Showing that $int_0^pifrac{cos ntheta}{costheta-costheta_0}dtheta=pifrac{sin ntheta_0}{sintheta_0}$
$begingroup$
I am reading Debnath & Bhatta "Integral Transforms and Their Applications, 3rd". They cited one example from Zayed "Handbook of Function and Generalized Function Transformations" and stated an integral (Eq.(9.5.45)), for a non-negative integer n,
$$int_0^pi frac{cos(n theta)}{cos(theta)-cos(theta_0)}dtheta=pi frac{sin(n theta_0)}{sin(theta_0)}$$
It turns out many books on Hilbert transform use this relation for Airfoil Design example, e.g., Prederick W.King, Chapter 11.14 "Hilbert Transform-V1".
Interestingly, I remember the following one from Paul J. Nahin, Eq.(2.3.8) of "Inside Interesting Integrals"
$$int_0^pi frac{cos(n theta)-cos(n theta_0)}{cos(theta)-cos(theta_0)}dtheta=pi frac{sin(n theta_0)}{sin(theta_0)}.$$
You can find the proof in that book.
So, if both integrals are correct, then we should have
$$int_0^pi frac{1}{cos(theta)-cos(theta_0)}dtheta=0,$$ which I cannot see why. Mathmatica gives an pure imaginary result here. How shall I interpret these and how can I prove the first integral?
integration trigonometry
$endgroup$
add a comment |
$begingroup$
I am reading Debnath & Bhatta "Integral Transforms and Their Applications, 3rd". They cited one example from Zayed "Handbook of Function and Generalized Function Transformations" and stated an integral (Eq.(9.5.45)), for a non-negative integer n,
$$int_0^pi frac{cos(n theta)}{cos(theta)-cos(theta_0)}dtheta=pi frac{sin(n theta_0)}{sin(theta_0)}$$
It turns out many books on Hilbert transform use this relation for Airfoil Design example, e.g., Prederick W.King, Chapter 11.14 "Hilbert Transform-V1".
Interestingly, I remember the following one from Paul J. Nahin, Eq.(2.3.8) of "Inside Interesting Integrals"
$$int_0^pi frac{cos(n theta)-cos(n theta_0)}{cos(theta)-cos(theta_0)}dtheta=pi frac{sin(n theta_0)}{sin(theta_0)}.$$
You can find the proof in that book.
So, if both integrals are correct, then we should have
$$int_0^pi frac{1}{cos(theta)-cos(theta_0)}dtheta=0,$$ which I cannot see why. Mathmatica gives an pure imaginary result here. How shall I interpret these and how can I prove the first integral?
integration trigonometry
$endgroup$
$begingroup$
Note the denominator vanishes for $theta=theta_0+2kpi$. This is maybe a Cauchy principal value.
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 0:31
add a comment |
$begingroup$
I am reading Debnath & Bhatta "Integral Transforms and Their Applications, 3rd". They cited one example from Zayed "Handbook of Function and Generalized Function Transformations" and stated an integral (Eq.(9.5.45)), for a non-negative integer n,
$$int_0^pi frac{cos(n theta)}{cos(theta)-cos(theta_0)}dtheta=pi frac{sin(n theta_0)}{sin(theta_0)}$$
It turns out many books on Hilbert transform use this relation for Airfoil Design example, e.g., Prederick W.King, Chapter 11.14 "Hilbert Transform-V1".
Interestingly, I remember the following one from Paul J. Nahin, Eq.(2.3.8) of "Inside Interesting Integrals"
$$int_0^pi frac{cos(n theta)-cos(n theta_0)}{cos(theta)-cos(theta_0)}dtheta=pi frac{sin(n theta_0)}{sin(theta_0)}.$$
You can find the proof in that book.
So, if both integrals are correct, then we should have
$$int_0^pi frac{1}{cos(theta)-cos(theta_0)}dtheta=0,$$ which I cannot see why. Mathmatica gives an pure imaginary result here. How shall I interpret these and how can I prove the first integral?
integration trigonometry
$endgroup$
I am reading Debnath & Bhatta "Integral Transforms and Their Applications, 3rd". They cited one example from Zayed "Handbook of Function and Generalized Function Transformations" and stated an integral (Eq.(9.5.45)), for a non-negative integer n,
$$int_0^pi frac{cos(n theta)}{cos(theta)-cos(theta_0)}dtheta=pi frac{sin(n theta_0)}{sin(theta_0)}$$
It turns out many books on Hilbert transform use this relation for Airfoil Design example, e.g., Prederick W.King, Chapter 11.14 "Hilbert Transform-V1".
