A function $f(z,x)$ is tempered if all of the following are true:
- $f(z, x)$ is infinitely differentiable in $z$
- $f(z,x)$ is defined for all $z,x \in \mathbb{R}$
- Every derivative of $f(z,x)$ is defined for all $z,x \in \mathbb{R}$
- For $f(z,x)$ and all its derivatives: $\lim\limits_{z \to \infty} \left\vert f^{(m)}(z,x) \right\vert = 0$ $\forall x \in \mathbb{R}$
- For $f(z,x)$ and all its derivatives $\lim\limits_{z \to -\infty} \left\vert f^{(m)}(z,x) \right\vert = 0$ $\forall x \in \mathbb{R}$
- The function, $f(z,x)$, and all its derivatives, $f^{(m)}(z,x)$, are bounded and the value of that bound for a particular derivative is given by some function, $g(m)$ with the understanding that: $\lim\limits_{m \to \infty} g(m)$ may or may not converge to some finite value.
- $\sum\limits_{n \in \mathbb{Z}} f(n, x)$ converges
- $\lim\limits_{T \to \infty}\int\limits_{t = T}^{T} f(t, x) dt$ converges
Such well behaved functions occur in harmonic analysis surprisingly often.
The Euler-Maclaurin summation formula for a function that is $(2 m)$-differentiable is: \begin{align} \sum\limits_{n = a}^{b} f(n, x) &= \int\limits_{t = a}^{b} f(t,x) dt \\ &+ \frac{1}{2}\left[ f(b,x) - f(a,x) \right] \\ &+ \sum\limits_{k = 1}^{m} \frac{B_{2k}}{(2k)!}\left[ f^{(2 k -1 )}(b,x) - f^{(2 k -1 )}(a,x) \right] \\ &+ \int\limits_{t = a}^{b} f^{(2 m)}(t,x) \frac{P_{2m}(t)}{(2m)!} dt \end{align} where $B_{2k}$ is the Bernoulli number and $P_{2m}(t)$ is the Bernoulli polynomial.
Under the assumptions of tempered given above, the second and third terms of the Euler-MacLaurin summation formula are zero for all $m$ leaving:
\begin{equation} \sum\limits_{n \in \mathbb{Z}} f(n,x) - \lim\limits_{T \to \infty}\int\limits_{t = T}^{T} f(t,x) dt = \lim\limits_{T \to \infty}\int\limits_{t = T}^{T} f^{(2 m)}(t,x) \frac{P_{2m}(t)}{(2m)!} dt \end{equation}
Since the error term is independent of the number of derivatives taken, I can use $m=1$ to get: \begin{equation} \left\vert \sum\limits_{n \in \mathbb{Z}} f(n,x) - \lim\limits_{T \to \infty}\int\limits_{t = T}^{T} f(t,x) dt \right\vert \le \frac{1}{12} \lim\limits_{T \to \infty}\int\limits_{t = T}^{T} \left\vert f^{\prime\prime}(t,x) \right\vert dt \end{equation} by taking advantage of the fact that $\left\vert P_2(t) \right\vert \le B_2 = \frac{1}{6}$
My question is this:
Under these conditions for $f(z,x)$ can I do better than this regarding the error term?
One possible improvement is if the first derivative, $f^{\prime}(z,x)$ is an even function in $z$. Then the above formula for the Euler-MacLaurin formula with symmetric limits becomes: \begin{align} \sum\limits_{n = -a}^{a} f(n, x) &= \int\limits_{t = -a}^{a} f(t,x) dt \\ &+ \frac{1}{2}\left[ f(a,x) - f(-a,x) \right] \\ &+ \int\limits_{t = -a}^{a} f^{\prime}(t,x) P_{1}(t) dt \end{align}
Since, the first derivative, $f^{\prime}(t,x)$, is even and $P_1(t)$ is odd, the integral portion of the error term goes to zero; hint: pair up the 2 panels: $[-(t+1) , -t]$ and $[t , t+1]$. The assumption of tempered implies the second term is also zero, which yields: \begin{equation} \sum\limits_{n \in \mathbb{Z}} f(n, x) = \lim_{T \to \infty} \int\limits_{t = -T}^{T} f(t,x) dt \end{equation} provided $f^{\prime}(t,x)$ is even in $t$: i.e. $f^{\prime}(-t,x)=f^{\prime}(t,x)$ $\forall x$
