Using the Cauchy Integral Theorem for Derivatives to evaluate an integral











up vote
0
down vote

favorite












I ran into this question which hints me to use Cauchy's Integral Theorem for Derivatives, however I don't seem to be able to fit this integral into the form of the Integral Formula



$$displaystyle int_{|z|=2} frac{cos(z)}{z(z^2+8)}dz$$ I tried using the fact that $displaystyle int_gamma f(z)dz=int_a^b f(gamma(t))gamma'(t)dt$ for $gamma(t)$ where $t in [a, b]$. But got nowhere. Is there a way I can transform the given integral into the form in which I can use the Integral Formula as stated below?



$$f^{(k)}(w)=frac{1}{2pi i}int_gamma frac{f(z)}{(z-w)^{k+1}}dz$$ where $f^{(k)}(w)$ is the $k^{th}$ derivative of $f$






complex-analysis complex-integration







share|cite|improve this question


























    up vote
    0
    down vote

    favorite












    I ran into this question which hints me to use Cauchy's Integral Theorem for Derivatives, however I don't seem to be able to fit this integral into the form of the Integral Formula



    $$displaystyle int_{|z|=2} frac{cos(z)}{z(z^2+8)}dz$$ I tried using the fact that $displaystyle int_gamma f(z)dz=int_a^b f(gamma(t))gamma'(t)dt$ for $gamma(t)$ where $t in [a, b]$. But got nowhere. Is there a way I can transform the given integral into the form in which I can use the Integral Formula as stated below?



    $$f^{(k)}(w)=frac{1}{2pi i}int_gamma frac{f(z)}{(z-w)^{k+1}}dz$$ where $f^{(k)}(w)$ is the $k^{th}$ derivative of $f$






    complex-analysis complex-integration







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I ran into this question which hints me to use Cauchy's Integral Theorem for Derivatives, however I don't seem to be able to fit this integral into the form of the Integral Formula



      $$displaystyle int_{|z|=2} frac{cos(z)}{z(z^2+8)}dz$$ I tried using the fact that $displaystyle int_gamma f(z)dz=int_a^b f(gamma(t))gamma'(t)dt$ for $gamma(t)$ where $t in [a, b]$. But got nowhere. Is there a way I can transform the given integral into the form in which I can use the Integral Formula as stated below?



      $$f^{(k)}(w)=frac{1}{2pi i}int_gamma frac{f(z)}{(z-w)^{k+1}}dz$$ where $f^{(k)}(w)$ is the $k^{th}$ derivative of $f$






      complex-analysis complex-integration







      share|cite|improve this question













      I ran into this question which hints me to use Cauchy's Integral Theorem for Derivatives, however I don't seem to be able to fit this integral into the form of the Integral Formula



      $$displaystyle int_{|z|=2} frac{cos(z)}{z(z^2+8)}dz$$ I tried using the fact that $displaystyle int_gamma f(z)dz=int_a^b f(gamma(t))gamma'(t)dt$ for $gamma(t)$ where $t in [a, b]$. But got nowhere. Is there a way I can transform the given integral into the form in which I can use the Integral Formula as stated below?



      $$f^{(k)}(w)=frac{1}{2pi i}int_gamma frac{f(z)}{(z-w)^{k+1}}dz$$ where $f^{(k)}(w)$ is the $k^{th}$ derivative of $f$






      complex-analysis complex-integration




      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 22 at 16:49









      Paras Khosla

      449




      449






















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          1
          down vote



          accepted










          By Cauchy's integral theorem,$$oint_{lvert zrvert=2}frac{cos z}{z(z^2+8)},mathrm dz=oint_{lvert zrvert=2}frac{frac{cos z}{z^2+8}}z,mathrm dz=2pi itimesfrac{cos0}{0^2+8}=frac{pi i}4.$$






          share|cite|improve this answer




























            up vote
            1
            down vote













            The function $displaystyle f(z)=frac{cos(z)}{z^2+8}$ is analytic on $|z|=2$ then using Cauchy's Integral Theorem
            $$displaystyle int_{|z|=2} frac{cos(z)}{z(z^2+8)}dz=int_{|z|=2} dfrac{frac{cos(z)}{z^2+8}}{z}dz=2pi itimes f(0)=dfrac{pi i}{4}$$






