\section{Introduction}
Let
\begin{equation}
 s=\sigma+\ensuremath{\mathrm{i}} t, \quad \sigma \in \mathbf{R}, \quad t \in \mathbf{R}.
\end{equation}
The Riemann zeta function $\zeta(s)$ and the modified Hurwitz zeta function $\zeta_1(s,\alpha)$ are defined, respectively, by
\begin{equation}
\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}, \quad \sigma>1,
\end{equation}
\begin{equation}
\zeta_1(s,\alpha) = \sum_{n=1}^\infty \frac{1}{(n+\alpha)^s}, \quad \sigma>1, \quad \alpha \geq 1, \label{th2.1}
\end{equation}
and by analytic continuation for $s\in \mathbf{C}$, $s\neq 1$.

The modified Hurwitz zeta function is simply related to the Hurwitz zeta function, $\zeta(s,\alpha)$:
\begin{equation}
\zeta(s,\alpha) = \frac{1}{\alpha^s} + \zeta_1(s,\alpha), \quad s\in \mathbf{C}, \alpha>0.
\end{equation}

In this paper we present certain relations between $\zeta(s)$ and $\zeta_1(s,\alpha)$ as well as between products of these functions. In more detail, the following results are presented in sections 2--6:

In \S 2 it is shown that the modified Hurwitz zeta function satisfies the identity
\begin{multline} |\zeta_1(s,\alpha)|^2 = \zeta_1(2\sigma,\alpha) + \frac{1}{2\pi\ensuremath{\mathrm{i}}} \int_{(c)} \left( \frac{\Gamma(s+z)}{\Gamma(s)} + \frac{\Gamma(\bar{s}+z)}{\Gamma(\bar{s})} \right) \Gamma(-z) \zeta(-z) \zeta_1(2\sigma+z,\alpha)\, \ensuremath{\mathrm{d}} z  \\
\sigma>1, \quad t>0, \quad \max(-\sigma,1-2\sigma) < c < 1, \label{f1.5}
\end{multline}
where $\Gamma(s)$, $s\in \mathbf{C}$, denotes the Gamma function and $(c)$ denotes the vertical line in the complex $z$-plane on which $\Re(z)=c$.

It is shown in \cite{fokas2016} that \eqref{f1.5} yields a singular integral equation for $|\zeta(s)|^2$, $0<\sigma<1$, $t>0$, and this equation provides the starting point for the proof of the analogue of Lindel\"{o}f's hypothesis for a function which differs from the Riemann zeta function only in the occurrence of the $\log n$ term. We recall that Lindel\"{o}f's hypothesis concerns the growth of $\zeta(s)$ as $t\rightarrow \infty$ along the critical line $\sigma=1/2$ and states that $\zeta(\tfrac{1}{2}+\ensuremath{\mathrm{i}} t) = \mathcal{O}(t^\epsilon)$ for every positive $\epsilon$. The Riemann hypothesis implies Lindel\"of's hypothesis and conversely, Lindel\"of's hypothesis implies that very few zeros could disobey Riemann's hypothesis \cite{gelbartmiller2003}. Significant progress has been made by developing ingenuous ways of estimating exponential sums $\sum_n e^{2\pi\ensuremath{\mathrm{i}} f(n)}$ using generalisations of the Vinogradov method \cite{vinogradov1935new}. Until recently the best result in this direction had been obtained by Huxley \cite{huxley2005exponential}, where it is proved that $\zeta(\tfrac{1}{2}+\ensuremath{\mathrm{i}} t) = \mathcal{O} \left( t^{32/205 + \epsilon}\right)$. A short time ago Bourgain announced a further improvement \cite{bourgain2014decoupling} where the exponent $32/205$ was reduced to $53/342$.

In \S 3 it is shown that there exists the following asymptotic relation between the Riemann zeta function and the modified Hurwtiz zeta function:
\begin{equation}
\zeta(s) = \int_0^1 B_N(\alpha) \zeta_1(s,\alpha)\, \ensuremath{\mathrm{d}} \alpha + \chi(s) \int_0^1 B_N(-\alpha) \zeta_1(1-s,\alpha)\,\ensuremath{\mathrm{d}} \alpha + \mathcal{O}(t^{-\sigma/2} \log t), \quad 0<\sigma<1, \quad t\rightarrow \infty, \label{f1.6}
\end{equation}
where $B_N(\alpha)$ is defined by
\begin{equation}
B_N(\alpha) = \sum_{1\leq n \leq N} e^{2\pi\ensuremath{\mathrm{i}} n \alpha}, \quad \alpha>0.
\end{equation}
and $N=\sqrt{t/2\pi}$. A direct consequence of \eqref{f1.6} is the following theorem:
\begin{theorem}
Let $I_k(t)$ denote the $2k$-th power mean of the modified Hurwitz zeta function, namely
\begin{equation}
I_k(t) = \int_0^1 | \zeta_1(\tfrac{1}{2}+\ensuremath{\mathrm{i}} t, \alpha)|^{2k}\, \ensuremath{\mathrm{d}} \alpha, \quad k \in \mathbf{N}.
\end{equation}
Then
\begin{equation} |\zeta(\tfrac{1}{2}+\ensuremath{\mathrm{i}} t)| \lesssim   t^{1/4k} I_k(t)^{1/2k}, \qquad k\in \mathbf{N} \label{Fok5}\end{equation}
in which the implicit constant is independent of $k$.
\end{theorem}

This result immediately implies that Lindel\"{o}f's hypothesis is true provided that for each $\epsilon>0$ $I_k(t) \lesssim_{k,\epsilon} t^{\epsilon}$ for each $k\in \mathbf{N}$.

In connection with \eqref{Fok5} we recall that an equivalent formulation of the Lindel\"of hypothesis involves estimating the $2k$-th power mean of the Riemann-zeta function
\begin{equation} J_k(T) = \frac{1}{T} \int_0^T \left| \zeta(\tfrac{1}{2}+\ensuremath{\mathrm{i}} t) \right|^{2k}\, \ensuremath{\mathrm{d}} t. \label{zetameans}\end{equation}
It can be shown \cite[Th. 13.2]{titchmarsh1986theory} that the Lindel\"of hypothesis holds true if and only if $J_k(T) = \mathcal{O}\left( T^\epsilon\right)$ for each $\epsilon>0$ and for each $k\in \mathbf{N}$.

In \S 4 the following identities are presented:
\begin{equation}
\int_0^1 \zeta_1(u_1,\alpha)\zeta_1(u_2,\alpha)\, \ensuremath{\mathrm{d}} \alpha = \frac{1}{u_1+u_2-1} + \int_1^\infty \alpha^{-u_1} \zeta_1(u_2,\alpha)\, \ensuremath{\mathrm{d}} \alpha + \int_1^\infty \alpha^{-u_2} \zeta_1(u_1,\alpha)\,\ensuremath{\mathrm{d}} \alpha, \label{f1.11}
\end{equation}
\begin{multline}
\int_0^1 \zeta_1(u_1,\alpha)\zeta_1(u_2,\alpha) \zeta_1(u_3,\alpha)\, \ensuremath{\mathrm{d}} \alpha = \frac{1}{u_1+u_2+u_3-1} + \int_1^\infty \alpha^{-u_1} \zeta_1(u_2,\alpha) \zeta_1(u_3,\alpha)\, \ensuremath{\mathrm{d}} \alpha \\
+ \int_1^\infty \alpha^{-u_2} \zeta_1(u_1,\alpha) \zeta_1(u_3,\alpha)\, \ensuremath{\mathrm{d}} \alpha + \int_1^\infty \alpha^{-u_3} \zeta_1(u_1,\alpha) \zeta_1(u_2,\alpha)\, \ensuremath{\mathrm{d}} \alpha \\
+ \int_1^\infty \alpha^{-u_1-u_2} \zeta_1(u_3,\alpha)\, \ensuremath{\mathrm{d}} \alpha + \int_1^\infty \alpha^{-u_2-u_3} \zeta_1(u_1,\alpha)\,\ensuremath{\mathrm{d}}\alpha + \int_1^\infty \alpha^{-u_3-u_1} \zeta_1(u_2,\alpha)\, \ensuremath{\mathrm{d}}\alpha, \label{f1.12}
\end{multline}
and
\begin{multline}
\int_0^1 \prod_{i=1}^4 \zeta_1(u_i,\alpha)\, \ensuremath{\mathrm{d}} \alpha = \frac{1}{u_1+u_2+u_3+u_4-1} + \sum_{\mathrm{perms}} \int_1^\infty \alpha^{-(u_1+u_2+u_3)}\zeta_1(u_4,\alpha)\, \ensuremath{\mathrm{d}} \alpha \\
+\sum_{\mathrm{perms}} \int_1^\infty \alpha^{-(u_1+u_2)} \zeta_1(u_3,\alpha)\zeta_1(u_4,\alpha)\, \ensuremath{\mathrm{d}} \alpha + \sum_{\mathrm{perms}} \int_1^\infty \alpha^{-u_1} \zeta_1(u_2,\alpha) \zeta_1(u_3,\alpha) \zeta_1(u_4,\alpha)\, \ensuremath{\mathrm{d}} \alpha, \label{f1.13}
\end{multline}
where $\Re \{u_i\}_{i=1}^4 <1$. The above formulae can be generalised in a straightforward way.

