I need to find the Euler Maclaurin summation of a function that can be generalized as the sinc function. $$\frac{\sin(f(x))}{f(x)}$$
But the sinc's function derivative is undefined at zero, which is present across the values of $f(x)$, and so I am wondering if I utilized the central difference differential method to serve as a limiting continuous differentiation for the sinc function will the Euler Maclaurin value still be accurate?
$$\operatorname{sinc}(x)$$
Central difference:$$\frac{\operatorname{sinc}(f(x+0.0001)) - \operatorname{sinc}(f(x-0.0001))}{0.0002}$$
$\def\sinc{\operatorname{sinc}}$It's straight forward enough to fix the singularity of $\sinc$ at 0, starting with the McLaurin series of $\sin$:
$$\begin{align} \sinc x = \frac{\sin x}x &= \frac1x\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!} \\ &= \sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n+1)!} \tag 1\\ &= \sum_{n=0}^\infty \frac{\sinc^{(n)}(0)}{n!}x^n \tag 2\\ \end{align}$$ By comparing coefficients of $(1)$ and $(2)$ you will find the needed $n$-th derivative of $\sinc$ at 0:
$$\sinc^{(n)}(0) = \begin{cases} 0, & \text{ if } n \text{ is odd} \\ (-1)^m\dfrac{1}{n+1}, & \text{ if } n=2m \text{ is even} \\ \end{cases}$$