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Evaluating the integral $int_=pifraczrightzdz$, where the function is not holomorphic


Evaluating the integral $int_C textRe z,dz$ from $-4$ to $4$ via lower half of the circleThe meaning of the Imaginary value of the Residue while Evaluating a Real Improper IntegralSupremum of holomorphic function on the unit diskIntegral of $int_-infty^infty left(frac1alpha + ix + frac1alpha - ixright)^2 , dx$Calculate $int_leftz^nsinleft(zright)dz$ for $nin mathbbZ$Improper integral of the form $I=int_0^infty fracsin(ax)x^2+b^2dx$How to integrate the complex function $f(z) = xy$ over the circle $C = [0, r]$Integral of holomorphic function is again holomorphicEvaluating a Path Integral in the Complex PlaneEvaluating $int_-infty^inftyfrace^axcoshxdx $ using contour integration













2












$begingroup$


I need to evaluate on the circle $left|zright|=pi$ the integral
$$int_=pifraczrightzdz.$$
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











$endgroup$







  • 1




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










  • $begingroup$
    Couple things: 1) when making a post like this please show your work 2) You say the function is not holomorphic "there" - just to be clear, it's holomorphic on the circle you're integrating on, just not inside the circle
    $endgroup$
    – Brevan Ellefsen
    2 days ago











  • $begingroup$
    @BrevanEllefsen why do you say it's holomorphic on the circle $|z|=pi$ can you show me how you got this? Using Cauchy Riemann equation in polar coordinates it seems it's holomorphic only on the circle $|z|=1$
    $endgroup$
    – Dac0
    2 days ago











  • $begingroup$
    @Dac0 your function is a scalar multiple of $int_D z^-1$ where $D$ is the circle of radius $pi$. The integrand is holomorphic everywhere but at the origin, which does not lie on your circle (it's inside it... In the exact middle)
    $endgroup$
    – Brevan Ellefsen
    2 days ago










  • $begingroup$
    @BrevanEllefsen I used for the Cauchy Riemann equation $fracpartial fpartial r=-fracirfracpartial fpartialtheta$ and then you have $fracpartial fpartial r=-e^-re^-itheta$ and $fracpartial fpartialtheta=-fracire^-re^-itheta$ so should be $r=1$. What am I doing wrong?
    $endgroup$
    – Dac0
    2 days ago
















2












$begingroup$


I need to evaluate on the circle $left|zright|=pi$ the integral
$$int_=pifraczrightzdz.$$
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











$endgroup$







  • 1




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










  • $begingroup$
    Couple things: 1) when making a post like this please show your work 2) You say the function is not holomorphic "there" - just to be clear, it's holomorphic on the circle you're integrating on, just not inside the circle
    $endgroup$
    – Brevan Ellefsen
    2 days ago











  • $begingroup$
    @BrevanEllefsen why do you say it's holomorphic on the circle $|z|=pi$ can you show me how you got this? Using Cauchy Riemann equation in polar coordinates it seems it's holomorphic only on the circle $|z|=1$
    $endgroup$
    – Dac0
    2 days ago











  • $begingroup$
    @Dac0 your function is a scalar multiple of $int_D z^-1$ where $D$ is the circle of radius $pi$. The integrand is holomorphic everywhere but at the origin, which does not lie on your circle (it's inside it... In the exact middle)
    $endgroup$
    – Brevan Ellefsen
    2 days ago










  • $begingroup$
    @BrevanEllefsen I used for the Cauchy Riemann equation $fracpartial fpartial r=-fracirfracpartial fpartialtheta$ and then you have $fracpartial fpartial r=-e^-re^-itheta$ and $fracpartial fpartialtheta=-fracire^-re^-itheta$ so should be $r=1$. What am I doing wrong?
    $endgroup$
    – Dac0
    2 days ago














2












2








2





$begingroup$


I need to evaluate on the circle $left|zright|=pi$ the integral
$$int_=pifraczrightzdz.$$
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
$$int_=pifraczrightzdz.$$
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 2 days ago