As a direct application of \eqref{f1.11} we present in \S 4 a new derivation of the following exact identity in \cite{katsurada1996explicit}:
\begin{multline} \int_0^1 \zeta_1(u,\alpha) \zeta_1(v,\alpha)\, \ensuremath{\mathrm{d}} \alpha = \frac{1}{u+v-1} + \Gamma(u+v-1) \zeta(u+v-1) \left[ \frac{\Gamma(1-u)}{\Gamma(u)} + \frac{\Gamma(1-v)}{\Gamma(v)} \right] \\
+ \frac{\zeta(u)-1}{v-1} + \frac{\zeta(v)-1}{u-1} + \frac{u}{v-1} \int_0^1 \alpha^{1-v}\zeta_1(\alpha,u+1)\, \ensuremath{\mathrm{d}} \alpha + \frac{v}{u-1} \int_0^1 \alpha^{1-u} \zeta_1(\alpha,v+1)\, \ensuremath{\mathrm{d}}\alpha. \label{I1} \end{multline}
This immediately yields the estimate
\begin{equation} I_1(t) = \log \left( \frac{t}{2\pi}\right) + \gamma + \mathcal{O} \left( \frac{1}{t^2}\right), \quad t\rightarrow \infty. \label{(1.5)} \end{equation}

It is shown in \cite{fokaskalim2016} that \eqref{f1.12} plays a crucial role for the derivation of an interesting identity between certain double exponential sums. Indeed, it is well known that if $0\leq \alpha < 1$, then the large $t$-asymptotics of $\zeta_1(s,\alpha)$ is dominated by the sum $S_1$, defined by
\begin{equation} S_1(\sigma,t,\alpha) = \sum_{1\leq n < t/2\pi} e^{-2\pi\ensuremath{\mathrm{i}} n \alpha} m^{s-1}, \quad 0<\sigma<1, t>0. \label{f1.16}
\end{equation}
However, if $1\leq \alpha <\infty$, the large $t$-asymptotics of $\zeta_1(s,\alpha)$ is dominated by the sum $S_1$ defined in \eqref{f1.16}, as well as by a different sum, $S_2'(\sigma,t,\alpha)$ \cite{fokasfern2016}. Thus, the large $t$-asymptotics of equation \eqref{f1.12} provides a relation between two double sums generated from $S_1$ and $S_2'$, and  the explicit formulae obtained from the large $t$-asymptotics of the linear and quadratic terms.

Similarly, equation \eqref{f1.13} yields novel relations between cubic exponential sums.

Before using equations \eqref{f1.12} and \eqref{f1.13} for the case that $\Re \{u_i\}_{i=1}^4<1$, it is necessary to regularize the terms involving $\alpha \rightarrow \infty$; this regularisation is discussed in \S 4.

Finally, in \S 5, by considering the Fourier series of the product $\zeta_1(u,\alpha)\zeta_1(v,\alpha)$ with complex numbers $u,v$ satisfying $\Re u>0, \Re v>0$, by using certain elementary estimates for the resulting coefficients, and by employing theorem 1 together with Parseval's identity, we obtain the following asymptotic result.
\begin{theorem}
For each $\eta>0$ we have
\[ |\zeta(\tfrac{1}{2}+\ensuremath{\mathrm{i}} t)|^4 \lesssim_\eta t^{1/2}  \sum_{|n|\leq t/\pi} \left| \int_1^{t/2\pi+\eta} \alpha^{-1/2+\ensuremath{\mathrm{i}} t} \zeta_1(\alpha,\tfrac{1}{2}+\ensuremath{\mathrm{i}} t) e^{-2\pi\ensuremath{\mathrm{i}} n \alpha}\, \ensuremath{\mathrm{d}} \alpha\right|^2. \]
In particular, if the sum is $\mathcal{O}_\epsilon(t^{\epsilon})$ for each $\epsilon>0$ then $|\zeta(\tfrac{1}{2}+\ensuremath{\mathrm{i}} t)| = \mathcal{O}_\epsilon (t^{1/8+\epsilon})$.
\end{theorem}

\section{An identity involving the Hurwitz function}
In order to derive \eqref{f1.5} we let
\begin{equation}
\alpha \geq 0, \quad u\in \mathbf{C}, \quad v \in \mathbf{C}, \quad \Re u>1, \quad \Re v>1. \label{th1.1}
\end{equation}
The definition of the modified Hurwitz zeta function, namely equation \eqref{th2.1}, implies
\begin{equation}
\zeta_1(u,\alpha) \zeta_1(v,\alpha) = \zeta_1(u+v,\alpha) + f(u,v,\alpha) + f(v,u,\alpha) \label{th1.2}
\end{equation}
where
\begin{equation}
f(u,v,\alpha) = \sum_{n=1}^\infty\sum_{m=1}^\infty (n+\alpha)^{-v} (n+m+\alpha)^{-u}.
\end{equation}
Assuming that
\begin{equation}
|\arg(-w)|<\pi, \quad -\Re a < b < 0,
\end{equation}
we observe the Mellin-Barnes type integral identity
\begin{equation}
\Gamma(a)(1-w)^{-a} = \frac{1}{2\pi\ensuremath{\mathrm{i}}} \int_{(b)} \Gamma(z+a) \Gamma(-z) (-w)^z\, \ensuremath{\mathrm{d}} z. \label{th1.4}
\end{equation}
Letting in equation \eqref{th1.4}
\[ w=- \frac{m}{n+\alpha}, \quad a=u, \]
we find
\[ \Gamma(u) (n+\alpha)^u (m+n+\alpha)^{-u} = \frac{1}{2\pi\ensuremath{\mathrm{i}}} \int_{(c)} \Gamma(u+z) \Gamma(-z) m^z (n+\alpha)^{-z}\, \ensuremath{\mathrm{d}} z. \]
Thus,
\[ (n+\alpha)^{-v} (n+m+\alpha)^{-u} = \frac{1}{2\pi\ensuremath{\mathrm{i}}} \int_{(c)} \frac{\Gamma(u+z)}{\Gamma(u)} \Gamma(-z) m^z (n+\alpha)^{-u-v-z}\, \ensuremath{\mathrm{d}} z. \]
Summing over $m$ and $n$ we obtain
\begin{equation}
f(u,v,\alpha) = \frac{1}{2\pi\ensuremath{\mathrm{i}}} \int_{(c)} \frac{\Gamma(u+z)}{\Gamma(u)} \Gamma(-z) \zeta(-z) \zeta_1(u+v+z,\alpha)\, \ensuremath{\mathrm{d}} z. \label{th1.5}
\end{equation}
Substituting \eqref{th1.5} into \eqref{th1.2} and then letting $u=s$, $v=\bar{s}$ in the resulting expression, we find \eqref{f1.5}.

\section{An asymptotic relation between the Riemann and Hurwitz functions}
The approximate functional equation for the Riemann zeta function provides the starting point for the estimation of the $\zeta(s)$ along the critical line. In this section we derive a weak analogue of this equation. Throughout we will set $N=\sqrt{t/2\pi}$ and refer to the sum
\[  B_N(\alpha) = \sum_{1\leq n \leq N} e^{2\pi\ensuremath{\mathrm{i}} n \alpha} \equiv \frac{e^{\ensuremath{\mathrm{i}} \pi (N+1)\alpha} \sin (N\pi \alpha)}{\sin (\pi \alpha)}. \]
This function is similar to the classical Dirichlet kernel that arises in Fourier analysis. As such, we have the following well-known estimates.
\begin{lemma}\label{dirichletest}
$\|B_N \|_{p} = \mathcal{O}(\log N)$ if $p=1$ and $\|B_N\|_{p} = \mathcal{O} (N^{1-1/p})$ for $p>1$.
\end{lemma}
Our first result expresses the approximate functional equation for $\zeta(s)$ as an integral equation involving the Hurwitz zeta function $\zeta_1(s,\alpha)$. The proof of Theorem 1 will follow directly from this result.

