Is there a simplification of the following function $$ f(N) \equiv \sum_{n=0}^N\text{sinc}(\pi n c) $$ where $c$ is a constant?

You could absorb $\pi$ in $c$. – Nov 07 '19 at 12:46

Since the Fourier transform of a sinc is a box, the Fourier transform of this function looks like a ziggurat. That would help understand the limiting behaviour, at least, because as $N$ becomes large the ziggurat tends to a triangle. – Nov 07 '19 at 12:52

Poisson summation would be a good method to solve this using Fourier as N approaches infinity, but unfortunately it does not apply for finite series. – Veselin Manev Nov 07 '19 at 12:55

Hint: The finite sum can be expressed in closed form using the Lerch Phi function $\Phi (z, 1, n) =\sum _{k=0}^{\infty } \frac{z^k}{k+n}$ and the $\log$function. – Dr. Wolfgang Hintze Nov 07 '19 at 13:46
1 Answers
Letting $x=c \pi$ the sum minus 1 can be written as imaginary parts of complex sums as follows
$$s_{1}(n,x) = s(n,x)1 \\=\Im \sum _{k=1}^n \frac{\exp (i k x)}{k x} = \Im\left(\frac{\left(e^{i x}\right)^{n+1} \Phi \left(e^{i x},1,n+1\right)+\log \left(1e^{i x}\right)}{x}\right)$$
where the Lerch Phi function is defined as $\Phi (z, 1, n) =\sum _{k=0}^{\infty } \frac{z^k}{k+n}$.
The graph shows the sum for $n=1$, $n=4$, and $n\to\infty$
Notice that the limiting function
$$s_{1}(\infty,x) = \Im\left(\frac{1}{x} \log \left(1e^{i x}\right)\right)$$
has jumps at $x= 2\pi m, m =1,2,3...$, which can easily be shown to happen by the amount $\frac{1}{2 m}$ from $1\frac{1}{4 m}$ to $1+\frac{1}{4 m}$.
The limiting function and shows the overall behaviour also of the finite sum. In the finite case the value of the sum wiggles around the limiting function according to the $\Phi$function, which goes to $0$ for large $n$.
The utilizing of these relations requires more specified questions as in the OP.
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