Evaluating the integral $int_=pifraczrightzdz$, where the function is not holomorphicEvaluating 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 $int_=2z^nsinleft(zright)dz$ 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
OP Amp not amplifying audio signal
How to stretch the corners of this image so that it looks like a perfect rectangle?
Do creatures with a listed speed of "0 ft., fly 30 ft. (hover)" ever touch the ground?
Getting extremely large arrows with tikzcd
Unlock My Phone! February 2018
What do you call someone who asks many questions?
How can saying a song's name be a copyright violation?
Does int main() need a declaration on C++?
Why is the sentence "Das ist eine Nase" correct?
Why is it a bad idea to hire a hitman to eliminate most corrupt politicians?
What's the meaning of "Sollensaussagen"?
What does the same-ish mean?
How do conventional missiles fly?
Ambiguity in the definition of entropy
Mathematica command that allows it to read my intentions
How does a refinance allow a mortgage to be repaid?
How to show a landlord what we have in savings?
Sums of two squares in arithmetic progressions
Forgetting the musical notes while performing in concert
Does the Idaho Potato Commission associate potato skins with healthy eating?
Can a virus destroy the BIOS of a modern computer?
Should I tell management that I intend to leave due to bad software development practices?
One verb to replace 'be a member of' a club
How to travel to Japan while expressing milk?
Evaluating the integral $int_leftfrace^-leftzdz$, 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 $int_=2z^nsinleft(zright)dz$ 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
$$int_leftfrace^-leftzdz.$$
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
$endgroup$
|
show 3 more comments
$begingroup$
I need to evaluate on the circle $left|zright|=pi$ the integral
$$int_leftfrace^-leftzdz.$$
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
$endgroup$
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
$$int_leftfrace^-leftzdz.$$
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
$endgroup$
I need to evaluate on the circle $left|zright|=pi$ the integral
$$int_leftfrace^-leftzdz.$$
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
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
|
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 $z = pi e^itheta, thetain[0,2pi]$:
$$
int_=pifrace^zzdz =
int_0^2pifracpi e^-pipi e^ithetapi i e^itheta = 2pi^2 i e^-pi.
$$
But... Cauchy formula can be used:
$$
int_=pifrace^zzdz =
int_=pifracpi e^-pizdz = 2pi^2 i e^-pi.
$$
$endgroup$
$begingroup$
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
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3169258%2fevaluating-the-integral-int-leftz-right-pi-frac-leftz-righte-left%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let be $z = pi e^itheta, thetain[0,2pi]$:
$$
int_=pifrace^zzdz =
int_0^2pifracpi e^-pipi e^ithetapi i e^itheta = 2pi^2 i e^-pi.
$$
But... Cauchy formula can be used:
$$
int_=pifrace^zzdz =
int_=pifracpi e^-pizdz = 2pi^2 i e^-pi.
$$
$endgroup$
$begingroup$
thank you, was what I was looking for :)
$endgroup$
– Dac0
2 days ago
add a comment |
$begingroup$
Let be $z = pi e^itheta, thetain[0,2pi]$:
$$
int_=pifrace^zzdz =
int_0^2pifracpi e^-pipi e^ithetapi i e^itheta = 2pi^2 i e^-pi.
$$
But... Cauchy formula can be used:
$$
int_=pifrace^zzdz =
int_=pifracpi e^-pizdz = 2pi^2 i e^-pi.
$$
$endgroup$
$begingroup$
thank you, was what I was looking for :)
$endgroup$
– Dac0
2 days ago
add a comment |
$begingroup$
Let be $z = pi e^itheta, thetain[0,2pi]$:
$$
int_=pifrace^zzdz =
int_0^2pifracpi e^-pipi e^ithetapi i e^itheta = 2pi^2 i e^-pi.
$$
But... Cauchy formula can be used:
$$
int_=pifrace^zzdz =
int_=pifracpi e^-pizdz = 2pi^2 i e^-pi.
$$
$endgroup$
Let be $z = pi e^itheta, thetain[0,2pi]$:
$$
int_=pifrace^zzdz =
int_0^2pifracpi e^-pipi e^ithetapi i e^itheta = 2pi^2 i e^-pi.
$$
But... Cauchy formula can be used:
$$
int_=pifrace^zzdz =
int_=pifracpi e^-pizdz = 2pi^2 i e^-pi.
$$
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
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
$endgroup$
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
$endgroup$
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
$endgroup$
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
edited 2 days ago
answered 2 days ago
José Carlos SantosJosé Carlos Santos
172k23132240
172k23132240
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3169258%2fevaluating-the-integral-int-leftz-right-pi-frac-leftz-righte-left%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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