\begin{lemma}\label{foklemma1}
Let $s=\sigma+\ensuremath{\mathrm{i}} t$ and $B_N$ as previously defined. Then we have
\begin{align} \zeta(s) &= \int_0^1 B_N(\alpha)\zeta_1(s,\alpha)\, \ensuremath{\mathrm{d}} \alpha + \chi(s) \int_0^1 B_N(-\alpha) \zeta_1(1-s,\alpha)\, \ensuremath{\mathrm{d}} \alpha \nonumber \\
&\quad- \int_0^1 B_N(\alpha) \sum_{1\leq n\leq N} (n+\alpha)^{-s}\, \ensuremath{\mathrm{d}} \alpha - \chi(s) \int_0^1 B_N(-\alpha) \sum_{1\leq n\leq N} (n+\alpha)^{s-1}\, \ensuremath{\mathrm{d}}\alpha \nonumber \\
&\quad  + \mathcal{O}\left( t^{-\sigma/2} \log t\right), \label{weakapprox}
\end{align}
when $0<\sigma<1$. Furthermore.
\begin{equation} \zeta(s) = \int_0^1 B_N(\alpha)\zeta_1(s,\alpha)\, \ensuremath{\mathrm{d}} \alpha + \chi(s) \int_0^1 B_N(-\alpha) \zeta_1(1-s,\alpha)\, \ensuremath{\mathrm{d}} \alpha + \mathcal{O} \left( t^{-\sigma/2} \log t\right), \quad t\rightarrow \infty, \label{weakapprox2} \end{equation}
where $\chi(s) = 2^{s-1} \pi^s \sin(\pi s/2) \Gamma(1-s)$ with $\Gamma(s)$ denoting the Gamma function and $\zeta_1(s,\alpha)$ denoting the modified Hurwitz zeta function.
\end{lemma}
Let us first establish \eqref{weakapprox}. The identity in \eqref{weakapprox2} will be a consequence of this.
\begin{proof}
First we recall the approximate functional equations for $\zeta(s)$ and $\zeta_1(s,\alpha)$ (see \cite{rane1983hurwitz} and references therein)
\begin{equation} \zeta(\sigma+\ensuremath{\mathrm{i}} t) = \sum_{1\leq n \leq N} n^{-s} + \chi(s) \sum_{1\leq n \leq N} n^{s-1} + \mathcal{O}\left( t^{-\sigma/2}\right), \label{fok1} \end{equation}
and
\begin{equation} \zeta_1(s,\alpha) = \sum_{1\leq n \leq N} (n+\alpha)^{-s} + \chi(s) \sum_{1\leq n \leq N} e^{-2\pi\ensuremath{\mathrm{i}} n\alpha} n^{s-1} + \mathcal{O}\left( t^{-\sigma/2}\right), \label{fok2} \end{equation}
uniformly in $0<\alpha<1$. The following identity is valid
\begin{equation} \sum_{1\leq n \leq N} n^z = \int_0^1 B_N(\alpha) \left( \sum_{1\leq m \leq N} e^{-2\pi\ensuremath{\mathrm{i}} m\alpha} m^z \right) \ensuremath{\mathrm{d}} \alpha, \quad z\in \mathbf{C}. \label{fok3}\end{equation}
Indeed the left hand side of \eqref{fok3} can be rewritten in the form
\begin{align*}  \sum_{1\leq m,n\leq N} \delta_{mn} m^z &= \sum_{1\leq m,n\leq N} m^z \int_0^1 e^{2\pi\ensuremath{\mathrm{i}} (n-m)\alpha}\, \ensuremath{\mathrm{d}} \alpha \\
 &= \int_0^1 \left( \sum_{1\leq n\leq N} e^{2\pi\ensuremath{\mathrm{i}} n\alpha} \right) \left( \sum_{1\leq m\leq N} e^{-2\pi\ensuremath{\mathrm{i}} m\alpha} m^z\right) \ensuremath{\mathrm{d}} \alpha,
 \end{align*}
which is the right hand side of \eqref{fok3}. Using $z=s-1$ and employing \eqref{fok2} we find
\begin{equation} \chi(s) \sum_{1\leq n\leq N} n^{s-1} = \int_0^1 B_N(\alpha) \left[ \zeta_1(s,\alpha) - \sum_{1\leq n\leq N} (n+\alpha)^{-s} \right]\ensuremath{\mathrm{d}} \alpha  + \int_0^1 \left[ \mathcal{O} \left( t^{-\sigma/2}\right) B_N(\alpha) \right]\ensuremath{\mathrm{d}} \alpha. \label{fok4} \end{equation}
We note that
\[ \left|\int_0^1 \left[ \mathcal{O} \left( t^{-\sigma/2}\right) B_N(\alpha) \right] \ensuremath{\mathrm{d}}\alpha \right| \lesssim t^{-\sigma/2} \int_0^1 |B_N(\alpha)|\, \ensuremath{\mathrm{d}} \alpha = \mathcal{O} (t^{-\sigma/2}\log t). \]
Equation \eqref{fok3} implies
\begin{equation} \sum_{1\leq n\leq N} n^z = \int_0^1 B_N(-\alpha) \left( \sum_{1\leq m\leq N} e^{2\pi\ensuremath{\mathrm{i}} m\alpha} m^z \right) \ensuremath{\mathrm{d}} \alpha, \quad u\in\mathbf{C}. \label{fok5} \end{equation}
Replacing $s$ by $1-s$ in \eqref{fok2} and taking the complex conjugate of the resulting equation we find
\begin{equation} \zeta_1(1-s,\alpha) = \sum_{1\leq n\leq N} (n+\alpha)^{s-1} + \chi(1-s) \sum_{1\leq n\leq N} e^{2\pi\ensuremath{\mathrm{i}} n\alpha} n^{-s} + \mathcal{O} \left( t^{-(1-\sigma)/2} \right). \label{fok6} \end{equation}
Using \eqref{fok5} with $z=-s$ and employing \eqref{fok6} we find
\begin{equation} \sum_{1\leq n\leq N} n^{-s} = \int_0^1 B_N(-\alpha) \left[ \chi(s) \zeta_1(1-s,\alpha) - \chi(s) \sum_{1\leq n\leq N} (n+\alpha)^{s-1} \right]\ensuremath{\mathrm{d}}\alpha + \mathcal{O} \left( t^{-(1-\sigma)/2} \log t\right). \label{fok7} \end{equation}
Here we have used $\chi(s) = \mathcal{O}\left( t^{1/2-\sigma} \right)$ and a similar estimate as before
\[ \left| \chi(s) \int_0^1 \mathcal{O}\left(t^{-(1-\sigma)/2}\right) B_N(-\alpha)\, \ensuremath{\mathrm{d}}\alpha \right| \lesssim t^{-\sigma/2} \int_0^1 |B_N(-\alpha)|\, \ensuremath{\mathrm{d}} \alpha = \mathcal{O}\left(t^{-\sigma/2} \log t\right). \]
Using equations \eqref{fok4} and \eqref{fok5} in \eqref{fok1} we arrive at the result in the lemma.
\end{proof}

\begin{lemma}\label{foklemma3}
With $B_N(\alpha)$ defined as before we have
\[ \int_0^1 B_N(\alpha) \sum_{1\leq n \leq N} (n+\alpha)^{-s}\, \ensuremath{\mathrm{d}} \alpha = \frac{\ensuremath{\mathrm{i}}}{2\pi N^s} \sum_{1\leq n\leq N} \frac{1}{\frac{t}{2\pi N} - n} + \frac{1}{2\pi\ensuremath{\mathrm{i}}} \sum_{1\leq n \leq N} \frac{1}{ \frac{t}{2\pi} -n } + \mathcal{O}\left( t^{-(1+\sigma)/2} \right). \]
\end{lemma}
\begin{proof}
Using the periodicity of $B_N(\alpha)$ on $\mathbf{R}/\mathbf{Z}$ we find
\begin{align} \int_0^1 B_N(\alpha) \sum_{1\leq n \leq N} (n+\alpha)^{-s}\, \ensuremath{\mathrm{d}} \alpha &= \sum_{1\leq n\leq N} \int_n^{n+1} B_N(\alpha) \alpha^{-s}\, \ensuremath{\mathrm{d}} \alpha \nonumber \\
&= \int_1^N B_N(\alpha) \alpha^{-s}\, \ensuremath{\mathrm{d}} \alpha + \int_{N}^{N+1} B_N(\alpha) \alpha^{-s}\, \ensuremath{\mathrm{d}}\alpha. \label{fok11}
\end{align}
We next estimate the first integral on the right hand side of \eqref{fok11}, which we denote by $I_N(s)$:
\begin{equation} I_N(s) = \sum_{1\leq n\leq N} \int_1^N e^{2\pi\ensuremath{\mathrm{i}} n \alpha - \ensuremath{\mathrm{i}} t \log \alpha} \alpha^{-\sigma}\, \ensuremath{\mathrm{d}} \alpha, \qquad 0<\sigma < 1. \label{fok12} \end{equation}
The integral in the above sum \emph{does not} have any stationary points. Indeed, candidates for stationary points are the points $\alpha^*$ where
\[ \alpha^* = \frac{t}{2\pi n} > \frac{N^2}{n} \]
thus, since $1\leq n \leq N$, $\alpha^*$ is outside the range of integration. Hence the above integral can be estimated using integration by parts:
\begin{align} \int_1^N e^{2\pi\ensuremath{\mathrm{i}} n\alpha - \ensuremath{\mathrm{i}} t \log \alpha} \alpha^{-\sigma}\, \ensuremath{\mathrm{d}} \alpha &= \ensuremath{\mathrm{i}} \int_1^N \frac{\ensuremath{\mathrm{d}}}{\ensuremath{\mathrm{d}} \alpha} \left( e^{2\pi \ensuremath{\mathrm{i}} n\alpha - \ensuremath{\mathrm{i}} t \log \alpha} \right) \frac{\alpha^{1-\sigma}}{t-2\pi n \alpha}\, \ensuremath{\mathrm{d}} \alpha \nonumber \\
&= -\frac{1}{2\pi\ensuremath{\mathrm{i}}} \left( \frac{N^{-s}}{\frac{t}{2\pi N} - n} - \frac{1}{\frac{t}{2\pi} -n} \right) + E_n(s), \label{fok13}
\end{align}
where
\[ E_n(s) = \int_1^N \frac{\ensuremath{\mathrm{d}}}{\ensuremath{\mathrm{d}} \alpha} \left( e^{2\pi \ensuremath{\mathrm{i}} n\alpha - \ensuremath{\mathrm{i}} t \log \alpha} \right) \left[ \frac{(1-\sigma) \alpha^{1-\sigma}}{(t-2\pi n\alpha)^2} + \frac{2\pi n \alpha^{2-\sigma}}{(t-2\pi n \alpha)^3} \right] \ensuremath{\mathrm{d}} \alpha. \]
There error term $E_n(s)$ can be evaluated using the second mean value theorem for integrals. For instance, for some $\xi \in (1,N)$ we have
\begin{align*} \left| \Re E_n(s) \right| &= \left| \left[\frac{(1-\sigma) N^{1-\sigma}}{(t-2\pi nN)^2} + \frac{2\pi n N^{2-\sigma}}{(t-2\pi n N)^3}\right] \Re \int_\xi^N \frac{\ensuremath{\mathrm{d}}}{\ensuremath{\mathrm{d}} \alpha} \left( e^{2\pi \ensuremath{\mathrm{i}} n\alpha - \ensuremath{\mathrm{i}} t \log \alpha} \right)\ensuremath{\mathrm{d}} \alpha \right| \\
&= \mathcal{O}\left( \frac{ N^{1-\sigma}}{(t-2\pi nN)^2} \right) + \mathcal{O}\left( \frac{ n N^{2-\sigma}}{(t-2\pi n N)^3} \right),
\end{align*}
and similarly for $\Im E_n(s)$. It is now straightforward to show
\begin{equation} \sum_{1\leq n \leq N} E_n(s) = \mathcal{O} \left(t^{-(1+\sigma)/2}\right). \label{est1} \end{equation}
We also have the elementary estimate
\[ \left| \int_{N}^{N+1} B_N(\alpha) \alpha^{-s}\, \ensuremath{\mathrm{d}} \alpha \right| \lesssim t^{-\sigma/2} \int_0^1 |B_N(\alpha)|\, \ensuremath{\mathrm{d}} \alpha = \mathcal{O}\left( t^{-\sigma/2} \log t\right). \]
This combined with \eqref{est1} and \eqref{fok13} gives the desired result.
\end{proof}
The leading order terms in the above expansion are $\mathcal{O}\left( t^{-\sigma/2}\log t\right)$, thus they can be absorbed into the error term. Indeed, using Lemma \ref{foklemma3} it is now straightforward to see that
\[ \int_0^1 B_N(\alpha) \sum_{1\leq n\leq N} (n+\alpha)^{-s}\, \ensuremath{\mathrm{d}} \alpha = \mathcal{O} \left( t^{-\sigma/2} \log t\right). \]
Using similar arguments we also find
\[ \chi(s) \int_0^1 B_N(-\alpha) \sum_{1\leq n\leq N} (n+\alpha)^{s-1}\, \ensuremath{\mathrm{d}}\alpha  = \mathcal{O} \left( t^{-\sigma/2} \log t\right). \]
Combining this observation with the result of Lemma \ref{foklemma1} we conclude that
\[ \zeta(s) = \int_0^1 B_N(\alpha)\zeta_1(s,\alpha)\, \ensuremath{\mathrm{d}} \alpha + \chi(s) \int_0^1 B_N(-\alpha) \zeta_1(1-s,\alpha)\, \ensuremath{\mathrm{d}} \alpha + \mathcal{O} \left( t^{-\sigma/2} \log t\right), \quad t\rightarrow \infty \]
for $0<\sigma<1$. The proof to Theorem 1 now follows from Lemmas \ref{foklemma1}--\ref{foklemma3} with $s=1/2$ and the application of H\"older's inequality with exponents
\[ p=2k, \quad q= \frac{2k}{2k-1}. \]
In particular, using the estimates in Lemma \ref{dirichletest} we have
\begin{align*} \left| \int_0^1 B_N(\alpha) \zeta_1(s,\alpha)\, \ensuremath{\mathrm{d}} \alpha \right| &\leq \left( \int_0^1 |B_N(\alpha)|^{2k/(2k-1)}\, \ensuremath{\mathrm{d}} \alpha \right)^{1-1/2k} \left( \int_0^1 |\zeta_1(s,\alpha)|^{2k}\, \ensuremath{\mathrm{d}} \alpha \right)^{1/2k} \\
&\lesssim N^{1/2k} I_k(t)^{1/2k} \\
&\lesssim t^{1/4k} I_k(t)^{1/2k}.
\end{align*}
This gives rise to the result in Theorem 1.

