An analytic-strip error bound for the infinite trapezoid sum

Define $K_h(z)=\frac{\pi i}{h}\frac{e^{2\pi iz\mathrm{/}h}+1}{e^{2\pi iz\mathrm{/}h}-1}$. By [F2] and [F3], its poles are exactly $z=kh$, $k\in \mathbb{Z}$: if the denominator vanishes, its modulus gives $\mathrm{Im}z=0$, and Euler's formula then gives $2\pi z\mathrm{/}h=2j\pi$. At each such point the denominator has derivative $2\pi i\mathrm{/}h$ and the numerator is $2$, so $K_h$ has residue $1$. By [F8], $F{K}_h$ is holomorphic away from these points and has a simple pole with residue $F(kh)$ when $F(kh)\mathrm{\ne }0$, or a removable singularity with residue zero when $F(kh)=0$; fill in each removable value. Thus it is meromorphic on the strip.

Fix $N\ge 0$ and $R=(N+\frac{1}{2})h$. Let ${\mathrm{\Gamma }}_R$ traverse the rectangle $[-R,R]+i[-a,a]$ counterclockwise, beginning at $R-ia$. It is a closed piecewise linear contour inside $S_d$. No $kh$ lies on its boundary: the horizontal edges have nonzero imaginary part and $R\mathrm{/}h=N+1\mathrm{/}2$ is not an integer. Every exterior point belongs to the unbounded component of the complement of this rectangle boundary, since it can be joined to infinity along an outward horizontal or vertical ray; hence its index is zero by [F6]. For an interior point $p$, each ray from $p$ meets the boundary exactly once, and along a positive edge $z(t)$ one has $\mathrm{Im}(z^{\mathrm{\prime }}(t)\mathrm{/}(z(t)-p))>0$, because $p$ lies strictly to the left of that edge. The continuous argument therefore increases through exactly one full turn, so [F6] gives index one. In particular ${\mathrm{\Gamma }}_R$ is null-homologous in $S_d$. Together with the meromorphicity and boundary check above, this is precisely the admissibility required in [F4]. Its enclosed possible poles are exactly $kh$, $-N\le k\le N$, each of index one; removable points contribute zero. Thus ${\int }_{{\mathrm{\Gamma }}_R}F{K}_{h}dz=2\pi i{\sum }_{k=-N}^{N}F(kh)$.

On the upper edge, $\mathrm{\mid }{e}^{2\pi i(x+ia)\mathrm{/}h}\mathrm{\mid }=\rho :=e^{-2\pi a\mathrm{/}h}<1$ and $\mid K_h(x+ia)+\frac{\pi i}{h}\mid =\frac{2\pi }{h}\mid \frac{e^{2\pi i(x+ia)\mathrm{/}h}}{e^{2\pi i(x+ia)\mathrm{/}h}-1}\mid \le \frac{2\pi \mathrm{/}h}{e^{2\pi a\mathrm{/}h}-1}\mathrm{.}$ On the lower edge, the analogous estimate is $\mid K_h(x-ia)-\pi i\mathrm{/}h\mid \le (2\pi \mathrm{/}h)\mathrm{/}(e^{2\pi a\mathrm{/}h}-1)$. On either vertical edge, $e^{2\pi iz\mathrm{/}h}$ is negative real and $\mathrm{\mid }{K}_h(z)\mathrm{\mid }\le \pi \mathrm{/}h$. The vertical integrals therefore tend to zero as $N\to \mathrm{\infty }$ by the uniform strip bound in the statement.

[F7] applies to $F$ on each filled rectangle $[-R,R]+i[0,a]$ and $[-R,R]+i[-a,0]$, since both compact rectangles lie inside the open strip $S_d$ where $F$ is holomorphic. Their positively oriented boundaries have no poles at all, including their real edges. Their zero contour integrals and the same vanishing vertical-side bound for $F$ show ${\int }_{\mathbb{R}}F(x\pm ia)dx=I$. Replacing $K_h$ on the two long edges of step 2.1 by its constants $\mp \pi i\mathrm{/}h$, the residue identity becomes $h{\sum }_{k=-N}^{N}F(kh)=\frac{1}{2}({\int }_{-R}^{R}F(x-ia)dx+{\int }_{-R}^{R}F(x+ia)dx)+E_N,$ where $\mathrm{\mid }{E}_N\mathrm{\mid }\le ({\int }_{-R}^R\mathrm{\mid }F(x-ia)\mathrm{\mid }dx+{\int }_{-R}^R\mathrm{\mid }F(x+ia)\mathrm{\mid }dx)\mathrm{/}(e^{2\pi a\mathrm{/}h}-1)+o(1)$ as $N\to \mathrm{\infty }$.

The real and horizontal sums and integrals converge absolutely by the exponential bound. Letting $N\to \mathrm{\infty }$ in step 4.1, the average tends to $I$ and the error term is at most $(I_{-}+I_{+})\mathrm{/}(e^{2\pi a\mathrm{/}h}-1)$. This proves the assertion.