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I am wondering is there a closed form for this infinite sum. $$R(\beta)=\beta^2\sum_{k=1}^\infty \frac{\sin^2(\pi kn/\beta)}{(k^2-\beta^2)^2},$$ where $n\in\mathbb{N},\beta\in\mathbb{R}.$

LRDPRDX
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1 Answers1

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We may start with the expression obtained using Fourier series for $x\in(-\pi,\pi)$ and $z$ not a positive integer : $$\tag{1}\cos(x\,z)=\frac{2\,z\sin(\pi z)}{\pi}\;\left(\frac 1{2z^2}-\sum_{k=1}^\infty \;(-1)^k\frac{\cos(k\,x)}{k^2-z^2}\right)$$ that I'll rewrite as : $$\cos((\pi-a)z)=\frac{2\,z\sin(\pi z)}{\pi}\;\left(\frac 1{2z^2}-\sum_{k=1}^\infty \;(-1)^k\frac{\cos(k\,(\pi-a))}{k^2-z^2}\right)$$ or simply : $$\tag{2}\frac{\pi}2\frac{\cos((\pi-a)z)}{z\;\sin(\pi z)}=\frac 1{2z^2}-\sum_{k=1}^\infty \;\frac{\cos(k\,a)}{k^2-z^2}$$ Since $\;\displaystyle\sin^2(x)=\frac {1-\cos(2x)}2\;$ and $\;\displaystyle \frac d{dz}\frac 1{k^2-z^2}=\frac {2z}{(k^2-z^2)^2}\;$ we deduce that for : \begin{align} R(a,b)&:=2\sum_{k=1}^\infty \frac{\sin^2(k\,a)}{(k^2-b^2)^2}\\ &=\sum_{k=1}^\infty \frac 1{(k^2-b^2)^2}-\sum_{k=1}^\infty \frac {\cos(k\,2a)}{(k^2-b^2)^2}\\ &=\frac 1{2b}\frac d{db}\left[\sum_{k=1}^\infty \frac 1{k^2-b^2}-\sum_{k=1}^\infty \frac {\cos(k\,2a)}{k^2-b^2}\right]\\ &=\frac{\pi}2\frac 1{2b}\frac d{db}\left[-\frac{\cos(\pi b)}{b\;\sin(\pi b)}+\frac{\cos((\pi-2a)b)}{b\;\sin(\pi b)}\right]\\ \end{align} and thus the wished closed form for $b$ not a positive integer and $\,a\in(0,\pi)\,$ at least : $$\sum_{k=1}^\infty \frac{\sin^2(k\,a)}{(k^2-b^2)^2}=\frac {\pi}{8\,b^3}\left[\cot(\pi b)+\pi b\csc^2(\pi b)-(1+\pi b\cot(\pi b))\csc(\pi b)\cos((\pi-2a)b)-\\(\pi-2a)b\,\csc(\pi b)\sin((\pi-2a)b)\right]$$

pari/gp scripts used for numerical confirmation :

