\appendix
\section{Bounding $\sum_{n\leq a} n^{-s} - \int_0^a \frac{du}{u^s}$, or approximating $\zeta(s)$}
We want a good explicit estimate on 
$$\sum_{n\leq a} \frac{1}{n^s} - \int_0^{a} \frac{du}{u^s},$$
for $a$ a half-integer. As it turns out, this is the same problem as that of approximating
$\zeta(s)$ by a sum $\sum_{n\leq a} n^{-s}$. This is one of the two\footnote{The other one is the approximate functional equation.} main, standard ways of approximating $\zeta(s)$.

The non-explicit version of the result was first proved in \cite[Lemmas 1 and 2]{zbMATH02601353}. The proof there uses first-order Euler-Maclaurin combined with a decomposition of $\lfloor x\rfloor - x +1/2$ that turns out to be equivalent to Poisson summation.
The exposition in \cite[\S 4.7--4.11]{MR882550} uses first-order Euler-Maclaurin and 
van de Corput's Process B; the main idea of the latter is Poisson summation.


There are already several explicit versions of the result in the literature. In \cite{MR1687658}, \cite{MR3105334} and \cite{MR4114203}, what we have is successively sharper
explicit versions of Hardy and Littlewood's original proof. The proof in \cite[Lemma 2.10]{zbMATH07557592} proceeds simply by a careful estimation of the terms in
high-order Euler-Maclaurin; it does not use Poisson summation. Finally,
\cite{delaReyna} is an explicit version of \cite[\S 4.7--4.11]{MR882550}; it
gives a weaker bound than \cite{MR4114203} or \cite{zbMATH07557592}. The strongest of these results is \cite{MR4114203}.

We will give another version here, in part because we wish to relax conditions -- we will work with $\left|\Im s\right| < 2\pi a$ rather than $\left|\Im s\right| \leq a$ -- and in part to show that one can prove an asymptotically optimal result easily and concisely. We will use first-order Euler-Maclaurin and Poisson summation. We assume that $a$ is a half-integer; if one inserts the same assumption into 
\cite[Lemma 2.10]{zbMATH07557592}, one can improve the result there, yielding an error term closer to the one here.

{\em Notation.}
We recall that $e(\alpha) = e^{2\pi i \alpha}$, and $O^*(R)$ means a quantity of absolute value at most $R$.
\subsection{The decay of a Fourier transform}
Our first objective will be to estimate the Fourier transform of $t^{-s} \mathds{1}_{[a,b]}$. In particular, we will show that, if $a$ and $b$ are half-integers, the Fourier cosine transform has quadratic decay {\em when evaluated at integers}. In general, for real arguments, the Fourier transform of a discontinuous function such as $t^{-s} \mathds{1}_{[a,b]}$ does not have quadratic decay.

\begin{lemma}\label{lem:aachIBP}
 Let $s = \sigma + i \tau$, $\sigma\geq 0$, $\tau\in \mathbb{R}$. Let $\nu\in \mathbb{R}\setminus \{0\}$,
 $b>a>\frac{|\tau|}{2\pi |\nu|}$.
 Then
\begin{equation}\label{eq:aachquno}\int_a^b t^{-s} e(\nu t) dt =
 \left. \frac{t^{-\sigma} e(\varphi_\nu(t))}{2\pi i \varphi_\nu'(t)}\right|_a^b
 + \sigma \int_a^b \frac{t^{-\sigma-1}}{2\pi i \varphi_\nu'(t)} e(\varphi_\nu(t)) dt + \int_a^b \frac{t^{-\sigma} \varphi_\nu''(t)}{2\pi i (\varphi_\nu'(t))^2}
 e(\varphi_\nu(t)) dt,
\end{equation}
 where $\varphi_\nu(t) = \nu t - \frac{\tau}{2\pi} \log t$.
\end{lemma}
\begin{proof}
We write $t^{-s} e(\nu t) = t^{-\sigma} e(\varphi_\nu(t))$ and integrate by parts with
$u = t^{-\sigma}/(2\pi i \varphi_\nu'(t))$, $v = e(\varphi_\nu(t))$.
Here $\varphi_\nu'(t) = \nu - \tau/(2\pi t)\ne 0$ for $t\in [a,b]$ because 
$t\geq a > |\tau|/(2\pi |\nu|)$ implies $|\nu|>|\tau|/(2\pi t)$.
Clearly 
\[u dv = \frac{ t^{-\sigma}}{2\pi i \varphi_\nu'(t)} \cdot 2\pi i \varphi_\nu'(t) e(\varphi_\nu(t)) dt = 
t^{-\sigma} e(\varphi_\nu(t)) dt,\]
while 
\[du = \left(\frac{-\sigma t^{-\sigma-1}}{2\pi i \varphi_\nu'(t)} - \frac{t^{-\sigma} \varphi_\nu''(t)}{2\pi i (\varphi_\nu'(t))^2}\right) dt.\]
\end{proof}

