In the paper I found the following error estimation for the discretization error $$\int_{-\infty}^\infty g(u) du=:I \approx I_h:=h \sum_{k\in \mathbb Z} g(kh),$$ for some complex valued function $g\in L^1(\mathbb R,\mathbb C)$.
Theorem: Let $w=u+iv$, with $u$ and $v$ real. Suppose $g(w)$ is holomorphic in the strip $-d<v<c$ for some $c,d>0$ and $g(w)\to 0$ uniformly as $|w|\to \infty$ in that strip. Suppose further that for some $M_+,M_->0$ the function satisfies $$\int_{-\infty}^\infty |g(u+ir)|du \leq M_+\qquad \text{and}\qquad \int_{-\infty}^\infty |g(u-is)|du,$$ for all $0<r<c$ and $0<s<d$. Then $$|I-I_h|\leq DE_++DE_-$$ with $$DE_+=\frac{M_+}{e^{2\pi c/h}-1}\qquad \text{and} \qquad DE_-=\frac{M_-}{e^{2\pi d/h}-1}.$$
However, in the paper they do not give a proof for this statement, it was only written that it can be proven by using the residue theorem or some Fourier techniques. They referred to some paper/book which I could not find. So, I tried to prove it by my own and failed. If someone has an idea to fix my attempt or has some good links I would be greatfull. The following is my attempt to prove this statement.
Lets pick some $0<r<c$ and $0<s<d$.
At first lets use Chauchy's integral theorem and the uniform decay of $g$ to obtain
$$\int_{-\infty}^\infty g(u)du = \int_{-\infty+iz}^{\infty +iz} g(u)du$$
for all $z\in (-d,c)$. Further, using the residue theorem we get
$$g(hk)=\frac{1}{2\pi i} \int_{\Gamma_R-hk}\frac{g(u)}{u-hk} du,$$
for $\Gamma_R$ describes a rectangular closed curve given by the points $(-R,ir)$ and $(R,-is)$. Then by increasing $R\to \infty$ and the uniform decay of $g$ we get
$$g(hk)=\frac 1 {2\pi i} \left( \int_{-\infty +ir}^{\infty +ir} \frac{g(u)}{u-hk}du +\int_{-\infty -is}^{\infty -is}\frac{g(u)}{u-hk}du\right).$$
It follows now
\begin{align*}|I-I_h|&= \left | \int_{-\infty}^\infty g(u)du-h \sum_{k\in \mathbb Z} g(hk) \right|\newline
&=\frac h {2 \pi}\left | \frac{\pi i}{h} \int_{-\infty +ir}^{\infty+ir} g(u) du+ \frac{\pi i}{h} \int_{-\infty-is}^{\infty-is}g(u)du-\sum_{k\in \mathbb Z}\int_{-\infty+ir}^{\infty+ir}\frac {g(u)}{u-hk} du+ \int_{-\infty-is}^{\infty-is} \frac{g(u)}{u-hk}du \right |.
\end{align*}
Then I tried to find a proper series $\{a_k\}$ with $\sum_{k\in \mathbb Z}a_k=1$ and continue with
\begin{align*}
|I-I_h| &=\frac h {2 \pi}\left | \sum_{k\in \mathbb Z}\int_{-\infty+ir}^{\infty+ir}\left(\frac {a_k\pi i}h-\frac {1}{u-hk} \right)g(u) du+ \int_{-\infty-is}^{\infty-is} \left(\frac {a_k\pi i}h-\frac {1}{u-hk} \right)g(u)du \right |\newline
^\text{triangle +Hölder}&=\frac h{2 \pi} \sum_{k\in \mathbb Z} \max_{u \in \mathbb R}\left | \frac{ia_k \pi}h- \frac 1 {u+ir-hk}\right| \cdot M_++\max_{u \in \mathbb R}\left | \frac{i a_k \pi }h- \frac 1 {u-is-hk}\right| \cdot M_-\newline
^\text{compute maxima}&=\frac h{2 \pi} \sum_{k\in \mathbb Z} \left | \frac{ra_k \pi +h}{rh}\right| \cdot M_++\left | \frac{sa_k \pi -h}{sh}\right| \cdot M_-.
