Representing an integral as a finite sum












4












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Question: If $a$ is in an arbitrary commensurable ratio to $pi$, that is $a=tfrac {mpi}n$, then if $m+n$ is odd$$intlimits_0^1mathrm dy,frac {y^xsin a}{1+2ycos a+y^2}=frac 12sumlimits_{i=1}^{n-1}(-1)^{i-1}sin ialeft[frac {mathrm dGammaleft(frac {x+n+i}{2n}right)}{mathrm dx}-frac {mathrm dGammaleft(frac {x+i}{2n}right)}{mathrm dx}right]$$and when $m+n$ is even$$intlimits_0^1mathrm dy,frac {y^xsin a}{1+2ycos a+y^2}=frac 12sumlimits_{i=1}^{(n-1)/2}(-1)^{i-1}sin ialeft[frac {mathrm dGammaleft(frac {x+n+i}nright)}{mathrm dx}-frac {mathrm dGammaleft(frac {x+i}nright)}{mathrm dx}right]$$




I’m just having difficulty finding out where to start. Since the integral equals an infinite sum, it might be wise to start off with a taylor expansion of some sort. However, which function to expand I’m not very sure.



If you guys have any idea, I would be happy to hear them. Thanks!






real-analysis calculus integration definite-integrals







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    4












    $begingroup$



    Question: If $a$ is in an arbitrary commensurable ratio to $pi$, that is $a=tfrac {mpi}n$, then if $m+n$ is odd$$intlimits_0^1mathrm dy,frac {y^xsin a}{1+2ycos a+y^2}=frac 12sumlimits_{i=1}^{n-1}(-1)^{i-1}sin ialeft[frac {mathrm dGammaleft(frac {x+n+i}{2n}right)}{mathrm dx}-frac {mathrm dGammaleft(frac {x+i}{2n}right)}{mathrm dx}right]$$and when $m+n$ is even$$intlimits_0^1mathrm dy,frac {y^xsin a}{1+2ycos a+y^2}=frac 12sumlimits_{i=1}^{(n-1)/2}(-1)^{i-1}sin ialeft[frac {mathrm dGammaleft(frac {x+n+i}nright)}{mathrm dx}-frac {mathrm dGammaleft(frac {x+i}nright)}{mathrm dx}right]$$




    I’m just having difficulty finding out where to start. Since the integral equals an infinite sum, it might be wise to start off with a taylor expansion of some sort. However, which function to expand I’m not very sure.



    If you guys have any idea, I would be happy to hear them. Thanks!






    real-analysis calculus integration definite-integrals







    share|cite|improve this question









    $endgroup$















      4












      4








      4


      3



      $begingroup$



      Question: If $a$ is in an arbitrary commensurable ratio to $pi$, that is $a=tfrac {mpi}n$, then if $m+n$ is odd$$intlimits_0^1mathrm dy,frac {y^xsin a}{1+2ycos a+y^2}=frac 12sumlimits_{i=1}^{n-1}(-1)^{i-1}sin ialeft[frac {mathrm dGammaleft(frac {x+n+i}{2n}right)}{mathrm dx}-frac {mathrm dGammaleft(frac {x+i}{2n}right)}{mathrm dx}right]$$and when $m+n$ is even$$intlimits_0^1mathrm dy,frac {y^xsin a}{1+2ycos a+y^2}=frac 12sumlimits_{i=1}^{(n-1)/2}(-1)^{i-1}sin ialeft[frac {mathrm dGammaleft(frac {x+n+i}nright)}{mathrm dx}-frac {mathrm dGammaleft(frac {x+i}nright)}{mathrm dx}right]$$




      I’m just having difficulty finding out where to start. Since the integral equals an infinite sum, it might be wise to start off with a taylor expansion of some sort. However, which function to expand I’m not very sure.



