Juni 2012 Die guatemaltekische Davis-Cup-Mannschaft ist die Tennisnationalmannschaft Guatemalas. Geschichte | 1990 nahm Guatemala erstmals am Davis Cup teil. Das beste Resultat erreichte die Mannschaft jeweils 1995, 2000 und 2001 mit dem Halbfinale in der Amerika-Gruppenzone II. Bester Spieler ist Daniel Chávez Morales mit 29 Siegen bei insgesamt 31 Teilnahmen. Rekord...
                    Read more