\end{align*}
I was hoping to find an appropriate sequence $a_k$ such that the sums on the right side become $\frac 1 {e^{2 \pi r/c}-1}$ and $\frac 1 {e^{2 \pi s/c}-1}$. But I failed. If we find one such sequence, we are done with the proof.
1. A correct computation for $g(kh)$ is:
$$ g(kh) = \frac{1}{2\pi i} \left( \color{red}{\bbox[border:1px dotted red; padding:2px;]{-}} \int_{-\infty + ir}^{\infty + ir} \frac{g(z)}{z-hk} \, \mathrm{d}z + \int_{-\infty - is}^{\infty - is} \frac{g(z)}{z-hk} \, \mathrm{d}z \right). $$
2. Recall the well-known formula
$$ \lim_{N\to\infty} \sum_{\substack{k \in \mathbb{Z} \\ |k| \leq N}} \frac{1}{z - hk} = \frac{\pi}{h} \cot \left(\frac{\pi z}{h} \right). $$
To make use of this formula, let $R_N = (N+\frac{1}{2})h$ and let $M \geq N$ be another integer. Then
\begin{align*} \sum_{\substack{k \in \mathbb{Z} \\ |k| \leq N}} g(kh) h &= \sum_{\substack{k \in \mathbb{Z} \\ |k| \leq M}} g(kh) h \cdot \mathbf{1}_{\{|k| \leq N\}} \\ &= \sum_{\substack{k \in \mathbb{Z} \\ |k| \leq M}} \frac{h}{2\pi i} \oint_{\Gamma_{R_N}} \frac{g(z)}{z - hk} \, \mathrm{d}z \\ &= \frac{h}{2\pi i} \oint_{\Gamma_{R_N}} g(z) \sum_{\substack{k \in \mathbb{Z} \\ |k| \leq M}} \frac{1}{z - hk} \, \mathrm{d}z. \end{align*}
Letting $M \to \infty$ and invoking the cotangent limit above, which converges uniformly over the rectangle $\Gamma_{R_N}$, we get
\begin{align*} \sum_{\substack{k \in \mathbb{Z} \\ |k| \leq N}} g(kh) h = \frac{1}{2i} \oint_{\Gamma_{R_N}} g(z) \cot\left(\frac{\pi z}{h}\right) \, \mathrm{d}z. \end{align*}
Now letting $N \to \infty$ and noting that $\cot\left(\frac{\pi z}{h}\right)$ is uniformly bounded over $\bigcup_{N\geq 1} \Gamma_{R_N}$, we get
\begin{align*} \sum_{\substack{k \in \mathbb{Z} \\ |k| \leq N}} g(kh) h &= \frac{1}{2} \left( i \int_{-\infty + ir}^{\infty + ir} g(z) \cot \left(\frac{\pi z}{h} \right) \, \mathrm{d}z - i \int_{-\infty - is}^{\infty - is} g(z) \cot \left(\frac{\pi z}{h} \right) \, \mathrm{d}z \right). \end{align*}
Consequently, the difference between $I$ and $I_h$ is given by
\begin{align*} I - I_h &= \frac{1}{2} \biggl( \int_{-\infty}^{\infty} g(x+ir)\left[ 1 - i \cot \left(\frac{\pi(x+ir)}{h} \right) \right] \, \mathrm{d}x \\ &\hspace{2.5em} + \int_{-\infty}^{\infty} g(x-is)\left[ 1 + i \cot \left(\frac{\pi(x-is)}{h} \right) \right] \, \mathrm{d}x \biggr). \end{align*}
3. Now a tedious computation shows that, for $x \in \mathbb{R}$ and $r, h > 0$,
$$ \left| 1 - i \cot \left(\frac{\pi(x+ir)}{h} \right) \right| = \sqrt{\frac{2 \exp(-2\pi r/h)}{\cosh(2\pi r/h) - \cos(2\pi x/h)}} \leq \frac{2}{e^{2\pi r/h} - 1}. $$
Taking complex conjugate to the expression inside the absolute value and replacing $r$ by $s$, we also obtain
$$ \left| 1 + i \cot \left(\frac{\pi(x-is)}{h} \right) \right| \leq \frac{2}{e^{2\pi s/h} - 1}. $$
Combining all these computations, we conclude the theorem.