\section{Relations among products of the Hurwitz zeta functions}

\subsection{Quadratic formula}
\begin{lemma} \label{lemma4}
Let $\zeta_1(u,\alpha)$,  $u \in \mathbf{C}$, $\alpha >0$, denote the modified Hurwitz function, i.e.,

\begin{equation} \zeta_1(u,\alpha)= \sum^\infty_{m=1} \frac{1}{(m+\alpha)^u}, \quad \alpha > 0 \quad \Re u>1.\label{(2.1)} \end{equation}
Then, for $\Re u>1, \Re v>1,$
\begin{equation} \int^1_0  \zeta_1(v,\alpha) \zeta_1(u,\alpha)\, \ensuremath{\mathrm{d}} \alpha = \frac{1}{u+v-1} + \int^\infty_1  \alpha^{-v}\zeta_1(u,\alpha)\, \ensuremath{\mathrm{d}} \alpha + \int^\infty_1  \alpha^{-u}\zeta_1(v,\alpha)\, \ensuremath{\mathrm{d}} \alpha.  \\
     \label{(2.2)}\end{equation} \end{lemma}

\begin{proof}  Let $q_0(v,u)$ denote the LHS of equation \eqref{(2.2)}.

Using the integral representation of the modified Hurwitz function, namely
\begin{equation}\zeta_1(s,\alpha) = \frac{1}{\Gamma(s)} \int^{\infty}_{0}  \frac{e^{-\alpha\rho}\rho^{s-1}}{e^\rho -1}\, \ensuremath{\mathrm{d}} \rho, \quad \alpha >0, \quad \Re s>1, \label{(2.3)}\end{equation}
we find

$$q_0(v,u) = \frac{1}{\Gamma(u)\Gamma(v)} \int^{\infty}_{0} \ensuremath{\mathrm{d}}\rho_1 \int^{\infty}_{0} \ensuremath{\mathrm{d}} \rho_2\, \frac{\rho^{v-1}_1 \rho^{u-1}_2}{\rho_1+\rho_2} \frac{1-e^{-(\rho_1+\rho_2)}}{(e^{\rho_1}-1)(e^{\rho_2}-1)}.$$
Inserting in this equation the identity

\begin{equation}\frac{1-e^{-(\rho_1+\rho_2)}}{(e^{\rho_1}-1)(e^{\rho_2}-1)} = e^{-(\rho_1+\rho_2)}\left[1 + \frac{1}{e^{\rho_2}-1} + \frac{1}{e^{\rho_1}-1}\right], $$
we find
$$q_0(v,u) =J_0(v,u) + I_0(v,u) + I_0(u,v), \label{(2.5)} \end{equation}
where
\begin{equation}J_0(v,u) = \frac{1}{\Gamma(u)\Gamma(v)} \int^{\infty}_{0} \ensuremath{\mathrm{d}}\rho_1 \int^{\infty}_{0} \ensuremath{\mathrm{d}} \rho_2\, \frac{\rho^{v-1}_1\rho^{u-1}_2e^{-(\rho_1+\rho_2)}}{\rho_1+\rho_2}, \label{(2.6)} \end{equation}
and
\begin{equation} I_0(v,u) = \frac{1}{\Gamma(u)\Gamma(v)} \int^{\infty}_{0} \ensuremath{\mathrm{d}}\rho_1 \int^{\infty}_{0} \ensuremath{\mathrm{d}} \rho_2\, \frac{\rho^{v-1}_1\rho^{u-1}_2e^{-(\rho_1+\rho_2)}}{(\rho_1+\rho_2)(e^{\rho_2}-1)}. \label{(2.7)}\end{equation}
We will next show that
\begin{equation} J_0(v,u) = \frac{1}{u+v-1}. \label{(2.8)}\end{equation}
Indeed, using the integral representations of $\Gamma(u)$ and $\Gamma(v)$, we find
\begin{equation}
\Gamma(u)\Gamma(v)=\int^{\infty}_{0} \ensuremath{\mathrm{d}}\rho_1 \int^{\infty}_{0} \ensuremath{\mathrm{d}} \rho_2\,  \rho_1^{u-1}\rho_2^{v-1} e^{-(\rho_1+\rho_2)}.   \label{(2.9)}
\end{equation}
Replacing in the RHS of \eqref{(2.9)} $\rho_1$ and $\rho_2$ by $ \alpha x_1$ and $\alpha x_2$, multiplying the resulting expression by $\alpha^{-(u+v)}$,
and  integrating with respect to $\alpha$ from
$\alpha=1$ to $\alpha=\infty$,
 we find
$$
\frac{\Gamma(u)\Gamma(v)}{u+v-1} =\int^{\infty}_{0} \ensuremath{\mathrm{d}} x_1 \int^{\infty}_{0} \ensuremath{\mathrm{d}} x_2\,\frac{ x_1^{u-1}x_2^{v-1} e^{-(x_1+x_2)}}{x_1+x_2},
$$
which gives \eqref{(2.8)}.

Finally, we will show that
\begin{equation}I_0(v,u) = \int^\infty_1 \alpha^{-v} \zeta_1(u,\alpha)\, \ensuremath{\mathrm{d}} \alpha. \label{(2.10)} \end{equation}
Indeed, using the integral representations of $\Gamma(v)$ and of $\zeta_1(u,\alpha)$ we find
\begin{equation} \Gamma(v)\zeta_1(u,\alpha) = \int^{\infty}_{0}   e^{-\rho} \rho^{v-1}\, \ensuremath{\mathrm{d}} \rho \times \frac{1}{\Gamma(u)} \int^{\infty}_{0}  \  \rho^{u-1}_2 \frac{e^{-\alpha\rho_2}}{e^{\rho_2}-1}\,\ensuremath{\mathrm{d}} \rho_2. \label{(2.11)} \end{equation}
Replacing in the RHS of the above equation $\rho$ by $\alpha\rho_1$, multiplying the resulting equation by $\alpha^{-v}/\Gamma(v)$, and then integrating with respect to $\alpha$ from $\alpha=1$ to $\alpha=\infty$, we find \eqref{(2.10)}.

Inserting in equation \eqref{(2.5)} the expressions for $J_0(v,u)$, for $I_0(v,u)$ and for $ I_0(u,v)$ from equations \eqref{(2.8)} and \eqref{(2.10)}, we find \eqref{(2.2)}. \end{proof}

Eq. \eqref{f1.12} can be derived following the approach used in Lemmas \ref{lemma4} and \ref{lemma5}, and thus it is omitted.

