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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
$begingroup$
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
edited 2 days ago
Asaf Karagila♦
307k33440773
asked 2 days ago
Dac0Dac0
6,0331937
$endgroup$
|
show 3 more comments
$begingroup$
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
edited 2 days ago
Asaf Karagila♦
307k33440773
asked 2 days ago
Dac0Dac0
6,0331937
$endgroup$
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
|
show 3 more comments
$begingroup$
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
edited 2 days ago
Asaf Karagila♦
307k33440773
asked 2 days ago
Dac0Dac0
6,0331937
$endgroup$
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
edited 2 days ago
Asaf Karagila♦
307k33440773
asked 2 days ago
Dac0Dac0
6,0331937
edited 2 days ago
Asaf Karagila♦
307k33440773
asked 2 days ago
Dac0Dac0
6,0331937
edited 2 days ago
Asaf Karagila♦
307k33440773
edited 2 days ago
Asaf Karagila♦
307k33440773
edited 2 days ago
Asaf Karagila♦
307k33440773
307k33440773
asked 2 days ago
Dac0Dac0
6,0331937
asked 2 days ago
Dac0Dac0
6,0331937
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
|
show 3 more comments
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
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
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:
$$
int_zfrace^-zdz =
int_zfracpi e^-pizdz = 2pi^2 i e^-pi.
$$
answered 2 days ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35k42971
$endgroup$
$begingroup$
thank you, was what I was looking for :)
$endgroup$
– Dac0
2 days ago
add a comment |
$begingroup$
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
answered 2 days ago
José Carlos SantosJosé Carlos Santos
172k23132240
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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:
$$
int_zfrace^-zdz =
int_zfracpi e^-pizdz = 2pi^2 i e^-pi.
$$
answered 2 days ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35k42971
$endgroup$
$begingroup$
thank you, was what I was looking for :)
$endgroup$
– Dac0
2 days ago
add a comment |
$begingroup$
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:
$$
int_zfrace^-zdz =
int_zfracpi e^-pizdz = 2pi^2 i e^-pi.
$$
answered 2 days ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35k42971
$endgroup$
$begingroup$
thank you, was what I was looking for :)
$endgroup$
– Dac0
2 days ago
add a comment |
$begingroup$
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:
$$
int_zfrace^-zdz =
int_zfracpi e^-pizdz = 2pi^2 i e^-pi.
$$
answered 2 days ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35k42971
$endgroup$
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:
$$
int_zfrace^-zdz =
int_zfracpi e^-pizdz = 2pi^2 i e^-pi.
$$
answered 2 days ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35k42971
edited 2 days ago
answered 2 days ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35k42971
answered 2 days ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35k42971
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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
answered 2 days ago
José Carlos SantosJosé Carlos Santos
172k23132240
$endgroup$
add a comment |
$begingroup$
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
answered 2 days ago
José Carlos SantosJosé Carlos Santos
172k23132240
$endgroup$
add a comment |
$begingroup$
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
answered 2 days ago
José Carlos SantosJosé Carlos Santos
172k23132240
$endgroup$
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
answered 2 days ago
José Carlos SantosJosé Carlos Santos
172k23132240
edited 2 days ago
answered 2 days ago
José Carlos SantosJosé Carlos Santos
172k23132240
answered 2 days ago
José Carlos SantosJosé Carlos Santos
172k23132240
answered 2 days ago
José Carlos SantosJosé Carlos Santos
172k23132240
172k23132240
add a comment |
add a comment |
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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