What is the sum of a sinc function series sampled periodically What is the sum of the following sinc series?
$$f(k,y)=\sum_{n=-\infty}^{\infty} \frac{\text{sin} \pi(kn-y)}{\pi(kn-y)}$$
where $k$ is an integer greater then zero
This question is a generalisation of What is the sum over a shifted sinc function? and not a duplicate. The former question only considers the case for $k=1$
 A: I formulate a general solution here to compute $\sum_{n=-\infty}^{n=\infty}\frac{\sin(N\pi(y-\frac{n}{T}))}{\pi(y-\frac{n}{T})}$
for $T<N,$ with $T,N\in\mathbb{\mathbb{R}}^{+}$, and then use the
case where $T=1/k$ for an integer $k>1$ and $N=1$ to show that
the sum requested in this problem, $f(k,y)\equiv\sum_{n=-\infty}^{\infty}g(kn-y)=\sum_{n=-\infty}^{\infty}g(y-kn)$
for the (even) sinc function $g(y)=\begin{cases}
\frac{sin\pi y}{y} & y\not\neq0\\
1 & y=0
\end{cases}$ is equal to:
$f(k,y)=\begin{cases}
k & \frac{y}{k}\in\mathbb{Z}\\
\cos(\pi y)+\frac{\sin(\pi y(1-1/k))}{\sin(\pi y/k)} & \frac{y}{k}\notin\mathbb{Z},\,k\,\mathrm{even}\\
\frac{\sin(\pi y)}{\sin(\pi y/k)} & \frac{y}{k}\notin\mathbb{Z},\,k\,\mathrm{odd}
\end{cases}$
Let $W_{N}(t)$ be the rectangular window of width $N$ defined in
continuous time given by $W_{N}(t)=\begin{cases}
1 & -N/2<t<N/2\\
1/2 & |t|=N/2\\
0 & \textrm{else}
\end{cases}$ where $t$ is in units of seconds. Its Fourier transform is then
the (continuous, aperiodic) sinc function: $\hat{W}_{N}(y)=\begin{cases}
\frac{\sin(N\pi y)}{\pi y} & y\neq0\\
N & y=0
\end{cases}$ where the units of $y$ is hz (cycles per second). Sampling $W_{N}(t)$
at discrete intervals of $T$ seconds, yields a discrete, aperiodic
signal, $w_{N}(t)=\begin{cases}
1 & -N/2<t<N/2\\
1/2 & |t|=N/2\\
0 & \textrm{else}
\end{cases},t\in\mathbb{Z}T.$ The (continuous, periodic) Discrete Time Fourier Transform of this,
$\hat{w}_{N}(y)$ can be derived as folllows:
Directly from the definition of DTFT, and making the change of variable
$n=t/T$, we have $w_{N}(n)=\begin{cases}
1 & -N/2T<n<N/2T\\
1/2 & |n|=N/2T\\
0 & \textrm{else}
\end{cases},n\in\mathbb{Z}.$ 
For $yT\notin\mathbb{Z}$ and $N/T$ is an even integer (implying
that $N/2T$ is an integer value, and so $w_{N}(n)$ is evaluated
at the endpoints), then:
\begin{eqnarray*}
\hat{w}_{N}(y) & = & \sum_{n=-\infty}^{\infty}w_{N}(n)e^{-i2\pi yTn}=\frac{1}{2}\left(e^{-iN\pi y}+e^{iN\pi y}\right)+\sum_{n=-\frac{N-2T}{2T}}^{\frac{N-2T}{2T}}e^{-i2\pi yTn}=\\
 & =\frac{1}{2}\left(e^{-iN\pi y}+e^{iN\pi y}\right)+ & \frac{e^{i\pi(N-2T)y}-e^{-i\pi Ny}}{1-e^{-i2\pi yT}}=\frac{e^{-i\pi yT}}{e^{-i\pi yT}}\left(\frac{e^{i\pi y(N-T)}-e^{-i\pi y(N-T)}}{e^{i\pi yT}-e^{-i\pi yT}}\right)\\
 & =\cos(\pi yN)+ & \frac{\cos(\pi y(N-T))+i\sin(\pi y(N-T))-(\cos(\pi y(N-T))-i\sin(\pi y(N-T)))}{\cos(\pi yT)+i\sin(\pi yT)-(\cos(\pi yT)-i\sin(\pi yT))}\\
