Sum and interpolation of hurwitz zeta functions $$f(a,x)=\sum_{\tau=-\infty}^{\infty}\frac{\exp\left(2\pi i\tau x\right)}{(\tau+a)^{p+1}}$$
Can I apply Euler-Maclauren formula to this sum?
where $a\in(0,0.5)$, p is a natural number, and $x$ is a real number
 A: The problem with using Euler-Maclaurin is that $e^{2\pi i \tau x}$ is oscillatory.  The remainder term in the Euler-Maclaurin formula will involve the integral of the absolute value of the derivative of the summand.  The oscillation of $e^{2 \pi i \tau x}$ means that this derivative will be roughly the same size as the summand itself so the remainder term is not helpful (unless $x$ is going to zero in some sense).
This is in contrast to the case with $x=0$ where each differentiation gives an extra saving factor of $(\tau +a)^{-1}$.
A: Well, if $p$ is an integer, you should realize that $\frac{1}{(\tau+a)^{p+1}}$ can be obtained by integrating $Q(x)e^{-2\pi iax}$ against $e^{-2\pi i\tau x}$ where $Q(x)$ is the (unique) polynomial of degree $p$ satisfying $Q^{(m)}(0)=e^{-2\pi ia}Q^{(m)}(1)$ for $m<p$ and $Q^{(p)}=\frac{(2\pi i)^{p+1}}{e^{-2\pi ia}-1}$.
So, your function is just $Q(x)e^{-2\pi iax}$ on $(0,1)$ extended by periodicity to the entire line. The polynomial $Q$ can be easily found for each particular $p$, so if you need some small range of $p$, you have an exact closed form formula. If you want to consider large $p$, then it is not so useful but the origianal series gives you a high precision approximation if you keep just the first few terms. Either way, you have an "expression one can work with", don't you?   
The thing that totally puzzles me is why you think that your series has any relation to the Hurwitz zeta function.
A: Sure - the wikipedia page on Euler-Maclaurin (the sub-section on asymptotic expansion of sums, linked to above) gives you what you need.  Armed with any decent computer algebra system, the rest is just lots of symbolic manipulation.
A: $$\sum _{k=1}^{\infty } e^{(2 \pi  i k) x} (a+k)^{1-p}=\sum _{n=1}^{\infty } \left(a^2 e^{-2 i \pi  a (n+x)} (-i (n+x))^p (-i a (n+x))^{-p} E_{p-1}(-2 i a \pi  (n+x))+(2 \pi )^{p-2} e^{2 i \pi  a (n-x)} (i (n-x))^{p-2} \Gamma (2-p,2 i a \pi  (n-x))\right)-\frac{a^{1-p}}{2}+a^2 e^{-2 i \pi  a x} (-i x)^p (-i a x)^{-p} E_{p-1}(-2 i a \pi  x)$$