\subsection{Quadruple formula}

\begin{lemma}\label{lemma5}
Let $\zeta_1(s,\alpha)$ be defined as in \eqref{(2.3)}. Then for $\Re u_i>1$, $i=1,2,3,4$, the following identity is valid:
\begin{multline} \int_0^1 \prod_{i=1}^4 \zeta_1(u_i,\alpha)\, \ensuremath{\mathrm{d}} \alpha = \frac{1}{u_1+u_2+u_3+u_4-1} + \sum_{\mathrm{perms}} \int_1^\infty \alpha^{-(u_1+u_2+u_3)}\zeta_1(u_4,\alpha)\, \ensuremath{\mathrm{d}} \alpha \\
+\sum_{\mathrm{perms}} \int_1^\infty \alpha^{-(u_1+u_2)} \zeta_1(u_3,\alpha)\zeta_1(u_4,\alpha)\, \ensuremath{\mathrm{d}} \alpha + \sum_{\mathrm{perms}} \int_1^\infty \alpha^{-u_1} \zeta_1(u_2,\alpha) \zeta_1(u_3,\alpha) \zeta_1(u_4,\alpha)\, \ensuremath{\mathrm{d}} \alpha ,\label{quad}
 \end{multline}
 where the sums run over permutations of $(1,2,3,4)$ so that the first and third sums contain $4$ terms whilst the second sum contains $6$ terms.
\end{lemma}
\begin{proof}
Employing the representation \eqref{(2.3)} for each Hurwitz function and integrating over $(0,1)$ we find that the left hand side of \eqref{quad}, which we denote by $Q(u_1,u_2,u_3,u_4)$, is given by
\begin{equation}
Q = \frac{1}{\prod_{i=1}^4 \Gamma(u_i)} \int_{(0,\infty)^4} \frac{\prod_{i=1}^4 \rho_i^{u_i-1}\, \ensuremath{\mathrm{d}} \rho_i}{\rho_1+\rho_2+\rho_3+\rho_4} \frac{1- e^{-(\rho_1+\rho_2+\rho_3+\rho_4)}}{\prod_{i=1}^4 R_i} \label{Q}
\end{equation}
where the functions $\{R_i\}$ are defined by
\begin{equation} R_i = e^{\rho_i}-1 , \quad i=1,2,3,4. \end{equation}
The following identity is valid:
\begin{multline} \frac{1 - e^{-(\rho_1+\rho_2+\rho_3+\rho_4)}}{\prod_{i=1}^4 R_i} = e^{-(\rho_1+\rho_2+\rho_3+\rho_4)} \left[ 1 + \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} \right. \\
\left. +\frac{1}{R_1R_2} + \frac{1}{R_2R_3} + \frac{1}{R_3R_4} + \frac{1}{R_1 R_3} + \frac{1}{R_2 R_4} +\frac{1}{R_1R_4} + \frac{1}{R_1 R_2 R_3} + \frac{1}{R_1 R_2 R_4} + \frac{1}{R_1 R_3 R_4} + \frac{1}{R_2 R_3R_4}\right].
\end{multline}
Using this in \eqref{Q} we find
\begin{multline}
Q = \frac{1}{\prod_{i=1}^4 \Gamma(u_i)} \int_{(0,\infty)^4} \frac{\prod_{i=1}^4 \rho_i^{u_i-1}\, \ensuremath{\mathrm{d}} \rho_i}{\rho_1+\rho_2+\rho_3+\rho_4} e^{-(\rho_1+\rho_2+\rho_3+\rho_4)} \left[ 1 + \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} \right. \\
\left. +\frac{1}{R_1R_2} + \frac{1}{R_2R_3} + \frac{1}{R_3R_4} + \frac{1}{R_1 R_3} + \frac{1}{R_2 R_4} +\frac{1}{R_1R_4}+ \frac{1}{R_1 R_2 R_3} + \frac{1}{R_1 R_2 R_4} + \frac{1}{R_1 R_3 R_4} + \frac{1}{R_2 R_3R_4}\right]. \label{mess}
\end{multline}

In order to simplify the right hand side of \eqref{mess} we first note the definition of the Gamma function, namely the equation
\begin{equation} \Gamma(u) = \int_0^\infty r^{u-1} e^{-r}\, \ensuremath{\mathrm{d}} r, \quad \Re u >0, \label{Gamma} \end{equation}
implies that
\[ \prod_{i=1}^4 \Gamma(u_i) = \int_{(0,\infty)^4} \prod_{i=1}^4 \left( x_i^{u_i-1}\, \ensuremath{\mathrm{d}} x_i\right) e^{-x_1-x_2-x_3-x_4}. \]
Using in the right hand side of this equation the transformations
\begin{equation} x_i = \alpha \rho_i, \quad i=1,2,3,4, \label{trans}\end{equation}
dividing by the product of the four Gamma functions, and multiplying the resulting expression by $\alpha^{-u_1-u_2-u_3-u_4}$, we find the identity
\[ \alpha^{-u_1-u_2-u_3-u_4} = \frac{1}{\prod_{i=1}^4 \Gamma(u_i)} \int_{(0,\infty)^4} \prod_{i=1}^4 \left( \rho_i^{u_i-1} \, \ensuremath{\mathrm{d}} \rho_i \right) e^{-\alpha(\rho_1+\rho_2+\rho_3+\rho_4)}. \]
Integrating this equation over $(1,\infty)$ with respect to $\alpha$ we obtain
\begin{equation} \frac{1}{u_1+u_2+u_3+u_4-1} = \frac{1}{\prod_{i=1}^4 \Gamma(u_i)} \int_{(0,\infty)^4} \frac{ \prod_{i=1}^4 \left( \rho_i^{u_i-1}\, \ensuremath{\mathrm{d}} \rho_i\right)}{\rho_1+\rho_2+\rho_3+\rho_4} e^{-\rho_1-\rho_2-\rho_3-\rho_4}.  \label{frac}\end{equation}

Employing for $\zeta_1(u,\alpha)$ and $\Gamma(u)$ equations \eqref{(2.3)} and \eqref{Gamma} respectively we find
\[ \zeta_1(u_1,\alpha) \Gamma(u_2) \Gamma(u_3) \Gamma(u_4) = \frac{1}{\Gamma(u_1)} \int_{(0,\infty)^4} \ensuremath{\mathrm{d}}\rho_1\, \ensuremath{\mathrm{d}} x_2\, \ensuremath{\mathrm{d}} x_3\, \ensuremath{\mathrm{d}} x_4\, \frac{\rho_1^{u_1-1}}{R_1} x_2^{u_2-1} x_3^{u_3-1} x_4^{u_4-1} e^{-\alpha \rho_1 - (x_2+x_3+x_4)}. \]
Using in the right hand side of this equation the transformations \eqref{trans} but restricted only to $i=2,3,4$, dividing by $\Gamma(u_2) \Gamma(u_3) \Gamma(u_4)$, multiplying by $\alpha^{-(u_2+u_3+u_4)}$ and integrating with respect to $\alpha$ over $(1,\infty)$ we obtain
\begin{equation} \int_1^\infty \alpha^{-(u_2+u_3+u_4)} \zeta_1(u_1,\alpha)\, \ensuremath{\mathrm{d}} \alpha = \frac{1}{\prod_{i=1}^4 \Gamma(u_i)} \int_{(0,\infty)^4} \frac{ \prod_{i=1}^4 \left( \rho_i^{u_i-1}\, \ensuremath{\mathrm{d}} \rho_i\right) e^{-(\rho_1+\rho_2+\rho_3+\rho_4)}}{R_1 (\rho_1+\rho_2+\rho_3+\rho_4)}. \end{equation}

Similarly equations \eqref{(2.3)} and \eqref{Gamma} imply
\begin{align*}
&\zeta_1(u_1,\alpha) \zeta_1(u_2,\alpha) \Gamma(u_3) \Gamma(u_4)\\
=& \frac{1}{\Gamma(u_1)\Gamma(u_2)} \int_{(0,\infty)^4} \ensuremath{\mathrm{d}} \rho_1\, \ensuremath{\mathrm{d}}\rho_2\, \ensuremath{\mathrm{d}} x_3\, \ensuremath{\mathrm{d}} x_4\, \rho_1^{u_1-1} \rho_2^{u_2-1} \frac{ x_3^{u_3-1} x_4^{u_4-1} e^{-\alpha(\rho_1+\rho_2)-(x_3+x_4)}}{R_1R_2}.
\end{align*}
Using in the right hand side of this equation the transformations \eqref{trans} but only for $i=3,4$, dividing by $\Gamma(u_3)\Gamma(u_4)$, multiplying by $\alpha^{-u_3-u_4}$ and integrating the resulting expression with respect to $\alpha$ over $(1,\infty)$, we obtain
\begin{equation} \int_1^\infty \alpha^{-u_3-u_4} \zeta_1(u_1,\alpha) \zeta_1(u_2,\alpha) \ensuremath{\mathrm{d}} \alpha = \frac{1}{\prod_{i=1}^4 }\int_{(0,\infty)^4} \frac{ \prod_{i=1}^4 \left( \rho_i^{u_i-1} \, \ensuremath{\mathrm{d}} \rho_i\right) e^{-\rho_1-\rho_2-\rho_3-\rho_4}}{R_1 R_2 (\rho_1+\rho_2+\rho_3+\rho_4)}. \end{equation}

A similar procedure yields the identity
\begin{equation}
\int_1^\infty \alpha^{-u_4} \zeta_1(u_1,\alpha) \zeta_1(u_2,\alpha) \zeta_1(u_3,\alpha)\,\ensuremath{\mathrm{d}} \alpha = \frac{1}{\prod_{i=1}^4 \Gamma(u_i)} \int_{(0,\infty)^4} \frac{ \left( \prod_{i=1}^4 \rho_i^{u_i-1}\, \ensuremath{\mathrm{d}}\rho_i\right) e^{-\rho_1-\rho_2-\rho_3-\rho_4}}{R_1R_2R_3(\rho_1+\rho_2+\rho_3+\rho_4)}. \label{endeq}\end{equation}
Employing in equation \eqref{mess} equations \eqref{frac}--\eqref{endeq} and appropriate permutations of $(1,2,3,4)$ we find equation \eqref{quad}.
\end{proof}

\subsection{A new derivation of the results of \cite{katsurada1996explicit}}
Here we rederive some of the results from \cite{katsurada1996explicit}.

