How transform this integral to the “Cauchy's integral”












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I would like to know is there any tricks to transform this
$$frac { 1 }{ 2pi } int _{ -pi }^{ pi }{ frac { fleft( t right) }{ { e }^{ -it }-{ e }^{ ix } } dt } $$ integral to the $$frac { 1 }{ 2pi i } int _{ gamma }^{ }{ frac { fleft( t right) }{ t-xi } dt } $$
Thank you in advance.






integration complex-analysis







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    $begingroup$


    I would like to know is there any tricks to transform this
    $$frac { 1 }{ 2pi } int _{ -pi }^{ pi }{ frac { fleft( t right) }{ { e }^{ -it }-{ e }^{ ix } } dt } $$ integral to the $$frac { 1 }{ 2pi i } int _{ gamma }^{ }{ frac { fleft( t right) }{ t-xi } dt } $$
    Thank you in advance.






    integration complex-analysis







    share|cite|improve this question









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      0












      0








      0





      $begingroup$


      I would like to know is there any tricks to transform this
      $$frac { 1 }{ 2pi } int _{ -pi }^{ pi }{ frac { fleft( t right) }{ { e }^{ -it }-{ e }^{ ix } } dt } $$ integral to the $$frac { 1 }{ 2pi i } int _{ gamma }^{ }{ frac { fleft( t right) }{ t-xi } dt } $$
      Thank you in advance.






      integration complex-analysis







      share|cite|improve this question









      $endgroup$




      I would like to know is there any tricks to transform this
      $$frac { 1 }{ 2pi } int _{ -pi }^{ pi }{ frac { fleft( t right) }{ { e }^{ -it }-{ e }^{ ix } } dt } $$ integral to the $$frac { 1 }{ 2pi i } int _{ gamma }^{ }{ frac { fleft( t right) }{ t-xi } dt } $$
      Thank you in advance.






      integration complex-analysis




      integration complex-analysis






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      asked Jan 22 at 21:14









      haqnaturalhaqnatural

      20.8k72457




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          $begingroup$

          Starting with



          $$dfrac{1}{2pi i} int_{gamma}{dfrac{gleft(zright)}{z-xi}dz}$$



          make the substitutions



          $$xi = e^{-ix}$$
          $$g(z) = -e^{-ix}fleft(-ilogleft(zright)-2kpiright)$$



          and specify the contour as a path on the unit circle from an angle of $-pi$ to an angle of $pi$



          $$z= e^{it} quad -pi le t lt pi$$
          $$dz = ie^{it}dt$$



          so



          $$begin{align*} dfrac{1}{2pi i} int_{gamma}{dfrac{gleft(zright)}{z-xi}dz} &= dfrac{1}{2pi i} int_{gamma}{dfrac{-e^{-ix}fleft(-ilogleft(zright)-2kpiright)}{z-e^{-ix}}dz}\
          \
          &= dfrac{1}{2pi i} int_{-pi}^{pi}{dfrac{-e^{-ix}fleft(-ilogleft(e^{it}right)-2kpiright)ie^{it}}{e^{it}-e^{-ix}}dt}\
          \
          &= dfrac{1}{2pi } int_{-pi}^{pi}{dfrac{fleft(-ilogleft(e^{it}right)-2kpiright)}{e^{-it}-e^{ix}}dt}\
          \
          &= dfrac{1}{2pi } int_{-pi}^{pi}{dfrac{fleft(-ileft[ln|1|+i(t+2kpi)right]-2kpiright)}{e^{-it}-e^{ix}}dt}\
          \
          &= dfrac{1}{2pi } int_{-pi}^{pi}{dfrac{fleft(tright)}{e^{-it}-e^{ix}}dt}\
          end{align*}$$



          If you elect to work with the principal branch of the complex logarithm, then you can just set $k=0$ and use $mathrm{Log}()$ instead of $log()$.






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            1 Answer
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            1 Answer
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            0












            $begingroup$

            Starting with



            $$dfrac{1}{2pi i} int_{gamma}{dfrac{gleft(zright)}{z-xi}dz}$$



            make the substitutions



            $$xi = e^{-ix}$$
            $$g(z) = -e^{-ix}fleft(-ilogleft(zright)-2kpiright)$$



            and specify the contour as a path on the unit circle from an angle of $-pi$ to an angle of $pi$



            $$z= e^{it} quad -pi le t lt pi$$
            $$dz = ie^{it}dt$$



            so



            $$begin{align*} dfrac{1}{2pi i} int_{gamma}{dfrac{gleft(zright)}{z-xi}dz} &= dfrac{1}{2pi i} int_{gamma}{dfrac{-e^{-ix}fleft(-ilogleft(zright)-2kpiright)}{z-e^{-ix}}dz}\
            \
            &= dfrac{1}{2pi i} int_{-pi}^{pi}{dfrac{-e^{-ix}fleft(-ilogleft(e^{it}right)-2kpiright)ie^{it}}{e^{it}-e^{-ix}}dt}\
            \
            &= dfrac{1}{2pi } int_{-pi}^{pi}{dfrac{fleft(-ilogleft(e^{it}right)-2kpiright)}{e^{-it}-e^{ix}}dt}\
            \
            &= dfrac{1}{2pi } int_{-pi}^{pi}{dfrac{fleft(-ileft[ln|1|+i(t+2kpi)right]-2kpiright)}{e^{-it}-e^{ix}}dt}\
            \
            &= dfrac{1}{2pi } int_{-pi}^{pi}{dfrac{fleft(tright)}{e^{-it}-e^{ix}}dt}\
            end{align*}$$



