Integral with 2 different answers using real and complex analysis












2














The integral is:



$ int_0^{2 pi} frac {1}{2-cos theta}$



Just to skip time, the answer of the indefinite integral is $ frac{2tan^-1( sqrt{3} tan( frac{ theta}{2}))}{ sqrt{3}}$



Evaluating it from $0$ to $ 2 pi$



$ frac{2tan^-1( sqrt{3} tan( pi))}{ sqrt{3}} - frac{2tan^-1( sqrt{3} tan( 0))}{ sqrt{3}} = 0 - 0 = 0$



Using complex analysis:



The integral $ int_0^{2 pi} frac {1}{2-cos theta}$



is transformed into



$ 2i int_C frac {dz}{z^2-4z+1} = 2i int_C frac {dz}{(z-2+ sqrt{3})(z-2- sqrt{3})}$



Where C is the boundary of the circle $ |z| = 1 ---> z = e^{i theta}$



Then by Cauchy's integral formula, since $ z = 2 - sqrt{3}$ is inside the domain of the region bounded by $C$ , then:



$2i int_C frac {dz}{(z-2+ sqrt{3})(z-2- sqrt{3})} = 2 pi i frac{2i}{2- sqrt{3} -2- sqrt{3}} = 2 pi i frac{2i}{-2 sqrt{3}} = frac {2 pi}{ sqrt{3}}$



Using the real system, I get $0$ , using complex analysis I get $ frac{2 pi}{ sqrt{3}}$ , what's wrong?






complex-analysis definite-integrals cauchy-integral-formula







share|cite|improve this question



























    2














    The integral is:



    $ int_0^{2 pi} frac {1}{2-cos theta}$



    Just to skip time, the answer of the indefinite integral is $ frac{2tan^-1( sqrt{3} tan( frac{ theta}{2}))}{ sqrt{3}}$



    Evaluating it from $0$ to $ 2 pi$



    $ frac{2tan^-1( sqrt{3} tan( pi))}{ sqrt{3}} - frac{2tan^-1( sqrt{3} tan( 0))}{ sqrt{3}} = 0 - 0 = 0$



    Using complex analysis:



    The integral $ int_0^{2 pi} frac {1}{2-cos theta}$



    is transformed into



    $ 2i int_C frac {dz}{z^2-4z+1} = 2i int_C frac {dz}{(z-2+ sqrt{3})(z-2- sqrt{3})}$



    Where C is the boundary of the circle $ |z| = 1 ---> z = e^{i theta}$



    Then by Cauchy's integral formula, since $ z = 2 - sqrt{3}$ is inside the domain of the region bounded by $C$ , then:



    $2i int_C frac {dz}{(z-2+ sqrt{3})(z-2- sqrt{3})} = 2 pi i frac{2i}{2- sqrt{3} -2- sqrt{3}} = 2 pi i frac{2i}{-2 sqrt{3}} = frac {2 pi}{ sqrt{3}}$



    Using the real system, I get $0$ , using complex analysis I get $ frac{2 pi}{ sqrt{3}}$ , what's wrong?






    complex-analysis definite-integrals cauchy-integral-formula







    share|cite|improve this question

























      2












      2








      2







      The integral is:



      $ int_0^{2 pi} frac {1}{2-cos theta}$



      Just to skip time, the answer of the indefinite integral is $ frac{2tan^-1( sqrt{3} tan( frac{ theta}{2}))}{ sqrt{3}}$



      Evaluating it from $0$ to $ 2 pi$



      $ frac{2tan^-1( sqrt{3} tan( pi))}{ sqrt{3}} - frac{2tan^-1( sqrt{3} tan( 0))}{ sqrt{3}} = 0 - 0 = 0$



      Using complex analysis:



      The integral $ int_0^{2 pi} frac {1}{2-cos theta}$



      is transformed into



      $ 2i int_C frac {dz}{z^2-4z+1} = 2i int_C frac {dz}{(z-2+ sqrt{3})(z-2- sqrt{3})}$