 & = & \cos(\pi yN)+\frac{2i\sin(\pi y(N-T))}{2i\sin(\pi yT)}=\cos(\pi yN)+\frac{\sin(\pi y(N-T))}{\sin(\pi yT)}
\end{eqnarray*}
(where the geometric series sum formula used above relies on the assumption
that $yT\notin\mathbb{Z}$ so that $e^{-i2\pi yT}\neq1.$ )
If $N/T$ is an even integer but we now assume $yT\in\mathbb{Z}$,
then we can write $y=\frac{l}{T}$ for some $l\in\mathbb{Z}$ and
in this case we have that 
\begin{eqnarray*}
\hat{w}_{N}(y) & = & \sum_{n=-\infty}^{\infty}w_{N}(n)e^{-i2\pi yTn}=\sum_{n=-\infty}^{\infty}w_{N}(n)e^{-i2\pi ln}=\sum_{n=-\infty}^{\infty}w_{N}(n)\\
 & = & \sum_{n=-\frac{N}{2T}}^{\frac{N}{2T}}w_{N}(n)=\frac{N}{T}
\end{eqnarray*}
If $yT\notin\mathbb{Z}$ and $N/T$ is not even, then, letting $F\equiv Floor\left[N/2T\right]$
and $R=N/Tmod2$, we have that
\begin{eqnarray*}
\hat{w}_{N}(y) & = & \sum_{n=-\infty}^{\infty}w_{N}(n)e^{-i2\pi yTn}=\sum_{n=-F}^{F}e^{-i2\pi yTn}=\sum_{t=-(N-RT)/2T}^{(N-RT)/2T}\left(e^{-i2\pi yT}\right)^{n}\\
 &  & =\frac{e^{i\pi(N-RT)y}-e^{-i\pi(N-RT+2T)y}}{1-e^{-i2\pi yT}}=\frac{e^{-i\pi yT}}{e^{-i\pi yT}}\left(\frac{e^{i\pi y(N-RT+T)}-e^{-i\pi y(N-RT+T)}}{e^{i\pi yT}-e^{-i\pi yT}}\right)\\
 &  & =\frac{\cos(\pi y(N+T(1-R)))+i\sin(\pi y(N+T(1-R)))-(\cos(\pi y(N+T(1-R)))-i\sin(\pi y(N+T(1-R)))}{\cos(\pi yT)+i\sin(\pi yT)-(\cos(\pi yT)-i\sin(\pi yT))}\\
 &  & =\frac{2i\sin(\pi y(N+T(1-R)))}{2i\sin(\pi yT)}=\frac{\sin(\pi y(N+T(1-R))}{\sin(\pi yT)}\\
\end{eqnarray*}
(Note that in the case of $N/T$ integer valued, this gives $R=1$
and so the above would simplify to $\frac{\sin(\pi yN)}{\sin(\pi yT)}$).
If $N/T$ is not an even integer but we now assume $yT\in\mathbb{Z}$,
then we again write $y=\frac{l}{T}$ and in this case we have that
\begin{eqnarray*}
\hat{w}_{N}(y) & = & \sum_{n=-\infty}^{\infty}w_{N}(n)e^{-i2\pi yTn}=\sum_{n=-\infty}^{\infty}w_{N}(n)e^{-i2\pi ln}=\sum_{n=-\infty}^{\infty}w_{N}(n)\\
 & = & \sum_{n=-\frac{N-RT}{2T}}^{\frac{N-RT}{2T}}w_{N}(n)=\frac{N-RT}{T}+1=\frac{N+T(1-R)}{T}
\end{eqnarray*}
We also know from the Poisson summation formula, that $\hat{w}_{N}(y)$
is equal to the periodic summation of $\hat{W}_{N}(y)$ at periods
of $1/T$:
$$
\hat{w}_{N}(y)=\sum_{n=-\infty}^{n=\infty}\hat{W}_{N}(y-\frac{n}{T})=\begin{cases}
N\sum_{n=-\infty}^{n=\infty}g\left(N(y-\frac{n}{T})\right)\end{cases}
$$
as the sum of a periodically sinc function was defined above. This
is precisely the definition of $f(k,y)$ above for $k=1/T$ and $N=1$,
so plugging these into the general solution of 
DTFT(y)=\begin{cases}
N/T & yT\in\mathbb{Z},\,N/T\,\mathrm{even}\\
\frac{N+T(1-R)}{T} & yT\in\mathbb{Z},\,N/T\,\mathrm{not\,even}\\
\cos(\pi yN)+\frac{\sin(\pi y(N-T))}{\sin(\pi yT)} & yT\notin\mathbb{Z},\,N/T\,\mathrm{even}\\
\frac{\sin(\pi y(N+T(1-R))}{\sin(\pi yT)} & yT\notin\mathbb{Z},\,N/T\,\textrm{not}\,\mathrm{even}
\end{cases}
where $R=NT/P\,\textrm{mod}\,2$ yields the summation is as given
above.
