Integral in polar coordinate on a circle where the function is non holomorphic












2












$begingroup$


I need to evaluate on the circle $left|zright|=pi$ the integral
$$int_{left|zright|=pi}frac{left|zright|e^{-left|zright|}}{z}dz.$$
The function is not holomorphic there. Anyway, I tried to integrate it using polar coordinates and simplyfing the modulo and I got $2pi e^{-pi}$ while the result should be $2pi^2 ie^{-pi}$.
I'm sure is trivial and I overlooked a stupid error. Can anybody tell me where?






complex-analysis contour-integration







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  • $begingroup$
    Possible mistake: forgotten $pi i$ in $dz$.
    $endgroup$
    – Martín-Blas Pérez Pinilla
    5 hours ago
















2












$begingroup$


I need to evaluate on the circle $left|zright|=pi$ the integral
$$int_{left|zright|=pi}frac{left|zright|e^{-left|zright|}}{z}dz.$$
The function is not holomorphic there. Anyway, I tried to integrate it using polar coordinates and simplyfing the modulo and I got $2pi e^{-pi}$ while the result should be $2pi^2 ie^{-pi}$.
I'm sure is trivial and I overlooked a stupid error. Can anybody tell me where?






complex-analysis contour-integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    Possible mistake: forgotten $pi i$ in $dz$.
    $endgroup$
    – Martín-Blas Pérez Pinilla
    5 hours ago














2












2








2





$begingroup$


I need to evaluate on the circle $left|zright|=pi$ the integral
$$int_{left|zright|=pi}frac{left|zright|e^{-left|zright|}}{z}dz.$$
The function is not holomorphic there. Anyway, I tried to integrate it using polar coordinates and simplyfing the modulo and I got $2pi e^{-pi}$ while the result should be $2pi^2 ie^{-pi}$.
I'm sure is trivial and I overlooked a stupid error. Can anybody tell me where?






complex-analysis contour-integration







share|cite|improve this question











$endgroup$




I need to evaluate on the circle $left|zright|=pi$ the integral
$$int_{left|zright|=pi}frac{left|zright|e^{-left|zright|}}{z}dz.$$
The function is not holomorphic there. Anyway, I tried to integrate it using polar coordinates and simplyfing the modulo and I got $2pi e^{-pi}$ while the result should be $2pi^2 ie^{-pi}$.
I'm sure is trivial and I overlooked a stupid error. Can anybody tell me where?






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








edited 5 hours ago









José Carlos Santos

171k23132240




171k23132240










asked 5 hours ago









Dac0Dac0

5,9861936




5,9861936












  • $begingroup$
    Possible mistake: forgotten $pi i$ in $dz$.
    $endgroup$
    – Martín-Blas Pérez Pinilla
    5 hours ago


















  • $begingroup$
    Possible mistake: forgotten $pi i$ in $dz$.
    $endgroup$
    – Martín-Blas Pérez Pinilla
    5 hours ago
















$begingroup$
Possible mistake: forgotten $pi i$ in $dz$.
$endgroup$
– Martín-Blas Pérez Pinilla
5 hours ago




$begingroup$
Possible mistake: forgotten $pi i$ in $dz$.
$endgroup$
– Martín-Blas Pérez Pinilla
5 hours ago










2 Answers
2






active

oldest

votes


















4












$begingroup$

Let be $z = pi e^{itheta}, thetain[0,2pi]$:
$$
int_{|z|=pi}frac{|z|e^{-|z|}}{z}dz =
int_0^{2pi}frac{pi e^{-pi}}{pi e^{itheta}}pi i e^{itheta} = 2pi^2 i e^{-pi}.
$$
But... Cauchy formula can be used:
$$
int_{|z|=pi}frac{|z|e^{-|z|}}{z}dz =
int_{|z|=pi}frac{pi e^{-pi}}{z}dz = 2pi^2 i e^{-pi}.
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thank you, was what I was looking for :)
    $endgroup$
    – Dac0
    5 hours ago



















5












$begingroup$

If $gamma(t)=pi e^{it}$ ($tin[0,2pi]$), thenbegin{align}int_{lvert zrvert=pi}frac{lvert zrvert e^{-lvert zrvert}}z,mathrm dz&=int_0^{2pi}frac{bigllvertgamma(t)bigrrvert e^{-lvertgamma(t)rvert}}{gamma(t)}gamma'(t),mathrm dt\&=int_0^{2pi}pi e^{-pi}i,mathrm dt\&=2pi^2ie^{-pi}.end{align}






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    Let be $z = pi e^{itheta}, thetain[0,2pi]$:
    $$
    int_{|z|=pi}frac{|z|e^{-|z|}}{z}dz =
    int_0^{2pi}frac{pi e^{-pi}}{pi e^{itheta}}pi i e^{itheta} = 2pi^2 i e^{-pi}.
    $$
    But... Cauchy formula can be used:
    $$
    int_{|z|=pi}frac{|z|e^{-|z|}}{z}dz =
    int_{|z|=pi}frac{pi e^{-pi}}{z}dz = 2pi^2 i e^{-pi}.
    $$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      thank you, was what I was looking for :)
      $endgroup$
      – Dac0
      5 hours ago
















