Showing that $int_0^pifrac{cos ntheta}{costheta-costheta_0}dtheta=pifrac{sin ntheta_0}{sintheta_0}$












3












$begingroup$


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?










share|cite|improve this question











$endgroup$












  • $begingroup$
    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


















3












$begingroup$


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?










share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















3












3








3


1



$begingroup$


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?










share|cite|improve this question











$endgroup$




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






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share|cite|improve this question













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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




















  • $begingroup$
    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












2 Answers
2






active

oldest

votes


















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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



















0












$begingroup$

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.



enter image description here






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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
















    0












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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














    0












    0








    0





    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$



    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.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    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


















    • $begingroup$
      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











    0












    $begingroup$

    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.



    enter image description here






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      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.



      enter image description here






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        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.



        enter image description here






        share|cite|improve this answer









        $endgroup$



        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.



        enter image description here







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 '18 at 3:41









        gouwangzhangdonggouwangzhangdong

        838




        838






























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