Using the Cauchy Integral Theorem for Derivatives to evaluate an integral












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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?



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






complex-analysis complex-integration







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    0












    $begingroup$


    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?



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






    complex-analysis complex-integration







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      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?



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






      complex-analysis complex-integration







      share|cite|improve this question











      $endgroup$




      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?



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






      complex-analysis complex-integration




      complex-analysis complex-integration






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      share|cite|improve this question













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      edited Jan 11 at 12:43







      Paras Khosla

















      asked Nov 22 '18 at 16:49









      Paras KhoslaParas Khosla

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






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            $begingroup$

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






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              2 Answers
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              2 Answers
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              $begingroup$

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






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                1












                $begingroup$

                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









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                  1












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                  1





                  $begingroup$

                  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









                  $endgroup$



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







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                  answered Nov 22 '18 at 16:52









                  José Carlos SantosJosé Carlos Santos

                  160k22126232




                  160k22126232























                      1












                      $begingroup$

                      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









                      $endgroup$


















                        1












                        $begingroup$

                        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









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          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









                          $endgroup$



                          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 '18 at 16:53









                          NosratiNosrati

                          26.5k62354




                          26.5k62354






























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