\begin{lemma}\label{lem:aachra}
Let $g:[a,b]\to \mathbb{R}$ be continuous, with $|g(t)|$ non-increasing. Then
$g$ is monotone, and $\|g\|_{\TV} = |g(a)|-|g(b)|$.
\end{lemma}
\begin{proof}
Suppose $g$ changed sign: $g(a')>0>g(b')$ or $g(a')<0<g(b')$ for some $a\leq a'< b'\leq b$. By IVT, there would be an $r\in [a',b']$ such that $g(r)=0$. Since $|g|$ is non-increasing, $g(b')=0$; contradiction. So, $g$ does not change sign: either $g\leq 0$ or $g\geq 0$.

Thus, there is an $\varepsilon\in \{-1,1\}$ such that  $g(t) = \varepsilon |g(t)|$ for all $t\in [a,b]$. Hence, $g$ is monotone. Then $\|g\|_{\TV} = |g(a)-g(b)|$. Since $|g(a)|\geq |g(b)|$ and $g(a)$, $g(b)$
are either both non-positive or non-negative, $|g(a)-g(b)| = |g(a)|-|g(b)|$.
\end{proof}

\begin{lemma}\label{lem:aachmonophase}
Let $\varphi:[a,b]\to \mathbb{R}$ be $C^1$ with $\varphi'(t)\ne 0$ for
all $t\in [a,b]$. Let $h:[a,b]\to \mathbb{R}$ be such that $g(t) = h(t)/\varphi'(t)$ 
is continuous and $|g(t)|$ is non-increasing. Then
\[\left|\int_a^b h(t) e(\varphi(t)) dt\right|\leq \frac{|g(a)|}{\pi}.\]
\end{lemma}
This is a statement of known type (``non-stationary phase'').
\begin{proof}
Since $\varphi$ is $C^1$, $e(\varphi(t))$ is $C^1$, and 
$h(t) e(\varphi(t)) = \frac{h(t)}{2\pi i \varphi'(t)} \frac{d}{dt} e(\varphi(t))$ everywhere.
By Lemma \ref{lem:aachra}, $g$ is of bounded variation. Hence, we can integrate by parts:
%Since $h(t)/\varphi'(t)$ is continuous and its absolute value $g(t)$ is non-increasing, it is of constant %sign, and also monotone, and thus of bounded variation.
\[\int_a^b h(t) e(\varphi(t)) dt = \left. \frac{h(t) e(\varphi(t))}{2\pi i \varphi'(t)}\right|_a^b -
\int_a^b e(\varphi(t))\, d\left(\frac{h(t)}{2\pi i \varphi'(t)}\right).
\]
The first term on the right has absolute value $\leq \frac{|g(a)|+|g(b)|}{2\pi}$.
Again by Lemma \ref{lem:aachra}, 
\[\left|
\int_a^b e(\varphi(t))\, d\left(\frac{h(t)}{2\pi i \varphi'(t)}\right)
\right|\leq \frac{1}{2\pi} \|g\|_{\TV} = \frac{|g(a)|-|g(b)|}{2\pi}.
\]
We are done by $\frac{|g(a)|+|g(b)|}{2\pi} + \frac{|g(a)|-|g(b)|}{2\pi} = \frac{|g(a)|}{\pi}$.
\end{proof}

