Verification for $int_{-infty}^infty frac{cos(az)}{z^{2n}+1}dz$
I decided to tackle this integral by integrating over a semicircular contour in the upper-half plane and the real line, where the path is denoted as $Gamma$.
$$int_{Gamma} frac{e^{iaz}}{z^{2n}+1}dz=(int_{C_R}+int_{-R}^R) frac{e^{iaz}}{z^{2n}+1}dz$$
The integral over $C_R$ should go to $0$ as $Rtoinfty$ by Jordan's Lemma. Thus
$$int_{Gamma} frac{e^{iaz}}{z^{2n}+1}dz=int_{-infty}^infty frac{e^{iaz}}{z^{2n}+1}dz=2pi isum Res$$ by Residue Theorem. The function has a pole at $z_0=e^{frac{pi i}{2n}}$ and calculating the Residue at $z_0$ I get
$$lim_{zto z_0} e^{iaz} frac{z-z_0}{z^{2n}+1}=-e^{iaz_0}frac{z_0}{2n}$$
So $$int_{-infty}^infty frac{e^{iaz}}{z^{2n}+1}dz=-frac{pi i}{n}z_0e^{iaz_0}$$
After equating the real and imaginary terms on both sides I get that $$int_{-infty}^infty frac{cos(az)}{z^{2n}+1}dz=frac{pi}{n}e^{-asin(frac{pi}{2n})}(cos(frac{pi}{2n})sin(acos(frac{pi}{2n}))+sin(frac{pi}{2n})cos(acos(frac{pi}{2n})))$$ I found that this answer agrees with the Wolfram for $n=1$ but appears to be off by a factor of $2$ for $n=2$. Could anyone explain if/where I made a mistake and what the final answer should be?
proof-verification improper-integrals contour-integration
asked Jan 4 at 18:57
aledenaleden
1,862411
|
show 2 more comments
I decided to tackle this integral by integrating over a semicircular contour in the upper-half plane and the real line, where the path is denoted as $Gamma$.
$$int_{Gamma} frac{e^{iaz}}{z^{2n}+1}dz=(int_{C_R}+int_{-R}^R) frac{e^{iaz}}{z^{2n}+1}dz$$
The integral over $C_R$ should go to $0$ as $Rtoinfty$ by Jordan's Lemma. Thus
$$int_{Gamma} frac{e^{iaz}}{z^{2n}+1}dz=int_{-infty}^infty frac{e^{iaz}}{z^{2n}+1}dz=2pi isum Res$$ by Residue Theorem. The function has a pole at $z_0=e^{frac{pi i}{2n}}$ and calculating the Residue at $z_0$ I get
$$lim_{zto z_0} e^{iaz} frac{z-z_0}{z^{2n}+1}=-e^{iaz_0}frac{z_0}{2n}$$
So $$int_{-infty}^infty frac{e^{iaz}}{z^{2n}+1}dz=-frac{pi i}{n}z_0e^{iaz_0}$$
After equating the real and imaginary terms on both sides I get that $$int_{-infty}^infty frac{cos(az)}{z^{2n}+1}dz=frac{pi}{n}e^{-asin(frac{pi}{2n})}(cos(frac{pi}{2n})sin(acos(frac{pi}{2n}))+sin(frac{pi}{2n})cos(acos(frac{pi}{2n})))$$ I found that this answer agrees with the Wolfram for $n=1$ but appears to be off by a factor of $2$ for $n=2$. Could anyone explain if/where I made a mistake and what the final answer should be?
proof-verification improper-integrals contour-integration
asked Jan 4 at 18:57
aledenaleden
1,862411
3
There are $n$ poles in the upper half-plane. You appear only to be considering one of them.
– Lord Shark the Unknown
Jan 4 at 18:59
@LordSharktheUnknown What would the other poles be?
– aleden
Jan 4 at 19:06
@aleden: all the roots of $z^{2n}+1$ in the upper half-plane.
– Jack D'Aurizio
Jan 4 at 19:08
1
@aleden: $z^{2n}+1$ is a divisor of $z^{4n}-1$.
– Jack D'Aurizio
Jan 4 at 19:11
1
don't believe too much on wolfram alpha, neither mathematica
– Masacroso
Jan 4 at 19:27
|
show 2 more comments
I decided to tackle this integral by integrating over a semicircular contour in the upper-half plane and the real line, where the path is denoted as $Gamma$.
$$int_{Gamma} frac{e^{iaz}}{z^{2n}+1}dz=(int_{C_R}+int_{-R}^R) frac{e^{iaz}}{z^{2n}+1}dz$$
The integral over $C_R$ should go to $0$ as $Rtoinfty$ by Jordan's Lemma. Thus
$$int_{Gamma} frac{e^{iaz}}{z^{2n}+1}dz=int_{-infty}^infty frac{e^{iaz}}{z^{2n}+1}dz=2pi isum Res$$ by Residue Theorem. The function has a pole at $z_0=e^{frac{pi i}{2n}}$ and calculating the Residue at $z_0$ I get
$$lim_{zto z_0} e^{iaz} frac{z-z_0}{z^{2n}+1}=-e^{iaz_0}frac{z_0}{2n}$$
So $$int_{-infty}^infty frac{e^{iaz}}{z^{2n}+1}dz=-frac{pi i}{n}z_0e^{iaz_0}$$
After equating the real and imaginary terms on both sides I get that $$int_{-infty}^infty frac{cos(az)}{z^{2n}+1}dz=frac{pi}{n}e^{-asin(frac{pi}{2n})}(cos(frac{pi}{2n})sin(acos(frac{pi}{2n}))+sin(frac{pi}{2n})cos(acos(frac{pi}{2n})))$$ I found that this answer agrees with the Wolfram for $n=1$ but appears to be off by a factor of $2$ for $n=2$. Could anyone explain if/where I made a mistake and what the final answer should be?