The identity in \eqref{I1} is a consequence of equation \eqref{(2.2)} and of the following exact formula.

\begin{lemma}  Let $\zeta_1(u,\alpha), u \in \mathbf{C}$, $\alpha >0$, denote the modified Hurwitz function, and let $\zeta(u), u\in \mathbf{C}$, denote Riemann's zeta function. Then

\begin{equation} \int^{\infty}_{0} \ \alpha^{-v}\zeta_1(u,\alpha)\, \ensuremath{\mathrm{d}} \alpha = \frac{\Gamma(1-v)}{\Gamma(u)} \Gamma(u+v-1) \zeta(u+v-1) \label{(2.12)} \end{equation}
and

\begin{equation} \int^1_0  \  \alpha^{-v}\zeta_1(u,\alpha)\,\ensuremath{\mathrm{d}}\alpha =\frac{ \zeta(u)-1}{1-v} +\frac{ u}{1-v} \int^1_0  \alpha^{1-v} \zeta_1(u+1,\alpha)\, \ensuremath{\mathrm{d}}\alpha, \quad \Re v<2, \quad u\in \mathbf{C}. \label{(2.13)} \end{equation} \end{lemma}

\begin{proof} In order to derive \eqref{(2.12)} we first assume that $\Re u>1$, so that we can use the sum representation of $J_1(u,\alpha)$.  Furthermore, we assume that $\Re v<1$, so that the relevant integral converges at $\alpha=0$.  Then,
$$\int^{\infty}_{0}  \alpha^{-v} \zeta_1(u,\alpha)\, \ensuremath{\mathrm{d}}\alpha = \sum^\infty_{m=1} \int^{\infty}_{0}  \alpha^{-v} (m+\alpha)^{-u}\, \ensuremath{\mathrm{d}}\alpha = \sum^\infty_{m=1} m^{1-(u+v)} \int^{\infty}_{0}  \beta^{-v} (1+\beta)^{-u}\, \ensuremath{\mathrm{d}} \beta,$$
where we have used the change of variables $\alpha =\beta m$ in the second equation.  Then, the definition of Riemann's zeta function, together with the identity
\begin{equation} \int^{\infty}_{0}  \frac{\beta^{-v}}{(1+\beta)^u}\, \ensuremath{\mathrm{d}}\beta = \frac{\Gamma(1-v)\Gamma(u+v-1)}{\Gamma(u)}, \label{(2.14)} \end{equation}
imply equation \eqref{(2.12)}.

In order to derive equation \eqref{(2.13)} we use integration by parts:
$$\int^1_0  \alpha^{-v}\zeta_1(u,\alpha)\, \ensuremath{\mathrm{d}}\alpha = \left. \frac{\alpha^{1-v}}{1-v} \zeta_1(u,\alpha)\right|^1_0 - \frac{1}{1-v} \int^1_0  \alpha^{1-v} \frac{\partial}{\partial\alpha} \zeta_1(u,\alpha)\, \ensuremath{\mathrm{d}} \alpha.$$
Thus, \eqref{(2.13)} follows.
\end{proof}

\begin{proof}[Proof of identity \eqref{I1}] Splitting the first integral in the RHS of \eqref{(2.2)} and then using equations \eqref{(2.12)} and \eqref{(2.13)}, as well as using the analogous equations where $u$ and $v$ are interchanged, we find equation \eqref{I1}. \end{proof}

\begin{remark} In order to derive equation \eqref{(1.5)} we let $\sigma = \frac{1}{2} + \frac{\varepsilon}{2}$ in the LHS of \eqref{(1.5)}, and employ the identities
\begin{equation} \Gamma(\varepsilon) = \frac{1}{\varepsilon} - \gamma + O(\varepsilon), \quad \varepsilon \rightarrow 0, \label{(2.15)} \end{equation}
\begin{equation}\frac{\Gamma'\left( \frac{1}{2} + it\right)}{ \Gamma\left( \frac{1}{2} + it\right)} = \ln t + \frac{i\pi}{2} + O \left( \frac{1}{t^2}\right), \quad t \rightarrow \infty, \label{(2.16)} \end{equation}
as well as the identity
\begin{equation} \zeta(\varepsilon) = -\frac{1}{2} \left[ 1 + \varepsilon\ln(2\pi) + O(\varepsilon^2)\right], \quad \varepsilon \rightarrow 0. \label{(2.17)} \end{equation}
\end{remark}

\begin{remark}  By proving a simple estimate for the integrals in the RHS of \eqref{I1}, it is shown in \cite{katsurada1996explicit} that these integrals do {\it not} contribute to the leading asymptotics of the LHS of equation \eqref{I1}.  Actually, it is straightforward to show that if
\[ v=\sigma_1-it, \quad u = \sigma_2+it, \quad \sigma_1<2, \quad \sigma_1 >0,  \]
then
\begin{equation} \int^1_0 \alpha^{1-v} \zeta_1(u+1,\alpha)\, \ensuremath{\mathrm{d}}\alpha = \frac{1}{it} \sum^\infty_{m=1} \frac{1}{m(m+1)^u} + O\left( \frac{1}{t^2}\right), \quad t\rightarrow\infty. \label{(2.19)} \end{equation}
Indeed,
\begin{equation}\int^1_0 \alpha^{1-v} \zeta_1(u+1,\alpha) = \sum^\infty_{m=1} \int^1_0 \alpha^{1-\sigma_1} (m+\alpha)^{-1-\sigma_2}e^{it[\ln \alpha - \ln (\alpha+m)]}\, \ensuremath{\mathrm{d}}\alpha. \label{(2.20)} \end{equation}
Noting that
$$\frac{d}{d\alpha} [\ln \alpha - \ln(\alpha+m)] \neq 0, \quad m \geq 1, \quad 0<\alpha <1,$$
it follows that integrals in the RHS of \eqref{(2.20)} do not possess any stationary points.  Then, straightforward integration by parts yields \eqref{(2.19)}. \end{remark}

