Real integrals with two poles in the complex plane
I'm studying the Cauchy Integral Theorem / Formula, but realised I have a misunderstanding.
Consider an integral over the function $f: mathbb{R} to mathbb{C}$
$$
I = int^infty_{-infty} f(x) , dx = int^infty_{-infty} frac{e^{ix}}{x^2 + 1} , dx quad.
$$
This can be considered in the complex plane, where $f(z)$ is holomorphic except at its two poles at $z = pm i$. Choosing to consider the positive case, we therefore factorise as
$$
I = int_C frac{frac{e^{iz}}{z+i}}{z-i} , dz
= 2pi i frac{e^{-1}}{2i} =pi e^{-1}
quad ,
$$
where the contour $C = C_R + C_+$ is taken anticlockwise, $C_R$ is the real axis between $pm R$ and $C_+$ is the positive semicircle in the complex plane with $left|zright| = R$. It's clear from inspection that $C_+$ does not contribute to the path integral if we let $R to infty$, so we can equate the integral in the complex plane to the real integral $I$.
Now my question: I could equally well have chosen the negative half of the complex plane ($C'=C_R + C_-$) to evaluate my integral, negating the answer since the anticlockwise integral would otherwise calculate $int_{infty}^{-infty}$. This would give
$$
I = - int_{C'} frac{frac{e^{iz}}{z-i}}{z+i} , dz
= - 2pi i frac{e^{+1}}{-2i}
=pi e^{+1}
quad .
$$
But clearly, that's different to my previous calculation. I thought these two methods should give the same answer, so what went wrong?
complex-analysis contour-integration cauchy-integral-formula
asked 2 days ago
CharlieB
1233
add a comment |
I'm studying the Cauchy Integral Theorem / Formula, but realised I have a misunderstanding.
Consider an integral over the function $f: mathbb{R} to mathbb{C}$
$$
I = int^infty_{-infty} f(x) , dx = int^infty_{-infty} frac{e^{ix}}{x^2 + 1} , dx quad.
$$
This can be considered in the complex plane, where $f(z)$ is holomorphic except at its two poles at $z = pm i$. Choosing to consider the positive case, we therefore factorise as
$$
I = int_C frac{frac{e^{iz}}{z+i}}{z-i} , dz
= 2pi i frac{e^{-1}}{2i} =pi e^{-1}
quad ,
$$
where the contour $C = C_R + C_+$ is taken anticlockwise, $C_R$ is the real axis between $pm R$ and $C_+$ is the positive semicircle in the complex plane with $left|zright| = R$. It's clear from inspection that $C_+$ does not contribute to the path integral if we let $R to infty$, so we can equate the integral in the complex plane to the real integral $I$.
Now my question: I could equally well have chosen the negative half of the complex plane ($C'=C_R + C_-$) to evaluate my integral, negating the answer since the anticlockwise integral would otherwise calculate $int_{infty}^{-infty}$. This would give
$$
I = - int_{C'} frac{frac{e^{iz}}{z-i}}{z+i} , dz
= - 2pi i frac{e^{+1}}{-2i}
=pi e^{+1}
quad .
$$
But clearly, that's different to my previous calculation. I thought these two methods should give the same answer, so what went wrong?
complex-analysis contour-integration cauchy-integral-formula
asked 2 days ago
CharlieB
1233
add a comment |
I'm studying the Cauchy Integral Theorem / Formula, but realised I have a misunderstanding.
Consider an integral over the function $f: mathbb{R} to mathbb{C}$
$$
I = int^infty_{-infty} f(x) , dx = int^infty_{-infty} frac{e^{ix}}{x^2 + 1} , dx quad.
$$
This can be considered in the complex plane, where $f(z)$ is holomorphic except at its two poles at $z = pm i$. Choosing to consider the positive case, we therefore factorise as
$$
I = int_C frac{frac{e^{iz}}{z+i}}{z-i} , dz
= 2pi i frac{e^{-1}}{2i} =pi e^{-1}
quad ,
$$
where the contour $C = C_R + C_+$ is taken anticlockwise, $C_R$ is the real axis between $pm R$ and $C_+$ is the positive semicircle in the complex plane with $left|zright| = R$. It's clear from inspection that $C_+$ does not contribute to the path integral if we let $R to infty$, so we can equate the integral in the complex plane to the real integral $I$.
Now my question: I could equally well have chosen the negative half of the complex plane ($C'=C_R + C_-$) to evaluate my integral, negating the answer since the anticlockwise integral would otherwise calculate $int_{infty}^{-infty}$. This would give
$$
I = - int_{C'} frac{frac{e^{iz}}{z-i}}{z+i} , dz
= - 2pi i frac{e^{+1}}{-2i}
=pi e^{+1}
quad .
$$
But clearly, that's different to my previous calculation. I thought these two methods should give the same answer, so what went wrong?
complex-analysis contour-integration cauchy-integral-formula
asked 2 days ago
CharlieB
1233
I'm studying the Cauchy Integral Theorem / Formula, but realised I have a misunderstanding.