\begin{lemma}\label{lem:aachdecre}
Let $\sigma\geq 0$, $\tau\in \mathbb{R}$, $\nu \in \mathbb{R}\setminus \{0\}$. Let $b>a>\frac{|\tau|}{2\pi |\nu|}$. Then, for any $k\geq 1$, $f(t) = t^{-\sigma-k} |2\pi \nu-\tau/t|^{-k-1}$ is decreasing on
$[a,b]$.
\end{lemma}
\begin{proof}
Let $a\leq t\leq b$. Since $\left|\frac{\tau}{t \nu}\right| < 2\pi$, we see that $2\pi-\frac{\tau}{\nu t} >0$, and so
$|2\pi \nu-\tau/t|^{-k-1} = |\nu|^{-k-1} \left(2\pi - \frac{\tau}{t \nu}\right)^{-k-1}$. Now
we take logarithmic derivatives:
\[t (\log f(t))' = - (\sigma+k) - (k+1) \frac{\tau/t}{2\pi \nu - \tau/t}
= -\sigma - \frac{2\pi k + \frac{\tau}{t \nu}}{2\pi - \frac{\tau}{t \nu}} < -\sigma \leq 0,\]
since, again by $\frac{|\tau|}{t |\nu|} < 2\pi$ and $k\geq 1$, we have
$2\pi k + \frac{\tau}{t \nu}>0$, and, as we said, $2\pi - \frac{\tau}{t \nu}>0$.
\end{proof}

    
\begin{lemma}\label{lem:aachfour}
 Let $s = \sigma + i \tau$, $\sigma\geq 0$, $\tau\in \mathbb{R}$.  Let
 $\nu \in \mathbb{R}\setminus \{0\}$, $b>a>\frac{|\tau|}{2\pi |\nu|}$.
 Then
\[\int_a^b t^{-s} e(\nu t) dt =
 \left. \frac{t^{-\sigma} e(\varphi_\nu(t))}{2\pi i \varphi_\nu'(t)}\right|_a^b + 
\frac{a^{-\sigma-1}}{2\pi^2} O^*\left(\frac{\sigma}{(\nu-\vartheta)^2} +
\frac{|\vartheta|}{|\nu-\vartheta|^3}\right),
\]
 where $\varphi_\nu(t) = \nu t - \frac{\tau}{2\pi} \log t$ and $\vartheta = \frac{\tau}{2\pi a}$.
\end{lemma}
\begin{proof}
Apply Lemma~\ref{lem:aachIBP}. Since $\varphi_\nu'(t) = \nu - \tau/(2\pi t)$, we know by
Lemma \ref{lem:aachdecre} (with $k=1$) that $g_1(t) = \frac{t^{-\sigma-1}}{(\varphi_\nu'(t))^2}$ is decreasing on
$[a,b]$. We know that $\varphi_\nu'(t)\ne 0$ for $t\geq a$ by $a>\frac{|\tau|}{2\pi |\nu|}$, and so
we also know that $g_1(t)$ is continuous for $t\geq a$. Hence, by Lemma \ref{lem:aachmonophase},
\[\left|\int_a^b \frac{t^{-\sigma-1}}{2\pi i \varphi_\nu'(t)} e(\varphi_\nu(t)) dt \right|\leq
\frac{1}{2\pi} \cdot \frac{|g_1(a)|}{\pi} = \frac{1}{2\pi^2} \frac{a^{-\sigma-1}}{\left|\nu - \vartheta\right|^2},\]
since $\varphi_\nu'(a) = \nu - \vartheta$. We remember to include the factor of $\sigma$
in front of an integral in \eqref{eq:aachquno}.

Since $\varphi_\nu'(t)$ is as above and $\varphi_\nu''(t) = \tau/(2\pi t^2)$,
we know
by Lemma \ref{lem:aachdecre} (with $k=2$) that 
$g_2(t) = \frac{t^{-\sigma} |\varphi_\nu''(t)|}{|\varphi_\nu'(t)|^3} = 
\frac{|\tau|}{2\pi}  \frac{t^{-\sigma-2}}{|\varphi_\nu'(t)|^3} $ is decreasing on $[a,b]$; we also know, as before,
that $g_2(t)$ is continuous. Hence, again by Lemma \ref{lem:aachmonophase},
\[\left|\int_a^b \frac{t^{-\sigma} \varphi_\nu''(t)}{2\pi i (\varphi_\nu'(t))^2}
 e(\varphi_\nu(t)) dt\right|\leq \frac{1}{2\pi} \frac{|g_2(a)|}{\pi} = \frac{1}{2\pi^2}
 \frac{a^{-\sigma-1} |\vartheta|}{\left|\nu - \vartheta\right|^3}.
 \]
\end{proof}