Asaf Karagila♦

307k33440773




307k33440773










asked 2 days ago









Dac0Dac0

6,0331937




6,0331937







  • 1




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










  • $begingroup$
    Couple things: 1) when making a post like this please show your work 2) You say the function is not holomorphic "there" - just to be clear, it's holomorphic on the circle you're integrating on, just not inside the circle
    $endgroup$
    – Brevan Ellefsen
    2 days ago











  • $begingroup$
    @BrevanEllefsen why do you say it's holomorphic on the circle $|z|=pi$ can you show me how you got this? Using Cauchy Riemann equation in polar coordinates it seems it's holomorphic only on the circle $|z|=1$
    $endgroup$
    – Dac0
    2 days ago











  • $begingroup$
    @Dac0 your function is a scalar multiple of $int_D z^-1$ where $D$ is the circle of radius $pi$. The integrand is holomorphic everywhere but at the origin, which does not lie on your circle (it's inside it... In the exact middle)
    $endgroup$
    – Brevan Ellefsen
    2 days ago










  • $begingroup$
    @BrevanEllefsen I used for the Cauchy Riemann equation $fracpartial fpartial r=-fracirfracpartial fpartialtheta$ and then you have $fracpartial fpartial r=-e^-re^-itheta$ and $fracpartial fpartialtheta=-fracire^-re^-itheta$ so should be $r=1$. What am I doing wrong?
    $endgroup$
    – Dac0
    2 days ago













  • 1




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










  • $begingroup$
    Couple things: 1) when making a post like this please show your work 2) You say the function is not holomorphic "there" - just to be clear, it's holomorphic on the circle you're integrating on, just not inside the circle
    $endgroup$
    – Brevan Ellefsen
    2 days ago











  • $begingroup$
    @BrevanEllefsen why do you say it's holomorphic on the circle $|z|=pi$ can you show me how you got this? Using Cauchy Riemann equation in polar coordinates it seems it's holomorphic only on the circle $|z|=1$
    $endgroup$
    – Dac0
    2 days ago











  • $begingroup$
    @Dac0 your function is a scalar multiple of $int_D z^-1$ where $D$ is the circle of radius $pi$. The integrand is holomorphic everywhere but at the origin, which does not lie on your circle (it's inside it... In the exact middle)
    $endgroup$
    – Brevan Ellefsen
    2 days ago










  • $begingroup$
    @BrevanEllefsen I used for the Cauchy Riemann equation $fracpartial fpartial r=-fracirfracpartial fpartialtheta$ and then you have $fracpartial fpartial r=-e^-re^-itheta$ and $fracpartial fpartialtheta=-fracire^-re^-itheta$ so should be $r=1$. What am I doing wrong?
    $endgroup$
    – Dac0
    2 days ago








1




1




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




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












$begingroup$
Couple things: 1) when making a post like this please show your work 2) You say the function is not holomorphic "there" - just to be clear, it's holomorphic on the circle you're integrating on, just not inside the circle
$endgroup$
– Brevan Ellefsen
2 days ago





$begingroup$
Couple things: 1) when making a post like this please show your work 2) You say the function is not holomorphic "there" - just to be clear, it's holomorphic on the circle you're integrating on, just not inside the circle
$endgroup$
– Brevan Ellefsen
2 days ago













$begingroup$
@BrevanEllefsen why do you say it's holomorphic on the circle $|z|=pi$ can you show me how you got this? Using Cauchy Riemann equation in polar coordinates it seems it's holomorphic only on the circle $|z|=1$
$endgroup$
– Dac0
2 days ago





$begingroup$
@BrevanEllefsen why do you say it's holomorphic on the circle $|z|=pi$ can you show me how you got this? Using Cauchy Riemann equation in polar coordinates it seems it's holomorphic only on the circle $|z|=1$
$endgroup$
– Dac0
2 days ago













$begingroup$
@Dac0 your function is a scalar multiple of $int_D z^-1$ where $D$ is the circle of radius $pi$. The integrand is holomorphic everywhere but at the origin, which does not lie on your circle (it's inside it... In the exact middle)
$endgroup$
– Brevan Ellefsen
2 days ago




