Solving a complex integral $oint_Lfrac{e^{1/(z-a)}}zdz$ using Cauchy's formula
$begingroup$
I am practicing complex integration using Cauchy's formula, and I ran into a problem. The following integral:
$$oint_{L}{frac{e^{frac{1}{z-a}}}{z}}dz$$ where $$L={zinmathbb{C}:|z|=r}$$ for some $rneq|a|$.
So the contour is a circle in the origin that can not pass through one of the singularities $a$. The other singularity being $0$. That means it's eather of a greater or lesser radius than $|a|$. If it is of a radius lesser than $|a|$, then by Cauchy's formula:
$$oint_{L}{frac{e^{frac{1}{z-a}}}{(z-0)}}dz = frac{2pi i}{e^{1/a}}$$
If, however, $r>|a|$ then the integral would be the sum of the integrals of two separate disjunct contours containing only the $0$ singularity and only the $a$ singularity. This is where I'm stuck. I'm not sure how to solve the integral of the contour containing the $a$ singularity (can't get it in Cauchy formula form).
integration complex-analysis complex-integration cauchy-integral-formula
$endgroup$
add a comment |
$begingroup$
I am practicing complex integration using Cauchy's formula, and I ran into a problem. The following integral:
$$oint_{L}{frac{e^{frac{1}{z-a}}}{z}}dz$$ where $$L={zinmathbb{C}:|z|=r}$$ for some $rneq|a|$.
So the contour is a circle in the origin that can not pass through one of the singularities $a$. The other singularity being $0$. That means it's eather of a greater or lesser radius than $|a|$. If it is of a radius lesser than $|a|$, then by Cauchy's formula:
$$oint_{L}{frac{e^{frac{1}{z-a}}}{(z-0)}}dz = frac{2pi i}{e^{1/a}}$$
If, however, $r>|a|$ then the integral would be the sum of the integrals of two separate disjunct contours containing only the $0$ singularity and only the $a$ singularity. This is where I'm stuck. I'm not sure how to solve the integral of the contour containing the $a$ singularity (can't get it in Cauchy formula form).
integration complex-analysis complex-integration cauchy-integral-formula
$endgroup$
add a comment |
$begingroup$
I am practicing complex integration using Cauchy's formula, and I ran into a problem. The following integral:
$$oint_{L}{frac{e^{frac{1}{z-a}}}{z}}dz$$ where $$L={zinmathbb{C}:|z|=r}$$ for some $rneq|a|$.
So the contour is a circle in the origin that can not pass through one of the singularities $a$. The other singularity being $0$. That means it's eather of a greater or lesser radius than $|a|$. If it is of a radius lesser than $|a|$, then by Cauchy's formula:
$$oint_{L}{frac{e^{frac{1}{z-a}}}{(z-0)}}dz = frac{2pi i}{e^{1/a}}$$
If, however, $r>|a|$ then the integral would be the sum of the integrals of two separate disjunct contours containing only the $0$ singularity and only the $a$ singularity. This is where I'm stuck. I'm not sure how to solve the integral of the contour containing the $a$ singularity (can't get it in Cauchy formula form).
integration complex-analysis complex-integration cauchy-integral-formula
$endgroup$
I am practicing complex integration using Cauchy's formula, and I ran into a problem. The following integral:
$$oint_{L}{frac{e^{frac{1}{z-a}}}{z}}dz$$ where $$L={zinmathbb{C}:|z|=r}$$ for some $rneq|a|$.
So the contour is a circle in the origin that can not pass through one of the singularities $a$. The other singularity being $0$. That means it's eather of a greater or lesser radius than $|a|$. If it is of a radius lesser than $|a|$, then by Cauchy's formula:
$$oint_{L}{frac{e^{frac{1}{z-a}}}{(z-0)}}dz = frac{2pi i}{e^{1/a}}$$
If, however, $r>|a|$ then the integral would be the sum of the integrals of two separate disjunct contours containing only the $0$ singularity and only the $a$ singularity. This is where I'm stuck. I'm not sure how to solve the integral of the contour containing the $a$ singularity (can't get it in Cauchy formula form).
integration complex-analysis complex-integration cauchy-integral-formula
integration complex-analysis complex-integration cauchy-integral-formula
edited Jan 2 at 15:59
Did
249k23226466
249k23226466
asked Jan 2 at 15:56
math101math101
597
597
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint
I would suggest to use Residue theorem here. This theorem is a consequence of Cauchy’s formula.