      Where C is the boundary of the circle $ |z| = 1 ---> z = e^{i theta}$



      Then by Cauchy's integral formula, since $ z = 2 - sqrt{3}$ is inside the domain of the region bounded by $C$ , then:



      $2i int_C frac {dz}{(z-2+ sqrt{3})(z-2- sqrt{3})} = 2 pi i frac{2i}{2- sqrt{3} -2- sqrt{3}} = 2 pi i frac{2i}{-2 sqrt{3}} = frac {2 pi}{ sqrt{3}}$



      Using the real system, I get $0$ , using complex analysis I get $ frac{2 pi}{ sqrt{3}}$ , what's wrong?






      complex-analysis definite-integrals cauchy-integral-formula







      share|cite|improve this question













      The integral is:



      $ int_0^{2 pi} frac {1}{2-cos theta}$



      Just to skip time, the answer of the indefinite integral is $ frac{2tan^-1( sqrt{3} tan( frac{ theta}{2}))}{ sqrt{3}}$



      Evaluating it from $0$ to $ 2 pi$



      $ frac{2tan^-1( sqrt{3} tan( pi))}{ sqrt{3}} - frac{2tan^-1( sqrt{3} tan( 0))}{ sqrt{3}} = 0 - 0 = 0$



      Using complex analysis:



      The integral $ int_0^{2 pi} frac {1}{2-cos theta}$



      is transformed into



      $ 2i int_C frac {dz}{z^2-4z+1} = 2i int_C frac {dz}{(z-2+ sqrt{3})(z-2- sqrt{3})}$



      Where C is the boundary of the circle $ |z| = 1 ---> z = e^{i theta}$



      Then by Cauchy's integral formula, since $ z = 2 - sqrt{3}$ is inside the domain of the region bounded by $C$ , then:



      $2i int_C frac {dz}{(z-2+ sqrt{3})(z-2- sqrt{3})} = 2 pi i frac{2i}{2- sqrt{3} -2- sqrt{3}} = 2 pi i frac{2i}{-2 sqrt{3}} = frac {2 pi}{ sqrt{3}}$



      Using the real system, I get $0$ , using complex analysis I get $ frac{2 pi}{ sqrt{3}}$ , what's wrong?






      complex-analysis definite-integrals cauchy-integral-formula




      complex-analysis definite-integrals cauchy-integral-formula






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 56 mins ago









      khaled014z

      737




      737






















          2 Answers
          2






          active

          oldest

          votes


















          5














          The problem with the real approach is that you make the change of variable $t=tanleft(dfrac{theta}{2}right)$ for $0 < theta < 2 pi$.



          This is problematic since your substitution need to be defined and continuous for all $theta$, but you have a problem when $theta=pi$.



          Edit: Note that if you split the integral into $int_0^pi+int_pi^{2 pi}$, you are going to get the right answer, as for one integral you are going to get $arctan(- infty)$ and for the other $arctan(+infty)$:



          $$int_0^{2 pi} frac{mathrm{d}θ}{2-cos theta}=int_0^pi frac{mathrm{d}θ}{2-cos theta}+int_pi ^{2 pi} frac{mathrm{d}θ}{2-cos theta}\
          = lim_{r to pi_-} int_0^r frac{mathrm{d}θ}{2-cos theta}+ lim_{w to pi_+} int_w^{2 pi} frac{mathrm{d}θ}{2-cos theta}\= lim_{r to pi_-} left(frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}-0right)+ lim_{w to pi_+}left(0- frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}right).$$






          share|cite|improve this answer























          • Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
            – khaled014z
            47 mins ago










          • @khaled014z See the edit. Let me know if you want more details.
            – N. S.
            47 mins ago












          • Brilliant, that was kind of a tricky substitution, thank you
            – khaled014z
            43 mins ago



















          1














          Note that that tangent function, $tan(x)$, is discontinuous when $x=npi$. So, the antiderivative $frac2{sqrt{3}} arctanleft(sqrt 3 tan(theta/2)right)$ is not valid over the interval $[0,2pi]$.