    4












    $begingroup$

    Let be $z = pi e^{itheta}, thetain[0,2pi]$:
    $$
    int_{|z|=pi}frac{|z|e^{-|z|}}{z}dz =
    int_0^{2pi}frac{pi e^{-pi}}{pi e^{itheta}}pi i e^{itheta} = 2pi^2 i e^{-pi}.
    $$
    But... Cauchy formula can be used:
    $$
    int_{|z|=pi}frac{|z|e^{-|z|}}{z}dz =
    int_{|z|=pi}frac{pi e^{-pi}}{z}dz = 2pi^2 i e^{-pi}.
    $$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      thank you, was what I was looking for :)
      $endgroup$
      – Dac0
      5 hours ago














    4












    4








    4





    $begingroup$

    Let be $z = pi e^{itheta}, thetain[0,2pi]$:
    $$
    int_{|z|=pi}frac{|z|e^{-|z|}}{z}dz =
    int_0^{2pi}frac{pi e^{-pi}}{pi e^{itheta}}pi i e^{itheta} = 2pi^2 i e^{-pi}.
    $$
    But... Cauchy formula can be used:
    $$
    int_{|z|=pi}frac{|z|e^{-|z|}}{z}dz =
    int_{|z|=pi}frac{pi e^{-pi}}{z}dz = 2pi^2 i e^{-pi}.
    $$






    share|cite|improve this answer











    $endgroup$



    Let be $z = pi e^{itheta}, thetain[0,2pi]$:
    $$
    int_{|z|=pi}frac{|z|e^{-|z|}}{z}dz =
    int_0^{2pi}frac{pi e^{-pi}}{pi e^{itheta}}pi i e^{itheta} = 2pi^2 i e^{-pi}.
    $$
    But... Cauchy formula can be used:
    $$
    int_{|z|=pi}frac{|z|e^{-|z|}}{z}dz =
    int_{|z|=pi}frac{pi e^{-pi}}{z}dz = 2pi^2 i e^{-pi}.
    $$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 5 hours ago

























    answered 5 hours ago









    Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

    35k42971




    35k42971












    • $begingroup$
      thank you, was what I was looking for :)
      $endgroup$
      – Dac0
      5 hours ago


















    • $begingroup$
      thank you, was what I was looking for :)
      $endgroup$
      – Dac0
      5 hours ago
















    $begingroup$
    thank you, was what I was looking for :)
    $endgroup$
    – Dac0
    5 hours ago




    $begingroup$
    thank you, was what I was looking for :)
    $endgroup$
    – Dac0
    5 hours ago











    5












    $begingroup$

    If $gamma(t)=pi e^{it}$ ($tin[0,2pi]$), thenbegin{align}int_{lvert zrvert=pi}frac{lvert zrvert e^{-lvert zrvert}}z,mathrm dz&=int_0^{2pi}frac{bigllvertgamma(t)bigrrvert e^{-lvertgamma(t)rvert}}{gamma(t)}gamma'(t),mathrm dt\&=int_0^{2pi}pi e^{-pi}i,mathrm dt\&=2pi^2ie^{-pi}.end{align}






    share|cite|improve this answer











    $endgroup$


















      5












      $begingroup$

      If $gamma(t)=pi e^{it}$ ($tin[0,2pi]$), thenbegin{align}int_{lvert zrvert=pi}frac{lvert zrvert e^{-lvert zrvert}}z,mathrm dz&=int_0^{2pi}frac{bigllvertgamma(t)bigrrvert e^{-lvertgamma(t)rvert}}{gamma(t)}gamma'(t),mathrm dt\&=int_0^{2pi}pi e^{-pi}i,mathrm dt\&=2pi^2ie^{-pi}.end{align}






      share|cite|improve this answer











      $endgroup$
















        5












        5








        5





        $begingroup$

        If $gamma(t)=pi e^{it}$ ($tin[0,2pi]$), thenbegin{align}int_{lvert zrvert=pi}frac{lvert zrvert e^{-lvert zrvert}}z,mathrm dz&=int_0^{2pi}frac{bigllvertgamma(t)bigrrvert e^{-lvertgamma(t)rvert}}{gamma(t)}gamma'(t),mathrm dt\&=int_0^{2pi}pi e^{-pi}i,mathrm dt\&=2pi^2ie^{-pi}.end{align}






        share|cite|improve this answer











        $endgroup$



        If $gamma(t)=pi e^{it}$ ($tin[0,2pi]$), thenbegin{align}int_{lvert zrvert=pi}frac{lvert zrvert e^{-lvert zrvert}}z,mathrm dz&=int_0^{2pi}frac{bigllvertgamma(t)bigrrvert e^{-lvertgamma(t)rvert}}{gamma(t)}gamma'(t),mathrm dt\&=int_0^{2pi}pi e^{-pi}i,mathrm dt\&=2pi^2ie^{-pi}.end{align}







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 5 hours ago

























        answered 5 hours ago









        José Carlos SantosJosé Carlos Santos

        171k23132240




        171k23132240






























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