$begingroup$
@Dac0 your function is a scalar multiple of $int_D z^-1$ where $D$ is the circle of radius $pi$. The integrand is holomorphic everywhere but at the origin, which does not lie on your circle (it's inside it... In the exact middle)
$endgroup$
– Brevan Ellefsen
2 days ago












$begingroup$
@BrevanEllefsen I used for the Cauchy Riemann equation $fracpartial fpartial r=-fracirfracpartial fpartialtheta$ and then you have $fracpartial fpartial r=-e^-re^-itheta$ and $fracpartial fpartialtheta=-fracire^-re^-itheta$ so should be $r=1$. What am I doing wrong?
$endgroup$
– Dac0
2 days ago





$begingroup$
@BrevanEllefsen I used for the Cauchy Riemann equation $fracpartial fpartial r=-fracirfracpartial fpartialtheta$ and then you have $fracpartial fpartial r=-e^-re^-itheta$ and $fracpartial fpartialtheta=-fracire^-re^-itheta$ so should be $r=1$. What am I doing wrong?
$endgroup$
– Dac0
2 days ago











2 Answers
2






active

oldest

votes


















5












$begingroup$

Let be $z = pi e^itheta, thetain[0,2pi]$:
$$
int_zfrace^-zdz =
int_0^2pifracpi e^-pipi e^ithetapi i e^itheta = 2pi^2 i e^-pi.
$$
But... Cauchy formula can be used:
$$
int_zfrace^-zdz =
int_zfracpi e^-pizdz = 2pi^2 i e^-pi.
$$






share|cite|improve this answer











$endgroup$












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


















7












$begingroup$

If $gamma(t)=pi e^it$ ($tin[0,2pi]$), thenbeginalignint_lvert zrvert=pifraclvert zrvert e^-lvert zrvertz,mathrm dz&=int_0^2pifracbigllvertgamma(t)bigrrvert e^-lvertgamma(t)rvertgamma(t)gamma'(t),mathrm dt\&=int_0^2pipi e^-pii,mathrm dt\&=2pi^2ie^-pi.endalign






share|cite|improve this answer











$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    Let be $z = pi e^itheta, thetain[0,2pi]$:
    $$
    int_zfrace^-zdz =
    int_0^2pifracpi e^-pipi e^ithetapi i e^itheta = 2pi^2 i e^-pi.
    $$
    But... Cauchy formula can be used:
    $$
    int_zfrace^-zdz =
    int_zfracpi e^-pizdz = 2pi^2 i e^-pi.
    $$






    share|cite|improve this answer











    $endgroup$












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















    5












    $begingroup$

    Let be $z = pi e^itheta, thetain[0,2pi]$:
    $$
    int_zfrace^-zdz =
    int_0^2pifracpi e^-pipi e^ithetapi i e^itheta = 2pi^2 i e^-pi.
    $$
    But... Cauchy formula can be used:
    $$
    int_zfrace^-zdz =
    int_zfracpi e^-pizdz = 2pi^2 i e^-pi.
    $$






    share|cite|improve this answer











    $endgroup$












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













    5












    5








    5





    $begingroup$

    Let be $z = pi e^itheta, thetain[0,2pi]$:
    $$
    int_zfrace^-zdz =
    int_0^2pifracpi e^-pipi e^ithetapi i e^itheta = 2pi^2 i e^-pi.
    $$
    But... Cauchy formula can be used:
    $$
    int_zfrace^-zdz =
    int_zfracpi e^-pizdz = 2pi^2 i e^-pi.
    $$






    share|cite|improve this answer











    $endgroup$



    Let be $z = pi e^itheta, thetain[0,2pi]$:
    $$
    int_zfrace^-zdz =
    int_0^2pifracpi e^-pipi e^ithetapi i e^itheta = 2pi^2 i e^-pi.
    $$
    But... Cauchy formula can be used:
    $$
    int_zfrace^-zdz =
    int_zfracpi e^-pizdz = 2pi^2 i e^-pi.
    $$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 2 days ago

























    answered 2 days ago









    Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

    35k42971




    35k42971











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
















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















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




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











    7












    $begingroup$

    If $gamma(t)=pi e^it$ ($tin[0,2pi]$), thenbeginalignint_lvert zrvert=pifraclvert zrvert e^-lvert zrvertz,mathrm dz&=int_0^2pifracbigllvertgamma(t)bigrrvert e^-lvertgamma(t)rvertgamma(t)gamma'(t),mathrm dt\&=int_0^2pipi e^-pii,mathrm dt\&=2pi^2ie^-pi.endalign