      If you guys have any idea, I would be happy to hear them. Thanks!






      real-analysis calculus integration definite-integrals







      share|cite|improve this question









      $endgroup$





      Question: If $a$ is in an arbitrary commensurable ratio to $pi$, that is $a=tfrac {mpi}n$, then if $m+n$ is odd$$intlimits_0^1mathrm dy,frac {y^xsin a}{1+2ycos a+y^2}=frac 12sumlimits_{i=1}^{n-1}(-1)^{i-1}sin ialeft[frac {mathrm dGammaleft(frac {x+n+i}{2n}right)}{mathrm dx}-frac {mathrm dGammaleft(frac {x+i}{2n}right)}{mathrm dx}right]$$and when $m+n$ is even$$intlimits_0^1mathrm dy,frac {y^xsin a}{1+2ycos a+y^2}=frac 12sumlimits_{i=1}^{(n-1)/2}(-1)^{i-1}sin ialeft[frac {mathrm dGammaleft(frac {x+n+i}nright)}{mathrm dx}-frac {mathrm dGammaleft(frac {x+i}nright)}{mathrm dx}right]$$




      I’m just having difficulty finding out where to start. Since the integral equals an infinite sum, it might be wise to start off with a taylor expansion of some sort. However, which function to expand I’m not very sure.



      If you guys have any idea, I would be happy to hear them. Thanks!






      real-analysis calculus integration definite-integrals




      real-analysis calculus integration definite-integrals






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      asked Dec 17 '18 at 5:07









      CrescendoCrescendo

      2,2951625




      2,2951625






















          1 Answer
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          $begingroup$

          Hint: Expand the $dfrac{sin a}{1+2ycos a+y^2}$ and show
          $$frac {y}{1+2ycos a+y^2}=sum_{ngeq1}(-1)^{n-1}dfrac{sin na}{sin a}y^n$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Is the last line supposed to say$$frac 1{1+2ycos a+y^2}=sumlimits_{ngeq1}(-1)^{n-1}frac {sin na}{sin a}y^n$$
            $endgroup$
            – Frank W.
            Dec 17 '18 at 16:13










          • $begingroup$
            @Nosrati Okay I've replaced $2cos a$ with $e^{ia}+e^{-ia}$. Then, I factored the denominator and expanded to get the $a_n$ term as$$a_n=sumlimits_{k=0}^{n}(-1)^k e^{(n-2k)ia}=frac {sin a(n+1)}{sin a}$$Therefore, the sum equals$$intlimits_0^1mathrm dy,frac {y^xsin a}{1+2ycos a+y^2}=sumlimits_{ngeq1}frac {(-1)^{n-1}sin an}{x+n}$$I can see that the two are very close, but how do I arrive at the gamma functions? Additionally, how does the infinite sum which we have become finite?
            $endgroup$
            – Crescendo
            Dec 18 '18 at 5:21












          • $begingroup$
            Additionally, $sum_{i=1}^infty=sum_{i=1}^{n}+sum_{i=n+1}^{2n}+cdots=sum_{n=1}^{infty}sum_{i=1}^{n}$
            $endgroup$
            – Nosrati
            Dec 18 '18 at 5:41













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          $begingroup$

          Hint: Expand the $dfrac{sin a}{1+2ycos a+y^2}$ and show
          $$frac {y}{1+2ycos a+y^2}=sum_{ngeq1}(-1)^{n-1}dfrac{sin na}{sin a}y^n$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Is the last line supposed to say$$frac 1{1+2ycos a+y^2}=sumlimits_{ngeq1}(-1)^{n-1}frac {sin na}{sin a}y^n$$
            $endgroup$
            – Frank W.
            Dec 17 '18 at 16:13










          • $begingroup$
            @Nosrati Okay I've replaced $2cos a$ with $e^{ia}+e^{-ia}$. Then, I factored the denominator and expanded to get the $a_n$ term as$$a_n=sumlimits_{k=0}^{n}(-1)^k e^{(n-2k)ia}=frac {sin a(n+1)}{sin a}$$Therefore, the sum equals$$intlimits_0^1mathrm dy,frac {y^xsin a}{1+2ycos a+y^2}=sumlimits_{ngeq1}frac {(-1)^{n-1}sin an}{x+n}$$I can see that the two are very close, but how do I arrive at the gamma functions? Additionally, how does the infinite sum which we have become finite?
            $endgroup$
            – Crescendo
            Dec 18 '18 at 5:21












