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
edited Dec 3 '18 at 1:45
Blue
47.9k870152
asked Dec 3 '18 at 0:17
gouwangzhangdonggouwangzhangdong
638
$endgroup$
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
edited Dec 3 '18 at 1:45
Blue
47.9k870152
asked Dec 3 '18 at 0:17
gouwangzhangdonggouwangzhangdong
638
$endgroup$
$begingroup$
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
edited Dec 3 '18 at 1:45
Blue
47.9k870152
asked Dec 3 '18 at 0:17
gouwangzhangdonggouwangzhangdong
638
$endgroup$
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
47.9k870152
asked Dec 3 '18 at 0:17
gouwangzhangdonggouwangzhangdong
638
edited Dec 3 '18 at 1:45
Blue
47.9k870152
asked Dec 3 '18 at 0:17
gouwangzhangdonggouwangzhangdong
638
edited Dec 3 '18 at 1:45
Blue
47.9k870152
edited Dec 3 '18 at 1:45
Blue
47.9k870152
edited Dec 3 '18 at 1:45
Blue
47.9k870152
47.9k870152
asked Dec 3 '18 at 0:17
gouwangzhangdonggouwangzhangdong
638
asked Dec 3 '18 at 0:17
gouwangzhangdonggouwangzhangdong
638
asked Dec 3 '18 at 0:17
gouwangzhangdonggouwangzhangdong
638
638
$begingroup$
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.
answered Dec 3 '18 at 0:45
Ben WBen W
2,189615
$endgroup$
$begingroup$
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.
answered Dec 3 '18 at 3:41
gouwangzhangdonggouwangzhangdong
638
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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.
answered Dec 3 '18 at 0:45
Ben WBen W
2,189615
$endgroup$
$begingroup$
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.
answered Dec 3 '18 at 0:45
Ben WBen W
2,189615
$endgroup$
$begingroup$
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.
answered Dec 3 '18 at 0:45
Ben WBen W
2,189615
$endgroup$
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
2,189615
answered Dec 3 '18 at 0:45
Ben WBen W
2,189615
answered Dec 3 '18 at 0:45
Ben WBen W
2,189615
answered Dec 3 '18 at 0:45
Ben WBen W
2,189615
2,189615
$begingroup$
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.
answered Dec 3 '18 at 3:41
gouwangzhangdonggouwangzhangdong
638
$endgroup$
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.
answered Dec 3 '18 at 3:41
gouwangzhangdonggouwangzhangdong
638
$endgroup$
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.
answered Dec 3 '18 at 3:41
gouwangzhangdonggouwangzhangdong
638
$endgroup$
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
638
answered Dec 3 '18 at 3:41
gouwangzhangdonggouwangzhangdong
638
answered Dec 3 '18 at 3:41
gouwangzhangdonggouwangzhangdong
638
answered Dec 3 '18 at 3:41
gouwangzhangdonggouwangzhangdong
638
638
add a comment |
add a comment |
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$begingroup$
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