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