\begin{lemma}\label{lem:aachcanc}
Let $s = \sigma + i \tau$, $\sigma,\tau \in \mathbb{R}$. 
Let $n\in \mathbb{Z}_{>0}$. Let $a,b\in \mathbb{Z} + \frac{1}{2}$, $b>a>\frac{|\tau|}{2\pi n}$. Write $\varphi_\nu(t) = \nu t - \frac{\tau}{2\pi} \log t$.
Then
\[\frac{1}{2} \sum_{\nu = \pm n}\left. \frac{t^{-\sigma} e(\varphi_\nu(t))}{2\pi i \varphi_\nu'(t)}\right|_a^b =
\left. \frac{(-1)^n t^{-s} \cdot \frac{\tau}{2\pi t}}{2\pi i \left(n^2 - \left(\frac{\tau}{2\pi t}\right)^2\right)}\right|_a^b .
\]
\end{lemma}
It is this easy step that gives us quadratic decay on $n$. It is just as
in the proof of van der Corput's Process B in, say, \cite[I.6.3, Thm.~4]{zbMATH06471876}.
\begin{proof}
Since $e(\varphi_\nu(t)) = e(\nu t) t^{-i \tau} = (-1)^{\nu} t^{-i \tau}$ for any half-integer $t$ and any integer $\nu$,
\[\left. \frac{t^{-\sigma} e(\varphi_\nu(t))}{2\pi i \varphi_\nu'(t)}\right|_a^b = 
\left. \frac{(-1)^{\nu} t^{-s}}{2\pi i \varphi_\nu'(t)}\right|_a^b 
\]
for $\nu = \pm n$. Clearly $(-1)^{\nu} = (-1)^n$. Since $\varphi_\nu'(t) = \nu - \alpha$ for $\alpha = \frac{\tau}{2\pi t}$,
\[\frac{1}{2} \sum_{\nu = \pm n} \frac{1}{\varphi_\nu'(t)} = \frac{1/2}{n - \alpha} + 
\frac{1/2}{- n - \alpha} = \frac{-\alpha}{\alpha^2-n^2} = \frac{\alpha}{n^2-\alpha^2}.
\]
\end{proof}

\begin{proposition}\label{prop:applem} Let $s = \sigma + i \tau$, $\sigma\geq 0$, $\tau\in \mathbb{R}$. 
 Let $a,b\in \mathbb{Z} + \frac{1}{2}$, $b>a>\frac{|\tau|}{2\pi}$. Write $\vartheta = \frac{\tau}{2\pi a}$.  Then, for any integer $n\geq 1$,
$$\begin{aligned}\int_a^b t^{-s} \cos 2\pi n t\, dt &= 
\left. \left(\frac{(-1)^n t^{-s}}{2\pi i} \cdot\frac{\frac{\tau}{2\pi t}}{n^2 - \left(\frac{\tau}{2\pi t}\right)^2}\right)\right|_a^b \\ &\quad+ \frac{a^{-\sigma-1}}{4\pi^2}\, O^*\left(\frac{\sigma}{(n-\vartheta)^2} + \frac{\sigma}{(n+\vartheta)^2} + \frac{|\vartheta|}{|n-\vartheta|^3} + \frac{|\vartheta|}{|n+\vartheta|^3}\right).\end{aligned}$$
\end{proposition}
\begin{proof}
Write $\cos 2\pi n t = \frac{1}{2} (e(n t) + e(-n t))$. Since $n\geq 1$ and $a>\frac{|\tau|}{2\pi}$, we know that $a>\frac{|\tau|}{2 \pi n}$, and so we can apply
Lemma \ref{lem:aachfour} with $\nu = \pm n$. 
We then apply Lemma~\ref{lem:aachcanc} to combine the boundary contributions $\left.  \right|_a^b$
for $\nu=\pm n$.
\end{proof}