A: $$f(k,y)=\sum_{n=-\infty}^{\infty} \frac{\sin \pi(kn-y)}{\pi(kn-y)}=\cos(\pi y)\sum_{n=-\infty}^{\infty} \frac{\sin (\pi kn)}{\pi(kn-y)}-\sin(\pi y)\sum_{n=-\infty}^{\infty} \frac{\cos (\pi kn)}{\pi(kn-y)}$$
$\sin(\pi kn)=0\quad$ and $\quad\cos(\pi kn)=(-1)^{kn}$
$$f(k,y)=-\sin(\pi y)\sum_{n=-\infty}^{\infty} \frac{(-1)^{kn}}{\pi(kn-y)}=
\frac{\sin(\pi y)}{\pi k}\sum_{n=-\infty}^{\infty} \frac{(-1)^{kn}}{n-\frac{y}{k}}$$
$$\sum_{n=0}^{\infty} \frac{(-1)^{kn}}{n-\frac{y}{k}}=\Phi\left((-1)^k\:,\:1\:,\:-\frac{y}{k}\right)$$
$\Phi(z,s,a)$ is the Lerch function : http://mathworld.wolfram.com/LerchTranscendent.html
$$\sum_{n=-\infty}^{0} \frac{(-1)^{kn}}{n-\frac{y}{k}}=
\sum_{m=0}^{\infty} \frac{(-1)^{-km}}{-m+\frac{y}{k}}
=-\Phi\left((-1)^k\:,\:1\:,\:\frac{y}{k}\right)$$
and the term for $n=0$ is equal to$\quad\frac{1}{0-\frac{y}{k}}=-\frac{k}{y}$
$$\sum_{n=-\infty}^{\infty} \frac{(-1)^{kn}}{n-\frac{y}{k}}=\Phi\left((-1)^k\:,\:1\:,\:-\frac{y}{k}\right)-\Phi\left((-1)^k\:,\:1\:,\:\frac{y}{k}\right)-\left(-\frac{k}{y}\right)$$
$$f(y,k)=-\frac{\sin(\pi y)}{\pi k}\left(\Phi\left((-1)^k\:,\:1\:,\:-\frac{y}{k}\right)-\Phi\left((-1)^k\:,\:1\:,\:\frac{y}{k}\right)+\frac{k}{y}\right)$$
A: If $\phi(t)$ is arbitrary function and $\Phi(\omega)$ its Fourier transform, then the following identity, known as Poisson's formula is true (see [1] pg.47):
$$
\sum^{\infty}_{n=-\infty}\phi(t+nT)=\frac{1}{T}\sum^{\infty}_{n=-\infty}e^{2\pi i nt/T}\Phi\left(\frac{2\pi n}{T}\right)\textrm{, }T>0.
$$
In the case of your's, the function $\phi(t)=\frac{\sin(at)}{t}$ have Fourier transform $\Phi(t)=\pi X_{[-a,a]}(t)$ ($X_{[-a,a]}(t)$ is the characteristic function in $[-a,a]$, $a>0$). Hence we get
$$
\sum^{\infty}_{n=-\infty}\frac{\sin(a(t+n T))}{t+nT}=\frac{\pi}{T}\sum^{\infty}_{n=-\infty}e^{2\pi i n t/T}X_{[-a,a]}\left(\frac{2\pi n}{T}\right),
$$
where
$$
X_{[-a,a]}\left(\frac{2\pi n}{T}\right)=\left\{
\begin{array}{cc}
 1,\textrm{ if }\left|\frac{2\pi n}{T}\right|\leq a\textrm{, for a certain }n\in\textbf{Z}\\
 0,\textrm{ otherwise }
\end{array}\right\}.