Interestingly, I remember the following one from Paul J. Nahin, Eq.(2.3.8) of "Inside Interesting Integrals"
$$int_0^pi frac{cos(n theta)-cos(n theta_0)}{cos(theta)-cos(theta_0)}dtheta=pi frac{sin(n theta_0)}{sin(theta_0)}.$$
You can find the proof in that book.
So, if both integrals are correct, then we should have
$$int_0^pi frac{1}{cos(theta)-cos(theta_0)}dtheta=0,$$ which I cannot see why. Mathmatica gives an pure imaginary result here. How shall I interpret these and how can I prove the first integral?
integration trigonometry
integration trigonometry
edited Dec 3 '18 at 1:45
Blue
48.5k870154
48.5k870154
asked Dec 3 '18 at 0:17
gouwangzhangdonggouwangzhangdong
838
838
$begingroup$
Note the denominator vanishes for $theta=theta_0+2kpi$. This is maybe a Cauchy principal value.
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 0:31
add a comment |
$begingroup$
Note the denominator vanishes for $theta=theta_0+2kpi$. This is maybe a Cauchy principal value.
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 0:31
$begingroup$
Note the denominator vanishes for $theta=theta_0+2kpi$. This is maybe a Cauchy principal value.
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 0:31
$begingroup$
Note the denominator vanishes for $theta=theta_0+2kpi$. This is maybe a Cauchy principal value.
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 0:31
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Are those integrals even well-defined? Let $theta_0$ be such that $cos(theta_0)=1/2$. For instance let $theta_0=frac{pi}{3}$. Take $n=1$. Now
$$int_0^pifrac{cos(ntheta_0)}{costheta-cos(theta_0)};dtheta=int_0^pifrac{1/2}{costheta-1/2};dtheta.$$
This integral is actually an improper one, as $pi/3$ is a singularity. And it does not converge.
Similarly,
$$int_0^pifrac{cos(ntheta)}{cos(theta)-cos(ntheta_0)};dtheta=int_0^pifrac{cos(theta)}{cos(theta)-1/2};dtheta$$
fails to converge.
$endgroup$
$begingroup$
But it converges to $pi$ as a principal value integral. That is, $lim_{epsilonto0^+} int_0^{pi/3-epsilon}+int_{pi/3+epsilon}^pi$.
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 0:58
$begingroup$
Hey, I see your concern regarding $theta_0$. If you first let n=0,1,...into the integral and then there is no singularity at all.
$endgroup$
– gouwangzhangdong
Dec 3 '18 at 1:11
add a comment |
$begingroup$
I might have solved it. See my attached hand writing. Thanks go for Jean for solving singularity through PV integral.
Other comments are also appreciated. Let me know if my trick does not work.
$endgroup$
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%2f3023421%2fshowing-that-int-0-pi-frac-cos-n-theta-cos-theta-cos-theta-0d-theta-pi%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
$begingroup$
Are those integrals even well-defined? Let $theta_0$ be such that $cos(theta_0)=1/2$. For instance let $theta_0=frac{pi}{3}$. Take $n=1$. Now
$$int_0^pifrac{cos(ntheta_0)}{costheta-cos(theta_0)};dtheta=int_0^pifrac{1/2}{costheta-1/2};dtheta.$$
This integral is actually an improper one, as $pi/3$ is a singularity. And it does not converge.
Similarly,
$$int_0^pifrac{cos(ntheta)}{cos(theta)-cos(ntheta_0)};dtheta=int_0^pifrac{cos(theta)}{cos(theta)-1/2};dtheta$$
fails to converge.
$endgroup$
$begingroup$
But it converges to $pi$ as a principal value integral. That is, $lim_{epsilonto0^+} int_0^{pi/3-epsilon}+int_{pi/3+epsilon}^pi$.
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 0:58
$begingroup$
Hey, I see your concern regarding $theta_0$. If you first let n=0,1,...into the integral and then there is no singularity at all.
$endgroup$
– gouwangzhangdong
Dec 3 '18 at 1:11
add a comment |
$begingroup$
Are those integrals even well-defined? Let $theta_0$ be such that $cos(theta_0)=1/2$. For instance let $theta_0=frac{pi}{3}$. Take $n=1$. Now
$$int_0^pifrac{cos(ntheta_0)}{costheta-cos(theta_0)};dtheta=int_0^pifrac{1/2}{costheta-1/2};dtheta.$$
This integral is actually an improper one, as $pi/3$ is a singularity. And it does not converge.
Similarly,
$$int_0^pifrac{cos(ntheta)}{cos(theta)-cos(ntheta_0)};dtheta=int_0^pifrac{cos(theta)}{cos(theta)-1/2};dtheta$$
fails to converge.