          • $begingroup$
            Additionally, $sum_{i=1}^infty=sum_{i=1}^{n}+sum_{i=n+1}^{2n}+cdots=sum_{n=1}^{infty}sum_{i=1}^{n}$
            $endgroup$
            – Nosrati
            Dec 18 '18 at 5:41


















          3












          $begingroup$

          Hint: Expand the $dfrac{sin a}{1+2ycos a+y^2}$ and show
          $$frac {y}{1+2ycos a+y^2}=sum_{ngeq1}(-1)^{n-1}dfrac{sin na}{sin a}y^n$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Is the last line supposed to say$$frac 1{1+2ycos a+y^2}=sumlimits_{ngeq1}(-1)^{n-1}frac {sin na}{sin a}y^n$$
            $endgroup$
            – Frank W.
            Dec 17 '18 at 16:13










          • $begingroup$
            @Nosrati Okay I've replaced $2cos a$ with $e^{ia}+e^{-ia}$. Then, I factored the denominator and expanded to get the $a_n$ term as$$a_n=sumlimits_{k=0}^{n}(-1)^k e^{(n-2k)ia}=frac {sin a(n+1)}{sin a}$$Therefore, the sum equals$$intlimits_0^1mathrm dy,frac {y^xsin a}{1+2ycos a+y^2}=sumlimits_{ngeq1}frac {(-1)^{n-1}sin an}{x+n}$$I can see that the two are very close, but how do I arrive at the gamma functions? Additionally, how does the infinite sum which we have become finite?
            $endgroup$
            – Crescendo
            Dec 18 '18 at 5:21












          • $begingroup$
            Additionally, $sum_{i=1}^infty=sum_{i=1}^{n}+sum_{i=n+1}^{2n}+cdots=sum_{n=1}^{infty}sum_{i=1}^{n}$
            $endgroup$
            – Nosrati
            Dec 18 '18 at 5:41
















          3












          3








          3





          $begingroup$

          Hint: Expand the $dfrac{sin a}{1+2ycos a+y^2}$ and show
          $$frac {y}{1+2ycos a+y^2}=sum_{ngeq1}(-1)^{n-1}dfrac{sin na}{sin a}y^n$$






          share|cite|improve this answer









          $endgroup$



          Hint: Expand the $dfrac{sin a}{1+2ycos a+y^2}$ and show
          $$frac {y}{1+2ycos a+y^2}=sum_{ngeq1}(-1)^{n-1}dfrac{sin na}{sin a}y^n$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 17 '18 at 5:35









          NosratiNosrati

          26.5k62354




          26.5k62354












          • $begingroup$
            Is the last line supposed to say$$frac 1{1+2ycos a+y^2}=sumlimits_{ngeq1}(-1)^{n-1}frac {sin na}{sin a}y^n$$
            $endgroup$
            – Frank W.
            Dec 17 '18 at 16:13










          • $begingroup$
            @Nosrati Okay I've replaced $2cos a$ with $e^{ia}+e^{-ia}$. Then, I factored the denominator and expanded to get the $a_n$ term as$$a_n=sumlimits_{k=0}^{n}(-1)^k e^{(n-2k)ia}=frac {sin a(n+1)}{sin a}$$Therefore, the sum equals$$intlimits_0^1mathrm dy,frac {y^xsin a}{1+2ycos a+y^2}=sumlimits_{ngeq1}frac {(-1)^{n-1}sin an}{x+n}$$I can see that the two are very close, but how do I arrive at the gamma functions? Additionally, how does the infinite sum which we have become finite?
            $endgroup$
            – Crescendo
            Dec 18 '18 at 5:21












          • $begingroup$
            Additionally, $sum_{i=1}^infty=sum_{i=1}^{n}+sum_{i=n+1}^{2n}+cdots=sum_{n=1}^{infty}sum_{i=1}^{n}$
            $endgroup$
            – Nosrati
            Dec 18 '18 at 5:41




