          Instead, we have



          $$int_0^{2pi}frac{1}{2-cos(theta)},dtheta=2int_0^pifrac{1}{2-cos(theta)},dtheta=frac{4}{sqrt3}left.left(arctanleft(sqrt 3 tan(theta/2)right)right)right|_0^pi=frac{2pi}{sqrt3}$$






          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',
            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%2f3059020%2fintegral-with-2-different-answers-using-real-and-complex-analysis%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            5














            The problem with the real approach is that you make the change of variable $t=tanleft(dfrac{theta}{2}right)$ for $0 < theta < 2 pi$.



            This is problematic since your substitution need to be defined and continuous for all $theta$, but you have a problem when $theta=pi$.



            Edit: Note that if you split the integral into $int_0^pi+int_pi^{2 pi}$, you are going to get the right answer, as for one integral you are going to get $arctan(- infty)$ and for the other $arctan(+infty)$:



            $$int_0^{2 pi} frac{mathrm{d}θ}{2-cos theta}=int_0^pi frac{mathrm{d}θ}{2-cos theta}+int_pi ^{2 pi} frac{mathrm{d}θ}{2-cos theta}\
            = lim_{r to pi_-} int_0^r frac{mathrm{d}θ}{2-cos theta}+ lim_{w to pi_+} int_w^{2 pi} frac{mathrm{d}θ}{2-cos theta}\= lim_{r to pi_-} left(frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}-0right)+ lim_{w to pi_+}left(0- frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}right).$$






            share|cite|improve this answer























            • Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
              – khaled014z
              47 mins ago










            • @khaled014z See the edit. Let me know if you want more details.
              – N. S.
              47 mins ago












            • Brilliant, that was kind of a tricky substitution, thank you
              – khaled014z
              43 mins ago
















            5














            The problem with the real approach is that you make the change of variable $t=tanleft(dfrac{theta}{2}right)$ for $0 < theta < 2 pi$.



            This is problematic since your substitution need to be defined and continuous for all $theta$, but you have a problem when $theta=pi$.



            Edit: Note that if you split the integral into $int_0^pi+int_pi^{2 pi}$, you are going to get the right answer, as for one integral you are going to get $arctan(- infty)$ and for the other $arctan(+infty)$:



            $$int_0^{2 pi} frac{mathrm{d}θ}{2-cos theta}=int_0^pi frac{mathrm{d}θ}{2-cos theta}+int_pi ^{2 pi} frac{mathrm{d}θ}{2-cos theta}\
            = lim_{r to pi_-} int_0^r frac{mathrm{d}θ}{2-cos theta}+ lim_{w to pi_+} int_w^{2 pi} frac{mathrm{d}θ}{2-cos theta}\= lim_{r to pi_-} left(frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}-0right)+ lim_{w to pi_+}left(0- frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}right).$$






            share|cite|improve this answer























            • Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
              – khaled014z
              47 mins ago










            • @khaled014z See the edit. Let me know if you want more details.
              – N. S.
              47 mins ago












            • Brilliant, that was kind of a tricky substitution, thank you
              – khaled014z
              43 mins ago














            5












            5








            5






            The problem with the real approach is that you make the change of variable $t=tanleft(dfrac{theta}{2}right)$ for $0 < theta < 2 pi$.



            This is problematic since your substitution need to be defined and continuous for all $theta$, but you have a problem when $theta=pi$.



            Edit: Note that if you split the integral into $int_0^pi+int_pi^{2 pi}$, you are going to get the right answer, as for one integral you are going to get $arctan(- infty)$ and for the other $arctan(+infty)$:



            $$int_0^{2 pi} frac{mathrm{d}θ}{2-cos theta}=int_0^pi frac{mathrm{d}θ}{2-cos theta}+int_pi ^{2 pi} frac{mathrm{d}θ}{2-cos theta}\
            = lim_{r to pi_-} int_0^r frac{mathrm{d}θ}{2-cos theta}+ lim_{w to pi_+} int_w^{2 pi} frac{mathrm{d}θ}{2-cos theta}\= lim_{r to pi_-} left(frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}-0right)+ lim_{w to pi_+}left(0- frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}right).$$






            share|cite|improve this answer














            The problem with the real approach is that you make the change of variable $t=tanleft(dfrac{theta}{2}right)$ for $0 < theta < 2 pi$.