proof-verification improper-integrals contour-integration
asked Jan 4 at 18:57
aledenaleden
1,862411
I decided to tackle this integral by integrating over a semicircular contour in the upper-half plane and the real line, where the path is denoted as $Gamma$.
$$int_{Gamma} frac{e^{iaz}}{z^{2n}+1}dz=(int_{C_R}+int_{-R}^R) frac{e^{iaz}}{z^{2n}+1}dz$$
The integral over $C_R$ should go to $0$ as $Rtoinfty$ by Jordan's Lemma. Thus
$$int_{Gamma} frac{e^{iaz}}{z^{2n}+1}dz=int_{-infty}^infty frac{e^{iaz}}{z^{2n}+1}dz=2pi isum Res$$ by Residue Theorem. The function has a pole at $z_0=e^{frac{pi i}{2n}}$ and calculating the Residue at $z_0$ I get
$$lim_{zto z_0} e^{iaz} frac{z-z_0}{z^{2n}+1}=-e^{iaz_0}frac{z_0}{2n}$$
So $$int_{-infty}^infty frac{e^{iaz}}{z^{2n}+1}dz=-frac{pi i}{n}z_0e^{iaz_0}$$
After equating the real and imaginary terms on both sides I get that $$int_{-infty}^infty frac{cos(az)}{z^{2n}+1}dz=frac{pi}{n}e^{-asin(frac{pi}{2n})}(cos(frac{pi}{2n})sin(acos(frac{pi}{2n}))+sin(frac{pi}{2n})cos(acos(frac{pi}{2n})))$$ I found that this answer agrees with the Wolfram for $n=1$ but appears to be off by a factor of $2$ for $n=2$. Could anyone explain if/where I made a mistake and what the final answer should be?
proof-verification improper-integrals contour-integration
proof-verification improper-integrals contour-integration
asked Jan 4 at 18:57
aledenaleden
1,862411
asked Jan 4 at 18:57
aledenaleden
1,862411
asked Jan 4 at 18:57
aledenaleden
1,862411
asked Jan 4 at 18:57
aledenaleden
1,862411
asked Jan 4 at 18:57
aledenaleden
1,862411
1,862411
3
There are $n$ poles in the upper half-plane. You appear only to be considering one of them.
– Lord Shark the Unknown
Jan 4 at 18:59
@LordSharktheUnknown What would the other poles be?
– aleden
Jan 4 at 19:06
@aleden: all the roots of $z^{2n}+1$ in the upper half-plane.
– Jack D'Aurizio
Jan 4 at 19:08
1
@aleden: $z^{2n}+1$ is a divisor of $z^{4n}-1$.
– Jack D'Aurizio
Jan 4 at 19:11
1
don't believe too much on wolfram alpha, neither mathematica
– Masacroso
Jan 4 at 19:27
|
show 2 more comments
3
There are $n$ poles in the upper half-plane. You appear only to be considering one of them.
– Lord Shark the Unknown
Jan 4 at 18:59
@LordSharktheUnknown What would the other poles be?
– aleden
Jan 4 at 19:06
@aleden: all the roots of $z^{2n}+1$ in the upper half-plane.
– Jack D'Aurizio
Jan 4 at 19:08
1
@aleden: $z^{2n}+1$ is a divisor of $z^{4n}-1$.
– Jack D'Aurizio
Jan 4 at 19:11
1
don't believe too much on wolfram alpha, neither mathematica
– Masacroso
Jan 4 at 19:27
3
3
There are $n$ poles in the upper half-plane. You appear only to be considering one of them.
– Lord Shark the Unknown
Jan 4 at 18:59
There are $n$ poles in the upper half-plane. You appear only to be considering one of them.
– Lord Shark the Unknown
Jan 4 at 18:59
@LordSharktheUnknown What would the other poles be?
– aleden
Jan 4 at 19:06
@LordSharktheUnknown What would the other poles be?
– aleden
Jan 4 at 19:06
@aleden: all the roots of $z^{2n}+1$ in the upper half-plane.
– Jack D'Aurizio
Jan 4 at 19:08
@aleden: all the roots of $z^{2n}+1$ in the upper half-plane.
– Jack D'Aurizio
Jan 4 at 19:08
1
1
@aleden: $z^{2n}+1$ is a divisor of $z^{4n}-1$.
– Jack D'Aurizio
Jan 4 at 19:11
@aleden: $z^{2n}+1$ is a divisor of $z^{4n}-1$.
– Jack D'Aurizio
Jan 4 at 19:11
1
1
don't believe too much on wolfram alpha, neither mathematica
– Masacroso
Jan 4 at 19:27
don't believe too much on wolfram alpha, neither mathematica
– Masacroso
Jan 4 at 19:27
|
show 2 more comments
0
active
oldest
votes
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%2f3061971%2fverification-for-int-infty-infty-frac-cosazz2n1dz%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3061971%2fverification-for-int-infty-infty-frac-cosazz2n1dz%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
3
There are $n$ poles in the upper half-plane. You appear only to be considering one of them.
– Lord Shark the Unknown
Jan 4 at 18:59
@LordSharktheUnknown What would the other poles be?
– aleden
Jan 4 at 19:06
@aleden: all the roots of $z^{2n}+1$ in the upper half-plane.
– Jack D'Aurizio
Jan 4 at 19:08
1
@aleden: $z^{2n}+1$ is a divisor of $z^{4n}-1$.
– Jack D'Aurizio
Jan 4 at 19:11
1
don't believe too much on wolfram alpha, neither mathematica
– Masacroso
Jan 4 at 19:27