\section{A relation between quadratic products
of Hurwitz zeta functions and their Fourier series}
Theorem 2 will be proved by examining the Fourier series for the function
\[ \zeta_1(u,\alpha)\zeta_1(v,\alpha) \]
for $\Re u, \Re v >0$. Following Rane \cite{rane1997approx} we first construct the Fourier series for $\zeta_1(s,\alpha)$.
\begin{lemma}
Let $\sigma\in (0,1)$. Then the Fourier series
\[ \frac{1}{s-1} + \sum_{n\neq 0} a_n(s) e^{2\pi\ensuremath{\mathrm{i}} n\alpha}, \qquad a_n(s) = \int_1^\infty \alpha^{-s} e^{-2\pi\ensuremath{\mathrm{i}} n \alpha}\, \ensuremath{\mathrm{d}}\alpha \]
converges pointwise to $\zeta_1(s,\alpha)$ for each $\alpha\in(0,1)$.
\end{lemma}
\begin{proof}
Since $\zeta_1(s,\alpha)$ is a smooth function of $\alpha$ (for fixed $s$) its Fourier series converges pointwise for $\alpha\in (0,1)$. The Fourier coefficients are defined by
\[ a_n(s) = \int_0^1 e^{-2\pi\ensuremath{\mathrm{i}} n \alpha}\zeta_1(s,\alpha)\, \ensuremath{\mathrm{d}} \alpha. \]
Note that the Fourier series for $\zeta(\alpha,s) =\alpha^{-s} + \zeta_1(s,\alpha)$ is well known, and has Fourier coefficients $\Gamma(1-s) (2\pi\ensuremath{\mathrm{i}} n)^{s-1}$. Hence
\[ a_n(s) = \Gamma(1-s) (2\pi\ensuremath{\mathrm{i}} n)^{s-1} - \int_0^1 \alpha^{-s} e^{-2\pi\ensuremath{\mathrm{i}} n\alpha}\, \ensuremath{\mathrm{d}} \alpha. \]
Using Euler's integral representation of the Gamma function we arrive at the desired result.
\end{proof}
\begin{remark}
Since $a_n(s)$ is expressible in terms of the incomplete Gamma function, we conclude that it has an analytic extension to all complex $s\neq 1$.
\end{remark}
Note that for $\sigma>1$ we have
\[ \frac{1}{s-1} = \int_1^\infty \alpha^{-s}\, \ensuremath{\mathrm{d}} \alpha, \]
so we may write
\[ a_0(s) = \int_1^\infty \alpha^{-s}\, \ensuremath{\mathrm{d}} \alpha, \]
for $\sigma>1$, and by analytic continuation elsewhere. Now we write
\[ \zeta_1(s,\alpha) = \sum_n a_n(s) e^{2\pi\ensuremath{\mathrm{i}} n \alpha}, \qquad \sigma>0 \]
where the $a_n(s)$ are defined accordingly.
\begin{lemma}
Let $\Re u, \Re v>1$ and define the functions
\[ q_n(u,v) =  a_n(u+v)  + \int_1^\infty \zeta_1(u,\alpha) \alpha^{-v}e^{-2\pi\ensuremath{\mathrm{i}} n \alpha}\, \ensuremath{\mathrm{d}} \alpha+ \int_1^\infty \zeta_1(v,\alpha) \alpha^{-u}e^{-2\pi\ensuremath{\mathrm{i}} n \alpha}\, \ensuremath{\mathrm{d}} \alpha  \quad n \in \mathbf{Z}. \]
Then the Fourier series $\sum_n q_n(u,v) e^{2\pi \ensuremath{\mathrm{i}} n \alpha}$ converges pointwise to $\zeta_1(u,\alpha)\zeta_1(v,\alpha)$ for $\alpha \in (0,1)$.
\end{lemma}
\begin{proof}
Since the Fourier coefficients for $\zeta_1(u,\alpha)$ are $a_n(u)$ the Fourier coefficients of the product $\zeta_1(u,\alpha) \zeta_1(v,\alpha)$ are given by the convolution
\[ q_n(u,v) = \sum_m a_m(u) a_{n-m}(v). \]
For $\Re u, \Re v >1$ we have
\[ \sum_m a_m(u) a_{n-m}(v) = \sum_m \int_1^\infty \ensuremath{\mathrm{d}} \alpha \int_1^\infty \ensuremath{\mathrm{d}} \beta\, \alpha^{-u} \beta^{-v} e^{-2\pi\ensuremath{\mathrm{i}} m(\alpha-\beta)} e^{-2\pi\ensuremath{\mathrm{i}} n \beta},
 \]
the double integral being absolutely convergent. Now recall the distributional result
\[ \sum_m e^{-2\pi\ensuremath{\mathrm{i}} m(\alpha-\beta)} = \sum_m \delta(\beta-\alpha - m). \]
Using this in the above we find
\begin{multline} q_n(u,v) =  \sum_{m\geq 1} \int_1^\infty \alpha^{-u} (m+\alpha)^{-v}e^{-2\pi\ensuremath{\mathrm{i}} n \alpha}\, \ensuremath{\mathrm{d}} \alpha  \\
+ \int_1^\infty \alpha^{-u-v} e^{-2\pi\ensuremath{\mathrm{i}} n \alpha}\, \ensuremath{\mathrm{d}} \alpha + \sum_{m\geq 1} \int_1^\infty \alpha^{-v} (m+\alpha)^{-u}e^{-2\pi\ensuremath{\mathrm{i}} n \alpha}\, \ensuremath{\mathrm{d}} \alpha. \label{convhalf} \end{multline}
Since $\Re u, \Re v>1$ the integrands of the first and third terms can be dominated by the integrable functions $\alpha^{-\Re u}$ and $\alpha^{-\Re v}$ respectively, allowing us to pass the sum inside the integral,
\[q_n(u,v) =  a_n(u+v)  + \int_1^\infty \zeta_1(u,\alpha) \alpha^{-v}e^{-2\pi\ensuremath{\mathrm{i}} n \alpha}\, \ensuremath{\mathrm{d}} \alpha+ \int_1^\infty \zeta_1(v,\alpha) \alpha^{-u}e^{-2\pi\ensuremath{\mathrm{i}} n \alpha}\, \ensuremath{\mathrm{d}} \alpha. \]
We note that both the integrals are absolutely convergent for $\Re u, \Re v>1$.
\end{proof}
To establish the main result in this section we must first perform an analytic continuation of the functions $q_n(u,v)$ valid for $\Re u, \Re v>0$. To this end, we recall the following result \cite{rane1997approx}:
\begin{equation} \zeta_1(s,\alpha) = \frac{\alpha^{1-s}}{s-1} - \frac{\alpha^{-s}}{2} + \lim_{N\rightarrow \infty}\sum_{0<|m|<N} \left( \int_\alpha^\infty x^{-s} e^{2\pi\ensuremath{\mathrm{i}} m x}\, \ensuremath{\mathrm{d}} x\right) e^{-2\pi\ensuremath{\mathrm{i}} m \alpha}, \label{Raneeq} \end{equation}
where $s=\sigma+\ensuremath{\mathrm{i}} t$ and $\sigma>0$. This result can be derived using the Euler-Maclaurin formula. We will need the following lemma to control the final term.
\begin{lemma}\label{intbyparts}
Let $s=\sigma+\ensuremath{\mathrm{i}} t$ with $\sigma>0$. Then if $\alpha\geq \eta > t/2\pi$ we have
\[ \left|\lim_{N\rightarrow \infty}\sum_{0<|m|<N}\left( \int_\alpha^\infty x^{-s} e^{2\pi\ensuremath{\mathrm{i}} m x}\, \ensuremath{\mathrm{d}} x\right) e^{-2\pi\ensuremath{\mathrm{i}} m \alpha}\right| \lesssim_\eta t \alpha^{-\sigma-1}. \]
\end{lemma}
\begin{proof}
For $\alpha$ in the stated range we can integrate by parts using
\begin{align*} \int_\alpha^\infty x^{-s} e^{2\pi\ensuremath{\mathrm{i}} m x}\,\ensuremath{\mathrm{d}} x &=  \int_{\alpha}^\infty \frac{x^{-\sigma}}{ \ensuremath{\mathrm{i}}( 2\pi m -  t/x)} \frac{\ensuremath{\mathrm{d}}}{\ensuremath{\mathrm{d}} x} \left( x^{-\ensuremath{\mathrm{i}} t} e^{2\pi\ensuremath{\mathrm{i}} m x} \right)\, \ensuremath{\mathrm{d}} x \\
&= - \frac{\alpha^{-s} e^{2\pi\ensuremath{\mathrm{i}} m \alpha}}{\ensuremath{\mathrm{i}}(2\pi m - t/x)} +\ensuremath{\mathrm{i}} \int_\alpha^\infty \frac{\ensuremath{\mathrm{d}}}{\ensuremath{\mathrm{d}} x} \left( \frac{x^{-\sigma}}{2\pi m-t/x}\right) x^{-\ensuremath{\mathrm{i}} t} e^{2\pi\ensuremath{\mathrm{i}} m x}\, \ensuremath{\mathrm{d}} x.
\end{align*}
We can estimate the sum arising from the first term
\[ \left| \sum_{0<|m|<N} \frac{\alpha^{-s}}{\ensuremath{\mathrm{i}}(2\pi m - t/\alpha)}\right| = \alpha^{-\sigma} \sum_{0<m<N} \frac{2t/\alpha}{(4\pi^2 m^2 - t^2/\alpha^2)} \lesssim_\eta t \alpha^{-\sigma-1}. \]
Computing the derivative and applying integration by parts again, the second term becomes
\[  \int_{\alpha}^\infty \frac{ x^{-\sigma-1}( 2\pi\sigma m + (1-\sigma)t/x))}{\ensuremath{\mathrm{i}}(2\pi m - t/x)^3} \frac{\ensuremath{\mathrm{d}}}{\ensuremath{\mathrm{d}} x}\left( x^{-\ensuremath{\mathrm{i}} t} e^{2\pi\ensuremath{\mathrm{i}} m x}\right) \ensuremath{\mathrm{d}} x. \]
An application of the second mean value theorem for integrals on the real and imaginary parts of this term show it to be $\mathcal{O}_\eta(\alpha^{-\sigma-1} m^{-2}) + \mathcal{O}_\eta(t \alpha^{-\sigma-2} m^{-3} )$. In particular
\[ \left| \sum_{0<|m|<N}\int_\alpha^\infty \frac{\ensuremath{\mathrm{d}}}{\ensuremath{\mathrm{d}} x} \left( \frac{x^{-\sigma}}{2\pi m-t/x}\right) x^{-\ensuremath{\mathrm{i}} t} e^{2\pi\ensuremath{\mathrm{i}} m x}\, \ensuremath{\mathrm{d}} x \right| \lesssim_\eta  \alpha^{-\sigma-1} \quad \textrm{for $\alpha\geq \eta > t/2\pi$}\]
so we have established our estimate.
\end{proof}
Now we return to the analytic continuation of $q_n(u,v)$ for $\Re u, \Re v>0$. The previous lemma establishes that
\begin{equation} \left|\alpha^{-v} \zeta_1(u,\alpha) - \frac{\alpha^{1-u-v}}{u-1} + \frac{\alpha^{-u-v}}{2}\right| \leq_\eta t \alpha^{-\Re u - \Re v -1} , \quad \textrm{for $\alpha\geq \eta> t/2\pi$}. \label{est} \end{equation}