For a radius $r > vert a vert$, you get
$$oint_{L}{frac{e^{frac{1}{z-a}}}{z}}dz =2ipi left(frac{1}{e^{1/a}}+mbox{Res}(f,a)right)$$
$endgroup$
$begingroup$
Yes, I tried to, but I'm not sure how to get $Res(f,a)$ to be equal $frac{1}{a}$.
$endgroup$
– math101
Jan 2 at 16:37
$begingroup$
I edited because I may have done an error in The computation of the residue. You can compute it by finding $a_{-1}$ where $sum_{-infty}^infty a_n (z-a)^n$ is Laurent expansion of $f$ at $a$.
$endgroup$
– mathcounterexamples.net
Jan 2 at 16:51
$begingroup$
I see. I'm struggling with that too a bit, but from what I've done so far, I don't think the series is going to have any negative terms ?
$endgroup$
– math101
Jan 2 at 17:07
add a comment |
$begingroup$
See this answer. Instead of taking the residue at $z = a$, take the residue at $z = infty$, which is the value of $-e^{1/(z - a)}$ at $z = infty$.
$endgroup$
$begingroup$
This is the better approach.
$endgroup$
– Szeto
Jan 3 at 13:32
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%2f3059644%2fsolving-a-complex-integral-oint-l-frace1-z-azdz-using-cauchys-formula%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
$begingroup$
Hint
I would suggest to use Residue theorem here. This theorem is a consequence of Cauchy’s formula.
For a radius $r > vert a vert$, you get
$$oint_{L}{frac{e^{frac{1}{z-a}}}{z}}dz =2ipi left(frac{1}{e^{1/a}}+mbox{Res}(f,a)right)$$
$endgroup$
$begingroup$
Yes, I tried to, but I'm not sure how to get $Res(f,a)$ to be equal $frac{1}{a}$.
$endgroup$
– math101
Jan 2 at 16:37
$begingroup$
I edited because I may have done an error in The computation of the residue. You can compute it by finding $a_{-1}$ where $sum_{-infty}^infty a_n (z-a)^n$ is Laurent expansion of $f$ at $a$.
$endgroup$
– mathcounterexamples.net
Jan 2 at 16:51
$begingroup$
I see. I'm struggling with that too a bit, but from what I've done so far, I don't think the series is going to have any negative terms ?
$endgroup$
– math101
Jan 2 at 17:07
add a comment |
$begingroup$
Hint
I would suggest to use Residue theorem here. This theorem is a consequence of Cauchy’s formula.
For a radius $r > vert a vert$, you get
$$oint_{L}{frac{e^{frac{1}{z-a}}}{z}}dz =2ipi left(frac{1}{e^{1/a}}+mbox{Res}(f,a)right)$$
$endgroup$
$begingroup$
Yes, I tried to, but I'm not sure how to get $Res(f,a)$ to be equal $frac{1}{a}$.
$endgroup$
– math101
Jan 2 at 16:37
$begingroup$
I edited because I may have done an error in The computation of the residue. You can compute it by finding $a_{-1}$ where $sum_{-infty}^infty a_n (z-a)^n$ is Laurent expansion of $f$ at $a$.
$endgroup$
– mathcounterexamples.net
Jan 2 at 16:51
$begingroup$
I see. I'm struggling with that too a bit, but from what I've done so far, I don't think the series is going to have any negative terms ?
$endgroup$
– math101
Jan 2 at 17:07
add a comment |
$begingroup$
Hint
I would suggest to use Residue theorem here. This theorem is a consequence of Cauchy’s formula.
For a radius $r > vert a vert$, you get
$$oint_{L}{frac{e^{frac{1}{z-a}}}{z}}dz =2ipi left(frac{1}{e^{1/a}}+mbox{Res}(f,a)right)$$
$endgroup$
Hint
I would suggest to use Residue theorem here. This theorem is a consequence of Cauchy’s formula.
For a radius $r > vert a vert$, you get
$$oint_{L}{frac{e^{frac{1}{z-a}}}{z}}dz =2ipi left(frac{1}{e^{1/a}}+mbox{Res}(f,a)right)$$
edited Jan 2 at 16:47
answered Jan 2 at 16:27
mathcounterexamples.netmathcounterexamples.net
27k22158
27k22158
$begingroup$
Yes, I tried to, but I'm not sure how to get $Res(f,a)$ to be equal $frac{1}{a}$.