r(a,b)=2*sumalt(k=0,sin(k*a)^2/(k^2-b^2)^2)
f(a,b)=Pi/(4*b^3)*(1/tan(Pi*b)+Pi*b/sin(Pi*b)^2-(1+Pi*b/tan(Pi*b))/sin(Pi*b)*cos((Pi-2*a)*b)-(Pi-2*a)*b/sin(Pi*b)*sin((Pi-2*a)*b))
Raymond Manzoni
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  • For $,a\not \in(0,\pi),$ simply replace $a$ by $(a \bmod{\pi})$ in the RHS. – Raymond Manzoni Apr 29 '17 at 14:55
  • In my question $a=\frac{1}{b}$ I suspect that in this case the step when you take a derivative would be other, wouldn't it? – LRDPRDX Apr 30 '17 at 17:13
  • @Wolfgang: yes setting $,a=\dfrac{\pi kn}b,$ at the start would be counterproductive and that's why I searched a more general expression. Your substitution should be done at the end in the expression I obtained or in : $$\sum_{k=1}^\infty \frac{\sin^2(k,a)}{(k^2-b^2)^2}=\frac {\pi}8\frac{\frac{\cos(\pi b)}b+\frac{\pi}{\sin(\pi b)}-\left(\frac 1b+\frac{\pi}{\tan(\pi b)}\right)\cos((\pi-2a)b)-(\pi-2a)\sin((\pi-2a)b)}{b^2,\sin(\pi b)}$$ – Raymond Manzoni Apr 30 '17 at 17:33
  • It seems I got it. Thanks. I will check every step again and hope will accept your answer. – LRDPRDX Apr 30 '17 at 17:56
  • A little bit strange. Let $\beta = n$ and $a = \pi$. Then sum is equal (the only non-zero summon ) to $\pi^2/4$. On the other hand, according to your answer it is $-\pi^2/8$. I think you should require $a\in(0,2\pi)$ this eliminates the minus. But I do not know how to get rid of extra 2 in the denominator. – LRDPRDX May 02 '17 at 13:42
  • Sorry. Requiring $a\in(0,2\pi)$ does not eliminate the minus. But this should be required if $x\in(-\pi, \pi)$ and $x = \pi - a$. – LRDPRDX May 02 '17 at 13:49
  • @Wolfgang: Well I explicitly excluded the case $b$ a positive integer since $R$ is not defined in this case. Anyway computing the limit I obtained $;\displaystyle\lim_{b\to n} g_n(b)=\dfrac {\pi^2}4$ for $g_n(b):=b^2 R(\pi n/b,b)$ (using the closed form). Of course the limit has to be evaluated with care (you can't simply substitute $b=n,$ and $,a=\pi$ !). Concerning the initial interval $a\in(0,2\pi)$ becoming $a\in(0,\pi)$ this comes from $\cos(k,a)$ in $(2)$ becoming $\cos(k,2a)$ in the derivation (I should have changed the name of the variable or replaced $a$ by $a/2$ at the right). – Raymond Manzoni May 03 '17 at 00:36
  • This sum defined wherever it is converged. So it is defined at $\beta = n$ and more over, at $\beta=m$ any integer. How did you obtain $\pi^2/4$ using the closed form? It is $-\pi^2/8$. – LRDPRDX May 03 '17 at 02:36
  • OK. It is either $-\pi^2/8$ or $\pi^2/8$ depending on how does $b$ approaches to $n$. I do not understand so. It is because of the last sine. – LRDPRDX May 03 '17 at 04:15
  • I am idiot. Sorry. Sign is always plus. But what about extra 2? – LRDPRDX May 03 '17 at 04:23
  • @Wolfgang: Your series is not defined for $\beta$ a positive integer (for the same reason that $\dfrac {\sin(x)}x$ is not defined at $0$ but may trivially be extended).
    There is no extra 2 factor, the last sin(e ;-)) adds indeed a contribution $\pi$ to the other $\pi$ giving at the end $\dfrac {\pi(\pi+\pi)}8$ as wished (in my previous comment the $R$ function was without the $2$ overall factor, sorry for the confusion).     – Raymond Manzoni May 03 '17 at 13:56
  • Firstly, I do not understand how sine could add a contribution larger than 1. – LRDPRDX May 03 '17 at 16:45
  • Secondly, I do not understand where do you take $\pi+\pi$ from. I tried to figure out. Honestly. – LRDPRDX May 03 '17 at 16:48
  • @Wolfgang: From the closed form I presented in the comment : $\lim_{b\to n}\frac 1{\sin((\pi b)}\left(\frac{\cos(\pi b)}b+\frac{\pi}{\sin(\pi b)}-\left(\frac 1b+\frac{\pi}{\tan(\pi b)}\right)\cos((\pi-2\pi n/b)b)\right)=\pi$ and (the "$\sin$ term") $\lim_{b\to n}\frac 1{\sin((\pi b)}\left(-(\pi-2\pi n/b)\sin((\pi-2\pi n/b)b)\right)=\pi$ (note btw that the substitution $b=n$ would give $-\pi$ instead of $\pi$ for the second limit). Hoping this clarified all the remaining points, – Raymond Manzoni May 03 '17 at 18:37
  • OMG. So sorry. I thought that all except the last sine term is zero. It needs to be more accurate. Thanks. – LRDPRDX May 04 '17 at 05:03
  • Glad it helped @Wolfgang. Excellent continuation, – Raymond Manzoni May 04 '17 at 07:40