    share|cite|improve this answer











    $endgroup$

















      7












      $begingroup$

      If $gamma(t)=pi e^it$ ($tin[0,2pi]$), thenbeginalignint_lvert zrvert=pifraclvert zrvert e^-lvert zrvertz,mathrm dz&=int_0^2pifracbigllvertgamma(t)bigrrvert e^-lvertgamma(t)rvertgamma(t)gamma'(t),mathrm dt\&=int_0^2pipi e^-pii,mathrm dt\&=2pi^2ie^-pi.endalign






      share|cite|improve this answer











      $endgroup$















        7












        7








        7





        $begingroup$

        If $gamma(t)=pi e^it$ ($tin[0,2pi]$), thenbeginalignint_lvert zrvert=pifraclvert zrvert e^-lvert zrvertz,mathrm dz&=int_0^2pifracbigllvertgamma(t)bigrrvert e^-lvertgamma(t)rvertgamma(t)gamma'(t),mathrm dt\&=int_0^2pipi e^-pii,mathrm dt\&=2pi^2ie^-pi.endalign






        share|cite|improve this answer











        $endgroup$



        If $gamma(t)=pi e^it$ ($tin[0,2pi]$), thenbeginalignint_lvert zrvert=pifraclvert zrvert e^-lvert zrvertz,mathrm dz&=int_0^2pifracbigllvertgamma(t)bigrrvert e^-lvertgamma(t)rvertgamma(t)gamma'(t),mathrm dt\&=int_0^2pipi e^-pii,mathrm dt\&=2pi^2ie^-pi.endalign







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 days ago

























        answered 2 days ago









        José Carlos SantosJosé Carlos Santos

        172k23132240




        172k23132240



























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            三茅街道 概览 国家   中华人民共和国 上级行政区 扬中市 行政区类型 街道办事处 行政区划代码 32 11 82 001 村级区划单位数 30 - 社区数 14 - 行政村数 16 地理、人口、经济 坐标 32°14′25″N 119°49′08″E  /  32.24028°N 119.81884°E  / 32.24028; 119.81884 坐标： 32°14′25″N 119°49′08″E  /  32.24028°N 119.81884°E  / 32.24028; 119.81884 电话区号 +86  时区 UTC+8（北京时间） 三茅街道 是中华人民共和国江苏省镇江市扬中市下辖的一个街道办事处 [1] 。西部毗邻新坝镇，东南方毗邻油坊镇、八桥镇，北隔长江主航道與泰州市高港区相望。 2000年以后，丰裕镇并入，三茅镇面积达到49.45平方千米，人口73284人。2005年，兴隆镇并入，面积133.19平方千米，人口13万，占扬中全市40%。 行政区划 [ 编辑 ] 三茅街道下辖以下村级行政区划单位 [2] ： 文化新村社区、扬子新村社区、江州路社区、英雄社区、广宁社区、中桥社区、建设社区、金星社区、同心社区、丰裕社区、新民社区、港联社区、新胜社区、新扬社区、民主村、中华村、友好村、锦程村、三桥村、大众村、滨江村、企东村、兴华村、指南村、兴阳村、明华村、营房村、永和村、永勤村和普济村。 参考资料 [ 编辑 ] ↑ 2017年统计用区划代码和城乡划分代码：扬中市. 中华人民共和国国家统计局. 2017.   ↑ 2017年统计用区划代码和城乡划分代码：三茅街道. 中华人民共和国国家统计局. 2017.     这是一个关于江苏行政区划的小作品。你可以通过 编辑或修订 扩充其内容。 查 论 编 扬中市行政区划 行政隶属：中华人民共和国江苏省镇江市 街道 三茅街道 兴隆街道 镇 新坝镇 油坊镇 八桥镇 西来桥镇 类似乡级单位 扬中市良种繁育场 扬中市种猪场 扬中市渔业社 雷公岛水产养殖场 扬...
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