$endgroup$
– math101
Jan 2 at 16:37
$begingroup$
I edited because I may have done an error in The computation of the residue. You can compute it by finding $a_{-1}$ where $sum_{-infty}^infty a_n (z-a)^n$ is Laurent expansion of $f$ at $a$.
$endgroup$
– mathcounterexamples.net
Jan 2 at 16:51
$begingroup$
I see. I'm struggling with that too a bit, but from what I've done so far, I don't think the series is going to have any negative terms ?
$endgroup$
– math101
Jan 2 at 17:07
add a comment |
$begingroup$
Yes, I tried to, but I'm not sure how to get $Res(f,a)$ to be equal $frac{1}{a}$.
$endgroup$
– math101
Jan 2 at 16:37
$begingroup$
I edited because I may have done an error in The computation of the residue. You can compute it by finding $a_{-1}$ where $sum_{-infty}^infty a_n (z-a)^n$ is Laurent expansion of $f$ at $a$.
$endgroup$
– mathcounterexamples.net
Jan 2 at 16:51
$begingroup$
I see. I'm struggling with that too a bit, but from what I've done so far, I don't think the series is going to have any negative terms ?
$endgroup$
– math101
Jan 2 at 17:07
$begingroup$
Yes, I tried to, but I'm not sure how to get $Res(f,a)$ to be equal $frac{1}{a}$.
$endgroup$
– math101
Jan 2 at 16:37
$begingroup$
Yes, I tried to, but I'm not sure how to get $Res(f,a)$ to be equal $frac{1}{a}$.
$endgroup$
– math101
Jan 2 at 16:37
$begingroup$
I edited because I may have done an error in The computation of the residue. You can compute it by finding $a_{-1}$ where $sum_{-infty}^infty a_n (z-a)^n$ is Laurent expansion of $f$ at $a$.
$endgroup$
– mathcounterexamples.net
Jan 2 at 16:51
$begingroup$
I edited because I may have done an error in The computation of the residue. You can compute it by finding $a_{-1}$ where $sum_{-infty}^infty a_n (z-a)^n$ is Laurent expansion of $f$ at $a$.
$endgroup$
– mathcounterexamples.net
Jan 2 at 16:51
$begingroup$
I see. I'm struggling with that too a bit, but from what I've done so far, I don't think the series is going to have any negative terms ?
$endgroup$
– math101
Jan 2 at 17:07
$begingroup$
I see. I'm struggling with that too a bit, but from what I've done so far, I don't think the series is going to have any negative terms ?
$endgroup$
– math101
Jan 2 at 17:07
add a comment |
$begingroup$
See this answer. Instead of taking the residue at $z = a$, take the residue at $z = infty$, which is the value of $-e^{1/(z - a)}$ at $z = infty$.
$endgroup$
$begingroup$
This is the better approach.
$endgroup$
– Szeto
Jan 3 at 13:32
add a comment |
$begingroup$
See this answer. Instead of taking the residue at $z = a$, take the residue at $z = infty$, which is the value of $-e^{1/(z - a)}$ at $z = infty$.
$endgroup$
$begingroup$
This is the better approach.
$endgroup$
– Szeto
Jan 3 at 13:32
add a comment |
$begingroup$
See this answer. Instead of taking the residue at $z = a$, take the residue at $z = infty$, which is the value of $-e^{1/(z - a)}$ at $z = infty$.
$endgroup$
See this answer. Instead of taking the residue at $z = a$, take the residue at $z = infty$, which is the value of $-e^{1/(z - a)}$ at $z = infty$.
answered Jan 3 at 1:34
MaximMaxim
6,0931221
6,0931221
$begingroup$
This is the better approach.
$endgroup$
– Szeto
Jan 3 at 13:32
add a comment |
$begingroup$
This is the better approach.
$endgroup$
– Szeto
Jan 3 at 13:32
$begingroup$
This is the better approach.
$endgroup$
– Szeto
Jan 3 at 13:32
$begingroup$
This is the better approach.
$endgroup$
– Szeto
Jan 3 at 13:32
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%2f3059644%2fsolving-a-complex-integral-oint-l-frace1-z-azdz-using-cauchys-formula%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