Consider an integral over the function $f: mathbb{R} to mathbb{C}$
$$
I = int^infty_{-infty} f(x) , dx = int^infty_{-infty} frac{e^{ix}}{x^2 + 1} , dx quad.
$$
This can be considered in the complex plane, where $f(z)$ is holomorphic except at its two poles at $z = pm i$. Choosing to consider the positive case, we therefore factorise as
$$
I = int_C frac{frac{e^{iz}}{z+i}}{z-i} , dz
= 2pi i frac{e^{-1}}{2i} =pi e^{-1}
quad ,
$$
where the contour $C = C_R + C_+$ is taken anticlockwise, $C_R$ is the real axis between $pm R$ and $C_+$ is the positive semicircle in the complex plane with $left|zright| = R$. It's clear from inspection that $C_+$ does not contribute to the path integral if we let $R to infty$, so we can equate the integral in the complex plane to the real integral $I$.
Now my question: I could equally well have chosen the negative half of the complex plane ($C'=C_R + C_-$) to evaluate my integral, negating the answer since the anticlockwise integral would otherwise calculate $int_{infty}^{-infty}$. This would give
$$
I = - int_{C'} frac{frac{e^{iz}}{z-i}}{z+i} , dz
= - 2pi i frac{e^{+1}}{-2i}
=pi e^{+1}
quad .
$$
But clearly, that's different to my previous calculation. I thought these two methods should give the same answer, so what went wrong?
complex-analysis contour-integration cauchy-integral-formula
complex-analysis contour-integration cauchy-integral-formula
asked 2 days ago
CharlieB
1233
asked 2 days ago
CharlieB
1233
asked 2 days ago
CharlieB
1233
asked 2 days ago
CharlieB
1233
asked 2 days ago
CharlieB
1233
1233
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
No, you cannot choose the the low half-plane of $mathbb C$. If $z=x+yi$ with $yleqslant0$, then$$lvert e^{iz}rvert=e^{operatorname{Re}(-y+xi)}=e^{-operatorname{Re}(y)}geqslant1$$and, in fact, as $y$ goes to $-infty$, $lvert e^{iz}rvert$ goes to $+infty$. So, the integral along the negative semicircle most definitely will contribute to the path integral. You will not have this problem if you work with the highest half-plane.
answered 2 days ago
José Carlos Santos
150k22120221
Ah, I'm so used to thinking that $| e^{ix} | = 1$ that I forgot it's not true for complex $x$. Thanks!
– CharlieB
2 days ago
add a comment |
If you choose the contour in the upper half-plane, the integral over the
semicircle tends to zero as $Rtoinfty$. This is because $|e^{iz}|le1$
on the upper half-plane. The bound $pi R/(R^2-1)$ easily follows
for the absolute value of the integral, when $R>1$.
But on the lower half-plane, the integral over the semicircle does not tend to
zero as $Rtoinfty$.
answered 2 days ago
Lord Shark the Unknown
101k958132
1
In hindsight, I should have suspected the part of my argument that was "clear from inspection"...
– CharlieB
2 days ago
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%2f3054798%2freal-integrals-with-two-poles-in-the-complex-plane%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
No, you cannot choose the the low half-plane of $mathbb C$. If $z=x+yi$ with $yleqslant0$, then$$lvert e^{iz}rvert=e^{operatorname{Re}(-y+xi)}=e^{-operatorname{Re}(y)}geqslant1$$and, in fact, as $y$ goes to $-infty$, $lvert e^{iz}rvert$ goes to $+infty$. So, the integral along the negative semicircle most definitely will contribute to the path integral. You will not have this problem if you work with the highest half-plane.
answered 2 days ago
José Carlos Santos
150k22120221
Ah, I'm so used to thinking that $| e^{ix} | = 1$ that I forgot it's not true for complex $x$. Thanks!
– CharlieB
2 days ago
add a comment |
No, you cannot choose the the low half-plane of $mathbb C$. If $z=x+yi$ with $yleqslant0$, then$$lvert e^{iz}rvert=e^{operatorname{Re}(-y+xi)}=e^{-operatorname{Re}(y)}geqslant1$$and, in fact, as $y$ goes to $-infty$, $lvert e^{iz}rvert$ goes to $+infty$. So, the integral along the negative semicircle most definitely will contribute to the path integral. You will not have this problem if you work with the highest half-plane.
answered 2 days ago
José Carlos Santos
150k22120221
Ah, I'm so used to thinking that $| e^{ix} | = 1$ that I forgot it's not true for complex $x$. Thanks!