$$
Consequently we have
$$
\sum^{\infty}_{n=-\infty}\frac{\sin(a(t+n T))}{t+nT}=\frac{\pi}{T}\sum_{|n|\leq aT/(2\pi)}e^{2\pi i n t/T}.
$$
Hence if for example $a=5/2$, $T=3$, then $n=-1,0,1$ and
$$
\sum^{\infty}_{n=-\infty}\frac{\sin(a(t+n T))}{t+nT}=\sum^{\infty}_{n=-\infty}\frac{\sin(\frac{5}{2}(t+2n))}{t+2n}=
$$
$$
=e^{2\pi i (-1) t/3}+e^{2\pi i 0 t/3}+e^{2\pi i 1 t/3}=\frac{\pi}{3}\left(1+2\cos\left(\frac{2\pi t}{3}\right)\right).
$$
Note that the Fourier transform $\Phi(x)$ of $\phi(t)$ is
$$
\Phi(x)=\int^{\infty}_{-\infty}\phi(t)e^{-itx}dt
$$
References
[1]: Athanasios Papoulis. ''The Fourier Integral and its Applications''. McGraw-Hill Book Company, Inc. United States of America, 1962
A: *

*We have:
$$f(k,y)=\sum_{n=-\infty}^{\infty} \frac{\text{sin} \pi(kn-y)}{\pi(kn-y)}$$
We consider first that $k$ is even, that means $\sin{(\pi kn-\pi y)}=\sin{(-\pi y)}$ no matter $n$
$f(k,y)=\frac{\sin(-\pi y)}{\pi}\sum_{n=-\infty}^{\infty} \frac{1}{(kn-y)}$
Let $\delta(n)=\frac{1}{(kn-y)}$, when $n=0$, $\delta=\frac{-1}{y}$
When $n$ is different than zero, $\delta(n)=\frac{1}{kn-y}$ if $n>0$, otherwise $\delta(x)=\frac{-1}{kx+y}$ where $x$ is positive and $n=-x$ which means, $\sum_{n=-\infty}^{\infty} \delta(n)=\frac{-1}{y}+\sum_{n=1}^{\infty}(\frac{1}{kn-y}-\frac{1}{kn+y})$
$$\sum_{n=1}^{\infty}(\frac{1}{kn-y}-\frac{1}{kn+y})=\left [\frac{1}{k}(ln(n-y/k)-ln(n+y/k))\right ]^\infty_1=\left [  \frac{1}{k}ln(\frac{n-y/k}{n+y/k})\right ]^\infty_1=\frac{1}{k}(ln1-ln(1-y/k)+ln(1+y/k))$$
The quantity is positive since $ln(k+y)>ln(k-y)$, that leads to:
$$f(k,y)=\frac{\sin(-\pi y)}{\pi}(\frac{-1}{y}+\frac{1}{k}ln(1+y/k)-\frac{1}{k}ln(1-y/k))$$

When $k$ is odd that leaves us with two sub-cases, $n=2l$ and $n=2l+1$.
Cases where $n=2l$ we call $f_{even}(k,y)=\frac{\sin(-\pi y)}{\pi}\sum_{n=-\infty}^{\infty} \frac{1}{(2kn-y)}$ because $\sin{(\pi kn-\pi y)}=\sin{(-\pi y)}$, Which gives the same first formula with $n$ substitued by $2n$.
If  $n=2l+1$ , $\sin{(\pi kn-\pi y)}=\sin{(\pi y)}$ because $kn$ is odd.
In such case: $f_{odd}=\frac{\sin(\pi y)}{\pi}\sum_{n=1}^{\infty} (\frac{1}{k(2n+1)-y}-\frac{1}{k(2n+1)+y})$
Merging both cases; $f=f_{odd}+f_{even}$$$=\sin(-\pi y)\left [  \frac{-1}{y}+\frac{1}{2k}ln(\frac{n-y/2k}{n+y/2k})\right ]^\infty_1+\sin(\pi y)\left [ \frac{1}{2k}ln(\frac{n+(k-y)/2k}{n+(k+y)/2k})\right ]^\infty_0$$$$=\frac{\sin(-\pi y)}{\pi}(\frac{-1}{y}+\frac{1}{2k}ln(1+y/2k)-\frac{1}{2k}ln(1-y/2k))+\frac{\sin(\pi y)}{\pi}(\frac{1}{2k}ln(\frac{k+y}{2k})-\frac{1}{2k}ln(\frac{k-y}{2k}))$$
