Here I want to get the closed form solution of the following summation

$$ \sum_{n=1}^\infty \frac{\sin(n+a)}{n+a}, a\in \mathbb{R}^+ \qquad(1) $$

Or the more general form:

$$ f(a;x) =\sum_{n=1}^\infty\operatorname{sinc} [(n+a)x]= \sum_{n=1}^\infty \frac{\sin [(n+a)x]}{(n+a)x}, a,x\in \mathbb{R}^+ \qquad(2) $$

I looked up into several books and got a similar seires with its closed form:

  1. Equation (550) in Summation of Series 2nd ed. (written by Jolley). $$ f(1/2;x)= \sum_{n=1}^\infty \frac{\sin [(n+1/2)x]}{(n+1/2)x} = \frac1x\left(\frac{\pi}{2}-2\sin\frac x2\right),0<x<\pi \qquad $$

But 2 issues in this series: (A) the domain of $x$ is limited (B) the parameter $a$ is limited.

  1. Equation (551) in the same book

$$ \sum_{n=-\infty}^\infty \frac{\sin [(n-a)x]}{(n-a)x} = \frac{\pi}{x},0<x<2\pi \qquad $$ Also the domain of $x$ is limited. And the summation starts from negative infinity.

Anyone can help me?

Possible relating QUESTIONS:

Does there exist a closed form for the sinc function series $\sum_{n=1}^\infty \frac{\sin\sqrt{n^2+1}}{\sqrt{n^2+1}}$?

  • $\begingroup$ In equation 551, is the right hand side supposed to have that denominator? $\endgroup$
    – B. Mehta
    Nov 29, 2017 at 3:04
  • $\begingroup$ @B.Mehta The original equation in the book is written as $$\sum_{n=-\infty}^{\infty} \dfrac{\sin(n-a)\theta}{n-a}=-\pi, 0<\theta<2\pi$$ $\endgroup$ Nov 29, 2017 at 3:05
  • $\begingroup$ that's different to what's in your post? $\endgroup$
    – B. Mehta
    Nov 29, 2017 at 3:16
  • $\begingroup$ @B.Mehta I think the translation invariance property of the series only hold for $a$ being an integer? Am I right? Otherwise the author Jolley should have written $f(a;x) = \pi/(2x)$, rather than just a particular form as the equation (550). $\endgroup$ Nov 29, 2017 at 3:19
  • $\begingroup$ Ah true! My mistake, will delete. $\endgroup$
    – B. Mehta
    Nov 29, 2017 at 3:20

2 Answers 2


$$I=\sum_{n=1}^\infty \frac{e^{+i (n+a)x}}{ (n+a)x}=\frac{e^{+i (a+1) x} }{(a+1) x}\, _2F_1\left(1,a+1;a+2;e^{+i x}\right)$$ $$J=\sum_{n=1}^\infty \frac{e^{-i (n+a)x}}{ (n+a)x}=\frac{e^{-i (a+1) x} }{(a+1) x}\, _2F_1\left(1,a+1;a+2;e^{-i x}\right)$$ $$\sum_{n=1}^\infty\frac{\sin((n+a)x)}{(n+a)x}=\frac{I-J}{2i}$$


$$\begin{eqnarray*}\sum_{n\geq 1}\frac{\sin(a+n)}{a+n}&=&\text{Im}\int_{0}^{+\infty}\sum_{n\geq 1}\exp\left(ia+in-ax-nx\right)\,dx\\ &=& \int_{0}^{+\infty}\frac{\sin(a)e^{-x}-\sin(a+1)}{2\left(\cos(1)-\cosh(x)\right)}\,e^{-ax}\,dx\\&=&\int_{0}^{1}\frac{u^a\left(\sin(a+1)-u\sin(a)\right)}{1-2u\cos(1)+u^2}\,du\end{eqnarray*}$$ can be computed in a explicit way for any $a\in\mathbb{Q}^+$ by partial fraction decomposition.
The asymptotic behaviour of the LHS is simple to study through the integral representation, but a closed form for any $a\in\mathbb{R}^+$ is beyond reach, if not through hypergeometric ${}_2 F_1$ functions. This is reminiscent (and not by chance) of the Hurwitz zeta function and the Digamma function.


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