            This is problematic since your substitution need to be defined and continuous for all $theta$, but you have a problem when $theta=pi$.



            Edit: Note that if you split the integral into $int_0^pi+int_pi^{2 pi}$, you are going to get the right answer, as for one integral you are going to get $arctan(- infty)$ and for the other $arctan(+infty)$:



            $$int_0^{2 pi} frac{mathrm{d}θ}{2-cos theta}=int_0^pi frac{mathrm{d}θ}{2-cos theta}+int_pi ^{2 pi} frac{mathrm{d}θ}{2-cos theta}\
            = lim_{r to pi_-} int_0^r frac{mathrm{d}θ}{2-cos theta}+ lim_{w to pi_+} int_w^{2 pi} frac{mathrm{d}θ}{2-cos theta}\= lim_{r to pi_-} left(frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}-0right)+ lim_{w to pi_+}left(0- frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}right).$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 19 mins ago









            Saad

            19.7k92252




            19.7k92252










            answered 49 mins ago









            N. S.

            102k5109204




            102k5109204












            • Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
              – khaled014z
              47 mins ago










            • @khaled014z See the edit. Let me know if you want more details.
              – N. S.
              47 mins ago












            • Brilliant, that was kind of a tricky substitution, thank you
              – khaled014z
              43 mins ago


















            • Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
              – khaled014z
              47 mins ago










            • @khaled014z See the edit. Let me know if you want more details.
              – N. S.
              47 mins ago












            • Brilliant, that was kind of a tricky substitution, thank you
              – khaled014z
              43 mins ago
















            Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
            – khaled014z
            47 mins ago




            Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
            – khaled014z
            47 mins ago












            @khaled014z See the edit. Let me know if you want more details.
            – N. S.
            47 mins ago






            @khaled014z See the edit. Let me know if you want more details.
            – N. S.
            47 mins ago














            Brilliant, that was kind of a tricky substitution, thank you
            – khaled014z
            43 mins ago




            Brilliant, that was kind of a tricky substitution, thank you
            – khaled014z
            43 mins ago











            1














            Note that that tangent function, $tan(x)$, is discontinuous when $x=npi$. So, the antiderivative $frac2{sqrt{3}} arctanleft(sqrt 3 tan(theta/2)right)$ is not valid over the interval $[0,2pi]$.



            Instead, we have



            $$int_0^{2pi}frac{1}{2-cos(theta)},dtheta=2int_0^pifrac{1}{2-cos(theta)},dtheta=frac{4}{sqrt3}left.left(arctanleft(sqrt 3 tan(theta/2)right)right)right|_0^pi=frac{2pi}{sqrt3}$$






            share|cite|improve this answer


























              1














              Note that that tangent function, $tan(x)$, is discontinuous when $x=npi$. So, the antiderivative $frac2{sqrt{3}} arctanleft(sqrt 3 tan(theta/2)right)$ is not valid over the interval $[0,2pi]$.



              Instead, we have



              $$int_0^{2pi}frac{1}{2-cos(theta)},dtheta=2int_0^pifrac{1}{2-cos(theta)},dtheta=frac{4}{sqrt3}left.left(arctanleft(sqrt 3 tan(theta/2)right)right)right|_0^pi=frac{2pi}{sqrt3}$$






              share|cite|improve this answer
























                1












                1








                1






                Note that that tangent function, $tan(x)$, is discontinuous when $x=npi$. So, the antiderivative $frac2{sqrt{3}} arctanleft(sqrt 3 tan(theta/2)right)$ is not valid over the interval $[0,2pi]$.