          • $begingroup$
            Is the last line supposed to say$$frac 1{1+2ycos a+y^2}=sumlimits_{ngeq1}(-1)^{n-1}frac {sin na}{sin a}y^n$$
            $endgroup$
            – Frank W.
            Dec 17 '18 at 16:13










          • $begingroup$
            @Nosrati Okay I've replaced $2cos a$ with $e^{ia}+e^{-ia}$. Then, I factored the denominator and expanded to get the $a_n$ term as$$a_n=sumlimits_{k=0}^{n}(-1)^k e^{(n-2k)ia}=frac {sin a(n+1)}{sin a}$$Therefore, the sum equals$$intlimits_0^1mathrm dy,frac {y^xsin a}{1+2ycos a+y^2}=sumlimits_{ngeq1}frac {(-1)^{n-1}sin an}{x+n}$$I can see that the two are very close, but how do I arrive at the gamma functions? Additionally, how does the infinite sum which we have become finite?
            $endgroup$
            – Crescendo
            Dec 18 '18 at 5:21












          • $begingroup$
            Additionally, $sum_{i=1}^infty=sum_{i=1}^{n}+sum_{i=n+1}^{2n}+cdots=sum_{n=1}^{infty}sum_{i=1}^{n}$
            $endgroup$
            – Nosrati
            Dec 18 '18 at 5:41


















          $begingroup$
          Is the last line supposed to say$$frac 1{1+2ycos a+y^2}=sumlimits_{ngeq1}(-1)^{n-1}frac {sin na}{sin a}y^n$$
          $endgroup$
          – Frank W.
          Dec 17 '18 at 16:13




          $begingroup$
          Is the last line supposed to say$$frac 1{1+2ycos a+y^2}=sumlimits_{ngeq1}(-1)^{n-1}frac {sin na}{sin a}y^n$$
          $endgroup$
          – Frank W.
          Dec 17 '18 at 16:13












          $begingroup$
          @Nosrati Okay I've replaced $2cos a$ with $e^{ia}+e^{-ia}$. Then, I factored the denominator and expanded to get the $a_n$ term as$$a_n=sumlimits_{k=0}^{n}(-1)^k e^{(n-2k)ia}=frac {sin a(n+1)}{sin a}$$Therefore, the sum equals$$intlimits_0^1mathrm dy,frac {y^xsin a}{1+2ycos a+y^2}=sumlimits_{ngeq1}frac {(-1)^{n-1}sin an}{x+n}$$I can see that the two are very close, but how do I arrive at the gamma functions? Additionally, how does the infinite sum which we have become finite?
          $endgroup$
          – Crescendo
          Dec 18 '18 at 5:21






          $begingroup$
          @Nosrati Okay I've replaced $2cos a$ with $e^{ia}+e^{-ia}$. Then, I factored the denominator and expanded to get the $a_n$ term as$$a_n=sumlimits_{k=0}^{n}(-1)^k e^{(n-2k)ia}=frac {sin a(n+1)}{sin a}$$Therefore, the sum equals$$intlimits_0^1mathrm dy,frac {y^xsin a}{1+2ycos a+y^2}=sumlimits_{ngeq1}frac {(-1)^{n-1}sin an}{x+n}$$I can see that the two are very close, but how do I arrive at the gamma functions? Additionally, how does the infinite sum which we have become finite?
          $endgroup$
          – Crescendo
          Dec 18 '18 at 5:21














          $begingroup$
          Additionally, $sum_{i=1}^infty=sum_{i=1}^{n}+sum_{i=n+1}^{2n}+cdots=sum_{n=1}^{infty}sum_{i=1}^{n}$
          $endgroup$
          – Nosrati
          Dec 18 '18 at 5:41






          $begingroup$
          Additionally, $sum_{i=1}^infty=sum_{i=1}^{n}+sum_{i=n+1}^{2n}+cdots=sum_{n=1}^{infty}sum_{i=1}^{n}$
          $endgroup$
          – Nosrati
          Dec 18 '18 at 5:41




















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