What is the finite sum of sinc functions?

Question

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?

Answer 1

Letting $x=c \pi$ the sum minus 1 can be written as imaginary parts of complex sums as follows

$$s_{1}(n,x) = s(n,x)-1 \\=\Im \sum _{k=1}^n \frac{\exp (i k x)}{k x} = \Im\left(-\frac{\left(e^{i x}\right)^{n+1} \Phi \left(e^{i x},1,n+1\right)+\log \left(1-e^{i x}\right)}{x}\right)$$

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

$$s_{1}(\infty,x) = -\Im\left(\frac{1}{x} \log \left(1-e^{i x}\right)\right)$$

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.