            If you elect to work with the principal branch of the complex logarithm, then you can just set $k=0$ and use $mathrm{Log}()$ instead of $log()$.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Starting with



              $$dfrac{1}{2pi i} int_{gamma}{dfrac{gleft(zright)}{z-xi}dz}$$



              make the substitutions



              $$xi = e^{-ix}$$
              $$g(z) = -e^{-ix}fleft(-ilogleft(zright)-2kpiright)$$



              and specify the contour as a path on the unit circle from an angle of $-pi$ to an angle of $pi$



              $$z= e^{it} quad -pi le t lt pi$$
              $$dz = ie^{it}dt$$



              so



              $$begin{align*} dfrac{1}{2pi i} int_{gamma}{dfrac{gleft(zright)}{z-xi}dz} &= dfrac{1}{2pi i} int_{gamma}{dfrac{-e^{-ix}fleft(-ilogleft(zright)-2kpiright)}{z-e^{-ix}}dz}\
              \
              &= dfrac{1}{2pi i} int_{-pi}^{pi}{dfrac{-e^{-ix}fleft(-ilogleft(e^{it}right)-2kpiright)ie^{it}}{e^{it}-e^{-ix}}dt}\
              \
              &= dfrac{1}{2pi } int_{-pi}^{pi}{dfrac{fleft(-ilogleft(e^{it}right)-2kpiright)}{e^{-it}-e^{ix}}dt}\
              \
              &= dfrac{1}{2pi } int_{-pi}^{pi}{dfrac{fleft(-ileft[ln|1|+i(t+2kpi)right]-2kpiright)}{e^{-it}-e^{ix}}dt}\
              \
              &= dfrac{1}{2pi } int_{-pi}^{pi}{dfrac{fleft(tright)}{e^{-it}-e^{ix}}dt}\
              end{align*}$$



              If you elect to work with the principal branch of the complex logarithm, then you can just set $k=0$ and use $mathrm{Log}()$ instead of $log()$.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Starting with



                $$dfrac{1}{2pi i} int_{gamma}{dfrac{gleft(zright)}{z-xi}dz}$$



                make the substitutions



                $$xi = e^{-ix}$$
                $$g(z) = -e^{-ix}fleft(-ilogleft(zright)-2kpiright)$$



                and specify the contour as a path on the unit circle from an angle of $-pi$ to an angle of $pi$



                $$z= e^{it} quad -pi le t lt pi$$
                $$dz = ie^{it}dt$$



                so



                $$begin{align*} dfrac{1}{2pi i} int_{gamma}{dfrac{gleft(zright)}{z-xi}dz} &= dfrac{1}{2pi i} int_{gamma}{dfrac{-e^{-ix}fleft(-ilogleft(zright)-2kpiright)}{z-e^{-ix}}dz}\
                \
                &= dfrac{1}{2pi i} int_{-pi}^{pi}{dfrac{-e^{-ix}fleft(-ilogleft(e^{it}right)-2kpiright)ie^{it}}{e^{it}-e^{-ix}}dt}\
                \
                &= dfrac{1}{2pi } int_{-pi}^{pi}{dfrac{fleft(-ilogleft(e^{it}right)-2kpiright)}{e^{-it}-e^{ix}}dt}\
                \
                &= dfrac{1}{2pi } int_{-pi}^{pi}{dfrac{fleft(-ileft[ln|1|+i(t+2kpi)right]-2kpiright)}{e^{-it}-e^{ix}}dt}\
                \
                &= dfrac{1}{2pi } int_{-pi}^{pi}{dfrac{fleft(tright)}{e^{-it}-e^{ix}}dt}\
                end{align*}$$



                If you elect to work with the principal branch of the complex logarithm, then you can just set $k=0$ and use $mathrm{Log}()$ instead of $log()$.






                share|cite|improve this answer









                $endgroup$



                Starting with



                $$dfrac{1}{2pi i} int_{gamma}{dfrac{gleft(zright)}{z-xi}dz}$$



                make the substitutions



                $$xi = e^{-ix}$$
                $$g(z) = -e^{-ix}fleft(-ilogleft(zright)-2kpiright)$$



                and specify the contour as a path on the unit circle from an angle of $-pi$ to an angle of $pi$



                $$z= e^{it} quad -pi le t lt pi$$
                $$dz = ie^{it}dt$$



                so



                $$begin{align*} dfrac{1}{2pi i} int_{gamma}{dfrac{gleft(zright)}{z-xi}dz} &= dfrac{1}{2pi i} int_{gamma}{dfrac{-e^{-ix}fleft(-ilogleft(zright)-2kpiright)}{z-e^{-ix}}dz}\
                \
                &= dfrac{1}{2pi i} int_{-pi}^{pi}{dfrac{-e^{-ix}fleft(-ilogleft(e^{it}right)-2kpiright)ie^{it}}{e^{it}-e^{-ix}}dt}\
                \
                &= dfrac{1}{2pi } int_{-pi}^{pi}{dfrac{fleft(-ilogleft(e^{it}right)-2kpiright)}{e^{-it}-e^{ix}}dt}\
                \
                &= dfrac{1}{2pi } int_{-pi}^{pi}{dfrac{fleft(-ileft[ln|1|+i(t+2kpi)right]-2kpiright)}{e^{-it}-e^{ix}}dt}\
                \
                &= dfrac{1}{2pi } int_{-pi}^{pi}{dfrac{fleft(tright)}{e^{-it}-e^{ix}}dt}\
                end{align*}$$



                If you elect to work with the principal branch of the complex logarithm, then you can just set $k=0$ and use $mathrm{Log}()$ instead of $log()$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 23 at 1:15









                Andy WallsAndy Walls

                1,754139




                1,754139






























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