                Instead, we have



                $$int_0^{2pi}frac{1}{2-cos(theta)},dtheta=2int_0^pifrac{1}{2-cos(theta)},dtheta=frac{4}{sqrt3}left.left(arctanleft(sqrt 3 tan(theta/2)right)right)right|_0^pi=frac{2pi}{sqrt3}$$






                share|cite|improve this answer












                Note that that tangent function, $tan(x)$, is discontinuous when $x=npi$. So, the antiderivative $frac2{sqrt{3}} arctanleft(sqrt 3 tan(theta/2)right)$ is not valid over the interval $[0,2pi]$.



                Instead, we have



                $$int_0^{2pi}frac{1}{2-cos(theta)},dtheta=2int_0^pifrac{1}{2-cos(theta)},dtheta=frac{4}{sqrt3}left.left(arctanleft(sqrt 3 tan(theta/2)right)right)right|_0^pi=frac{2pi}{sqrt3}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 45 mins ago









                Mark Viola

                130k1274170




                130k1274170






























                    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.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • 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.


                    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%2f3059020%2fintegral-with-2-different-answers-using-real-and-complex-analysis%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

                    Eastern Orthodox Church

                    Image
                    Not to be confused with Oriental Orthodoxy, Eastern Christianity, or Eastern Catholic Churches. "Eastern Orthodoxy", "Orthodox Church", "Orthodox Catholic Church", and "Orthodox Christian Church" redirect here. For other uses, see Orthodox (disambiguation). This article may be too long to read and navigate comfortably . The readable prose size is 132 kilobytes. Please consider splitting content into sub-articles, condensing it, or adding or removing subheadings. ( January 2018 ) Part of a series on the Eastern Orthodox Church Mosaic of Christ Pantocrator, Hagia Sophia Overview Structure Theology (History of theology) Liturgy Church history Holy Mysteries View of salvation View of Mary View of icons Background Crucifixion / Resurrection / Ascension of Jesus Christianity Christian Church Apostolic succession Four Marks of the Church Orthodoxy Organization Autocephaly Patriarchate ...
                    Read more

                    Zagreb

                    Image
                    This article is about the Croatian capital city. For other uses, see Zagreb (disambiguation). City in City of Zagreb, Croatia Zagreb City Grad Zagreb City of Zagreb Clockwise, from top: St. Mark's Square, Croatian State Archives, Zagreb Tram, Art Pavilion, Cibona & HOTO towers and Croatian National Theatre. Flag Coat of arms Zagreb Location of Zagreb in Croatia Show map of Croatia Zagreb Zagreb (Europe) Show map of Europe Coordinates: 45°49′N 15°59′E  /  45.817°N 15.983°E  / 45.817; 15.983 Coordinates: 45°49′N 15°59′E  /  45.817°N 15.983°E  / 45.817; 15.983 Country   Croatia County City of Zagreb RC diocese 1094 Free royal city 1242 Unified 1850 Subdivisions 17 city districts 218 local committees (70 settlements) Government  • Type Mayor-Council  • Mayor Milan Bandić (BM 365)  • City Assembly 51 members BM 365, ZL, NS-R (14) SDP, GLAS, HNS, HSS, NH-PS (13) ...
                    Read more

                    Understanding the information contained in the Deep Space Network XML data?

                    Image
                    up vote 1 down vote favorite NASA's Deep Space Network activity can be viewed conveniently at https://eyes.nasa.gov/dsn/dsn.html but if you would like data in a more raw XML form, it can be accessed as described in this answer Below is a small Python script that reads the XML into a list of dictionaries, and saves it to disk as a JSON. The dictionaries are organized at the higher level by dish, then by spacecraft per dish. Question: What are the most-likely meanings of all the data field contained in the Deep Space Network XML data? This is the example url from the other answer, used in the script below: https://eyes.nasa.gov/dsn/data/1365107113.xml def dictify(r,root=True): # https://stackoverflow.com/questions/2148119/how-to-convert-an-xml-string-to-a-dictionary-in-python/30923963#30923963 if root: ret...
                    Read more