            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%2f3009345%2fusing-the-cauchy-integral-theorem-for-derivatives-to-evaluate-an-integral%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








              up vote
              1
              down vote



              accepted










              By Cauchy's integral theorem,$$oint_{lvert zrvert=2}frac{cos z}{z(z^2+8)},mathrm dz=oint_{lvert zrvert=2}frac{frac{cos z}{z^2+8}}z,mathrm dz=2pi itimesfrac{cos0}{0^2+8}=frac{pi i}4.$$






              share|cite|improve this answer

























                up vote
                1
                down vote



                accepted










                By Cauchy's integral theorem,$$oint_{lvert zrvert=2}frac{cos z}{z(z^2+8)},mathrm dz=oint_{lvert zrvert=2}frac{frac{cos z}{z^2+8}}z,mathrm dz=2pi itimesfrac{cos0}{0^2+8}=frac{pi i}4.$$






                share|cite|improve this answer























                  up vote
                  1
                  down vote



                  accepted







                  up vote
                  1
                  down vote



                  accepted






                  By Cauchy's integral theorem,$$oint_{lvert zrvert=2}frac{cos z}{z(z^2+8)},mathrm dz=oint_{lvert zrvert=2}frac{frac{cos z}{z^2+8}}z,mathrm dz=2pi itimesfrac{cos0}{0^2+8}=frac{pi i}4.$$






                  share|cite|improve this answer












                  By Cauchy's integral theorem,$$oint_{lvert zrvert=2}frac{cos z}{z(z^2+8)},mathrm dz=oint_{lvert zrvert=2}frac{frac{cos z}{z^2+8}}z,mathrm dz=2pi itimesfrac{cos0}{0^2+8}=frac{pi i}4.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 22 at 16:52









                  José Carlos Santos

                  147k22117217




                  147k22117217






















                      up vote
                      1
                      down vote













                      The function $displaystyle f(z)=frac{cos(z)}{z^2+8}$ is analytic on $|z|=2$ then using Cauchy's Integral Theorem
                      $$displaystyle int_{|z|=2} frac{cos(z)}{z(z^2+8)}dz=int_{|z|=2} dfrac{frac{cos(z)}{z^2+8}}{z}dz=2pi itimes f(0)=dfrac{pi i}{4}$$






                      share|cite|improve this answer

























                        up vote
                        1
                        down vote













                        The function $displaystyle f(z)=frac{cos(z)}{z^2+8}$ is analytic on $|z|=2$ then using Cauchy's Integral Theorem
                        $$displaystyle int_{|z|=2} frac{cos(z)}{z(z^2+8)}dz=int_{|z|=2} dfrac{frac{cos(z)}{z^2+8}}{z}dz=2pi itimes f(0)=dfrac{pi i}{4}$$






                        share|cite|improve this answer























                          up vote
                          1
                          down vote










                          up vote
                          1
                          down vote









                          The function $displaystyle f(z)=frac{cos(z)}{z^2+8}$ is analytic on $|z|=2$ then using Cauchy's Integral Theorem
                          $$displaystyle int_{|z|=2} frac{cos(z)}{z(z^2+8)}dz=int_{|z|=2} dfrac{frac{cos(z)}{z^2+8}}{z}dz=2pi itimes f(0)=dfrac{pi i}{4}$$






                          share|cite|improve this answer












                          The function $displaystyle f(z)=frac{cos(z)}{z^2+8}$ is analytic on $|z|=2$ then using Cauchy's Integral Theorem
                          $$displaystyle int_{|z|=2} frac{cos(z)}{z(z^2+8)}dz=int_{|z|=2} dfrac{frac{cos(z)}{z^2+8}}{z}dz=2pi itimes f(0)=dfrac{pi i}{4}$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 22 at 16:53









                          Nosrati

                          26.3k62353




                          26.3k62353






























                              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%2f3009345%2fusing-the-cauchy-integral-theorem-for-derivatives-to-evaluate-an-integral%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

                              Ellipse (mathématiques)

                              Image
                              Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
                              Read more

                              Quarter-circle Tiles

                              Image
                              up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
                              Read more

                              Mont Emei

                              Image
                              Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
                              Read more