– CharlieB
2 days ago
add a comment |
No, you cannot choose the the low half-plane of $mathbb C$. If $z=x+yi$ with $yleqslant0$, then$$lvert e^{iz}rvert=e^{operatorname{Re}(-y+xi)}=e^{-operatorname{Re}(y)}geqslant1$$and, in fact, as $y$ goes to $-infty$, $lvert e^{iz}rvert$ goes to $+infty$. So, the integral along the negative semicircle most definitely will contribute to the path integral. You will not have this problem if you work with the highest half-plane.
answered 2 days ago
José Carlos Santos
150k22120221
No, you cannot choose the the low half-plane of $mathbb C$. If $z=x+yi$ with $yleqslant0$, then$$lvert e^{iz}rvert=e^{operatorname{Re}(-y+xi)}=e^{-operatorname{Re}(y)}geqslant1$$and, in fact, as $y$ goes to $-infty$, $lvert e^{iz}rvert$ goes to $+infty$. So, the integral along the negative semicircle most definitely will contribute to the path integral. You will not have this problem if you work with the highest half-plane.
answered 2 days ago
José Carlos Santos
150k22120221
answered 2 days ago
José Carlos Santos
150k22120221
answered 2 days ago
José Carlos Santos
150k22120221
answered 2 days ago
José Carlos Santos
150k22120221
150k22120221
Ah, I'm so used to thinking that $| e^{ix} | = 1$ that I forgot it's not true for complex $x$. Thanks!
– CharlieB
2 days ago
add a comment |
Ah, I'm so used to thinking that $| e^{ix} | = 1$ that I forgot it's not true for complex $x$. Thanks!
– CharlieB
2 days ago
Ah, I'm so used to thinking that $| e^{ix} | = 1$ that I forgot it's not true for complex $x$. Thanks!
– CharlieB
2 days ago
Ah, I'm so used to thinking that $| e^{ix} | = 1$ that I forgot it's not true for complex $x$. Thanks!
– CharlieB
2 days ago
add a comment |
If you choose the contour in the upper half-plane, the integral over the
semicircle tends to zero as $Rtoinfty$. This is because $|e^{iz}|le1$
on the upper half-plane. The bound $pi R/(R^2-1)$ easily follows
for the absolute value of the integral, when $R>1$.
But on the lower half-plane, the integral over the semicircle does not tend to
zero as $Rtoinfty$.
answered 2 days ago
Lord Shark the Unknown
101k958132
1
In hindsight, I should have suspected the part of my argument that was "clear from inspection"...
– CharlieB
2 days ago
add a comment |
If you choose the contour in the upper half-plane, the integral over the
semicircle tends to zero as $Rtoinfty$. This is because $|e^{iz}|le1$
on the upper half-plane. The bound $pi R/(R^2-1)$ easily follows
for the absolute value of the integral, when $R>1$.
But on the lower half-plane, the integral over the semicircle does not tend to
zero as $Rtoinfty$.
answered 2 days ago
Lord Shark the Unknown
101k958132
1
In hindsight, I should have suspected the part of my argument that was "clear from inspection"...
– CharlieB
2 days ago
add a comment |
If you choose the contour in the upper half-plane, the integral over the
semicircle tends to zero as $Rtoinfty$. This is because $|e^{iz}|le1$
on the upper half-plane. The bound $pi R/(R^2-1)$ easily follows
for the absolute value of the integral, when $R>1$.
But on the lower half-plane, the integral over the semicircle does not tend to
zero as $Rtoinfty$.
answered 2 days ago
Lord Shark the Unknown
101k958132
If you choose the contour in the upper half-plane, the integral over the
semicircle tends to zero as $Rtoinfty$. This is because $|e^{iz}|le1$
on the upper half-plane. The bound $pi R/(R^2-1)$ easily follows
for the absolute value of the integral, when $R>1$.
But on the lower half-plane, the integral over the semicircle does not tend to
zero as $Rtoinfty$.
answered 2 days ago
Lord Shark the Unknown
101k958132
answered 2 days ago
Lord Shark the Unknown
101k958132
answered 2 days ago
Lord Shark the Unknown
101k958132
answered 2 days ago
Lord Shark the Unknown
101k958132
101k958132
1
In hindsight, I should have suspected the part of my argument that was "clear from inspection"...
– CharlieB
2 days ago
add a comment |
1
In hindsight, I should have suspected the part of my argument that was "clear from inspection"...
– CharlieB
2 days ago
1
1
In hindsight, I should have suspected the part of my argument that was "clear from inspection"...
– CharlieB
2 days ago
In hindsight, I should have suspected the part of my argument that was "clear from inspection"...
– CharlieB
2 days ago
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3054798%2freal-integrals-with-two-poles-in-the-complex-plane%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