$endgroup$
$begingroup$
But it converges to $pi$ as a principal value integral. That is, $lim_{epsilonto0^+} int_0^{pi/3-epsilon}+int_{pi/3+epsilon}^pi$.
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 0:58
$begingroup$
Hey, I see your concern regarding $theta_0$. If you first let n=0,1,...into the integral and then there is no singularity at all.
$endgroup$
– gouwangzhangdong
Dec 3 '18 at 1:11
add a comment |
$begingroup$
Are those integrals even well-defined? Let $theta_0$ be such that $cos(theta_0)=1/2$. For instance let $theta_0=frac{pi}{3}$. Take $n=1$. Now
$$int_0^pifrac{cos(ntheta_0)}{costheta-cos(theta_0)};dtheta=int_0^pifrac{1/2}{costheta-1/2};dtheta.$$
This integral is actually an improper one, as $pi/3$ is a singularity. And it does not converge.
Similarly,
$$int_0^pifrac{cos(ntheta)}{cos(theta)-cos(ntheta_0)};dtheta=int_0^pifrac{cos(theta)}{cos(theta)-1/2};dtheta$$
fails to converge.
$endgroup$
Are those integrals even well-defined? Let $theta_0$ be such that $cos(theta_0)=1/2$. For instance let $theta_0=frac{pi}{3}$. Take $n=1$. Now
$$int_0^pifrac{cos(ntheta_0)}{costheta-cos(theta_0)};dtheta=int_0^pifrac{1/2}{costheta-1/2};dtheta.$$
This integral is actually an improper one, as $pi/3$ is a singularity. And it does not converge.
Similarly,
$$int_0^pifrac{cos(ntheta)}{cos(theta)-cos(ntheta_0)};dtheta=int_0^pifrac{cos(theta)}{cos(theta)-1/2};dtheta$$
fails to converge.
answered Dec 3 '18 at 0:45
Ben WBen W
2,290615
2,290615
$begingroup$
But it converges to $pi$ as a principal value integral. That is, $lim_{epsilonto0^+} int_0^{pi/3-epsilon}+int_{pi/3+epsilon}^pi$.
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 0:58
$begingroup$
Hey, I see your concern regarding $theta_0$. If you first let n=0,1,...into the integral and then there is no singularity at all.
$endgroup$
– gouwangzhangdong
Dec 3 '18 at 1:11
add a comment |
$begingroup$
But it converges to $pi$ as a principal value integral. That is, $lim_{epsilonto0^+} int_0^{pi/3-epsilon}+int_{pi/3+epsilon}^pi$.
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 0:58
$begingroup$
Hey, I see your concern regarding $theta_0$. If you first let n=0,1,...into the integral and then there is no singularity at all.
$endgroup$
– gouwangzhangdong
Dec 3 '18 at 1:11
$begingroup$
But it converges to $pi$ as a principal value integral. That is, $lim_{epsilonto0^+} int_0^{pi/3-epsilon}+int_{pi/3+epsilon}^pi$.
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 0:58
$begingroup$
But it converges to $pi$ as a principal value integral. That is, $lim_{epsilonto0^+} int_0^{pi/3-epsilon}+int_{pi/3+epsilon}^pi$.
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 0:58
$begingroup$
Hey, I see your concern regarding $theta_0$. If you first let n=0,1,...into the integral and then there is no singularity at all.
$endgroup$
– gouwangzhangdong
Dec 3 '18 at 1:11
$begingroup$
Hey, I see your concern regarding $theta_0$. If you first let n=0,1,...into the integral and then there is no singularity at all.
$endgroup$
– gouwangzhangdong
Dec 3 '18 at 1:11
add a comment |
$begingroup$
I might have solved it. See my attached hand writing. Thanks go for Jean for solving singularity through PV integral.
Other comments are also appreciated. Let me know if my trick does not work.
$endgroup$
add a comment |
$begingroup$
I might have solved it. See my attached hand writing. Thanks go for Jean for solving singularity through PV integral.
Other comments are also appreciated. Let me know if my trick does not work.
$endgroup$
add a comment |
$begingroup$
I might have solved it. See my attached hand writing. Thanks go for Jean for solving singularity through PV integral.
Other comments are also appreciated. Let me know if my trick does not work.
$endgroup$
I might have solved it. See my attached hand writing. Thanks go for Jean for solving singularity through PV integral.
Other comments are also appreciated. Let me know if my trick does not work.
answered Dec 3 '18 at 3:41
gouwangzhangdonggouwangzhangdong
838
838
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%2f3023421%2fshowing-that-int-0-pi-frac-cos-n-theta-cos-theta-cos-theta-0d-theta-pi%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
$begingroup$
Note the denominator vanishes for $theta=theta_0+2kpi$. This is maybe a Cauchy principal value.
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 0:31