In particular, the left hand side is an absolutely integrable function of $\alpha$ on $(1,\infty)$ provided that $\Re u, \Re v>0$. This suggests the splitting
\begin{multline*} \int_1^\infty \zeta_1(u,\alpha) \alpha^{-v}e^{-2\pi\ensuremath{\mathrm{i}} n \alpha}\, \ensuremath{\mathrm{d}} \alpha \equiv \int_1^\infty \left(\alpha^{-v} \zeta_1(u,\alpha) - \frac{\alpha^{1-u-v}}{u-1} + \frac{\alpha^{-u-v}}{2} \right) e^{-2\pi\ensuremath{\mathrm{i}} n \alpha}\, \ensuremath{\mathrm{d}} \alpha \\
 +  \frac{a_n (u+v-1)}{u-1} - \frac{a_n(u+v)}{2},\end{multline*}
which is valid for $\Re u, \Re v>1$. This gives rise to the representation
\begin{multline*} q_n(u,v) =  \left[ \frac{1}{u-1} + \frac{1}{v-1} \right]a_n(u+v-1)  + \int_1^\infty \left(\alpha^{-v} \zeta_1(u,\alpha) - \frac{\alpha^{1-u-v}}{u-1} + \frac{\alpha^{-u-v}}{2} \right)e^{-2\pi\ensuremath{\mathrm{i}} n \alpha}\, \ensuremath{\mathrm{d}} \alpha \\+ \int_1^\infty \left(\alpha^{-u} \zeta_1(v,\alpha) - \frac{\alpha^{1-u-v}}{v-1} + \frac{\alpha^{-u-v}}{2} \right)e^{-2\pi\ensuremath{\mathrm{i}} n \alpha}\, \ensuremath{\mathrm{d}} \alpha, \end{multline*}
which provides an analytic continuation of $q_n(u,v)$ for $\Re u, \Re v>0$.
\begin{remark}
Using $\partial_\alpha \zeta_1(s,\alpha) = -s \zeta_1(\alpha,s+1)$ in \eqref{Raneeq} we see that
\[ \frac{\partial \zeta_1}{\partial \alpha} (\alpha,u) = - \alpha^{-s} + \frac{s \alpha^{-s-1}}{2} - \lim_{N\rightarrow \infty}\sum_{0<|m|<N} \left( s\int_\alpha^\infty x^{-s-1} e^{2\pi\ensuremath{\mathrm{i}} m x}\, \ensuremath{\mathrm{d}} x\right) e^{-2\pi\ensuremath{\mathrm{i}} m \alpha}. \]
Using the previous lemma, this then implies that for $\alpha\geq \eta > t/2\pi$
\begin{equation} \left| \frac{\partial}{\partial \alpha} \left( \zeta_1(s,\alpha) - \frac{\alpha^{1-s}}{s-1} + \frac{\alpha^{-s}}{2} \right) \right| \lesssim_\eta t^2 \alpha^{-\sigma - 2}. \label{derivestimate} \end{equation}
\end{remark}
\begin{lemma}
If $u=\sigma+\ensuremath{\mathrm{i}} t$ and $v=\sigma-\ensuremath{\mathrm{i}} t$ then for each $\eta>0$
\[ \left|\int_{t/2\pi+\eta}^\infty \left(\alpha^{-v} \zeta_1(u,\alpha) - \frac{\alpha^{1-u-v}}{u-1} + \frac{\alpha^{-u-v}}{2} \right)e^{-2\pi\ensuremath{\mathrm{i}} n \alpha}\, \ensuremath{\mathrm{d}} \alpha\right| \lesssim_\eta \frac{t^{1-2\sigma}}{n}  \]
where the implied constant is independent of $t$.
\end{lemma}
\begin{proof}
Integrating by parts we find the above integral can be rewritten as
\begin{multline*} \frac{1}{2\pi\ensuremath{\mathrm{i}} n}\left(\alpha^{-v} \zeta_1(u,\alpha) - \frac{\alpha^{1-u-v}}{u-1} + \frac{\alpha^{-u-v}}{2} \right)e^{-2\pi\ensuremath{\mathrm{i}} n \alpha}\Big|_{\alpha = t/2\pi + \eta} \\
+ \frac{1}{2\pi\ensuremath{\mathrm{i}} n} \int_{t/2\pi+\eta}^\infty  \frac{\partial}{\partial \alpha} \left(\alpha^{-v} \zeta_1(u,\alpha) - \frac{\alpha^{1-u-v}}{u-1} + \frac{\alpha^{-u-v}}{2} \right)e^{-2\pi\ensuremath{\mathrm{i}} n \alpha}\, \ensuremath{\mathrm{d}} \alpha. \end{multline*}
The first term can be estimated using \eqref{est}, giving
\[ \left| \frac{1}{2\pi\ensuremath{\mathrm{i}} n}\left(\alpha^{-v} \zeta_1(u,\alpha) - \frac{\alpha^{1-u-v}}{u-1} + \frac{\alpha^{-u-v}}{2} \right)e^{-2\pi\ensuremath{\mathrm{i}} n \alpha}\bigg|_{\alpha = t/2\pi + \eta} \right| \lesssim_\eta  \frac{t^{-2\sigma }}{n}. \]
And the second term can be estimated using \eqref{derivestimate}
\[ \left|\frac{1}{2\pi\ensuremath{\mathrm{i}} n} \int_{t/2\pi+\eta}^\infty  \frac{\partial}{\partial \alpha} \left(\alpha^{-v} \zeta_1(u,\alpha) - \frac{\alpha^{1-u-v}}{u-1} + \frac{\alpha^{-u-v}}{2} \right)e^{-2\pi\ensuremath{\mathrm{i}} n \alpha}\, \ensuremath{\mathrm{d}} \alpha \right| \lesssim_\eta \frac{t^2}{n} \int_{t/2\pi+\eta} \alpha^{-2\sigma-2}\, \ensuremath{\mathrm{d}} \alpha. \]
Performing the final integration shows that this term is $\mathcal{O}_\eta( t^{1-2\sigma}/n)$.
\end{proof}
The previous Lemma gives the following
\begin{multline*} q_n(u,v) =  \left[ \frac{1}{u-1} + \frac{1}{v-1} \right]a_n(u+v-1)  + \int_1^{t/2\pi+\eta} \left(\alpha^{-v} \zeta_1(u,\alpha) - \frac{\alpha^{1-u-v}}{u-1} + \frac{\alpha^{-u-v}}{2} \right)e^{-2\pi\ensuremath{\mathrm{i}} n \alpha}\, \ensuremath{\mathrm{d}} \alpha \\+ \int_1^{t/2\pi + \eta} \left(\alpha^{-u} \zeta_1(v,\alpha) - \frac{\alpha^{1-u-v}}{v-1} + \frac{\alpha^{-u-v}}{2} \right)e^{-2\pi\ensuremath{\mathrm{i}} n \alpha}\, \ensuremath{\mathrm{d}} \alpha + \mathcal{O}_\eta \left(\frac{t^{1-2\sigma}}{n}\right), \end{multline*}
valid for $u=\sigma+\ensuremath{\mathrm{i}} t, v=\sigma-\ensuremath{\mathrm{i}} t$ and $\sigma>0$. We note that for $n\neq 0$ a simple integration by parts argument provides an analytic continuation for $a_n(s)$ into $\Re s >-1$, i.e.
\begin{align*} a_n(s) &= \int_1^\infty \alpha^{-s} e^{-2\pi\ensuremath{\mathrm{i}} n \alpha}\, \ensuremath{\mathrm{d}} \alpha \\
 &= \frac{1}{2\pi\ensuremath{\mathrm{i}} n} - \frac{s}{2\pi\ensuremath{\mathrm{i}} n} \int_1^\infty \alpha^{-s-1} e^{-2\pi\ensuremath{\mathrm{i}} n \alpha}\, \ensuremath{\mathrm{d}} \alpha.
 \end{align*}
In particular
\[ a_n(2\sigma-1) = \frac{1}{2\pi\ensuremath{\mathrm{i}} n} - \frac{(2\sigma-1)}{2\pi\ensuremath{\mathrm{i}} n} \int_1^\infty \alpha^{-2\sigma} e^{-2\pi\ensuremath{\mathrm{i}} n \alpha}\, \ensuremath{\mathrm{d}}\alpha = \mathcal{O} \left( \frac{1}{n}\right). \]
Using integration by parts, we also have
\[ \frac{1}{u-1} \int_1^{t/2\pi + \eta} \alpha^{1-u-v} e^{-2\pi\ensuremath{\mathrm{i}} n \alpha}\, \ensuremath{\mathrm{d}} \alpha = \mathcal{O}\left( \frac{t^{1-2\sigma}}{n} \right), \quad \int_1^{t/2\pi + \eta} \alpha^{-u-v} e^{-2\pi\ensuremath{\mathrm{i}} n \alpha} = \mathcal{O} \left( \frac{1}{n}\right).  \]
So for $0<\sigma \leq 1/2$ and $(u,v)=(\sigma+\ensuremath{\mathrm{i}} t,\sigma-\ensuremath{\mathrm{i}} t)$ we have
\[ q_n(u,v) = \int_1^{t/2\pi+\eta} \alpha^{-v} \zeta_1(u,\alpha) e^{-2\pi\ensuremath{\mathrm{i}} n\alpha}\, \ensuremath{\mathrm{d}} \alpha + \int_1^{t/2\pi+\eta} \alpha^{-v} \zeta_1(u,\alpha) e^{-2\pi\ensuremath{\mathrm{i}} n\alpha}\, \ensuremath{\mathrm{d}} \alpha + \mathcal{O}_\eta \left( \frac{t^{1-2\sigma}}{n}\right). \]
Finally we show that terms with $|n|>t/2\pi$ are easily controllable. For this we once again use the approximate functional equation for the Hurwitz zeta function in the form
\[ \zeta_1(s,\alpha) = \sum_{1\leq m \leq N} (\alpha+m)^{-s} + \frac{(N+\alpha)^{1-s}}{s-1}  - \frac{1}{2} (N+\alpha)^{-s} - s\int_N^\infty (\alpha + x)^{-s-1} (x-[x]-\tfrac{1}{2})\, \ensuremath{\mathrm{d}} x \]
which holds for $N>1$. This follows directly from the Euler-Maclaurin formula when $\Re s>1$ and then by analytic continuation for $\Re s>0$. We will require the following lemma
\begin{lemma}
For $y>1$, $\sigma_1, \sigma_2>0$ and $|n|>t/2\pi$ we have
\[ \left|\int_1^{t/2\pi + \eta} \alpha^{-\sigma_1 +\ensuremath{\mathrm{i}} t} (\alpha + y)^{-\sigma_2 - \ensuremath{\mathrm{i}} t} e^{-2\pi\ensuremath{\mathrm{i}} n \alpha}\, \ensuremath{\mathrm{d}} \alpha\right| \lesssim \frac{y^{-\sigma_2}}{|n- t/2\pi|},   \]
\[ \left|\int_1^{t/2\pi + \eta} \alpha^{-\sigma_1 -\ensuremath{\mathrm{i}} t} (\alpha + y)^{-\sigma_2 + \ensuremath{\mathrm{i}} t} e^{-2\pi\ensuremath{\mathrm{i}} n \alpha}\, \ensuremath{\mathrm{d}} \alpha\right| \lesssim \frac{y^{-\sigma_2}}{|n + t/2\pi|} .  \]
\end{lemma}
\begin{proof}
The proof is essentially the same as that used for Lemma \ref{intbyparts}. The oscillatory term doesn't have stationary points if $|n|>t/2\pi$ so integrating by parts yields the desired estimate.
\end{proof}
Applying this lemma and using the approximate functional equation we find that for $|n|>t/2\pi$ the following estimate is valid
\begin{multline*}\int_1^{t/2\pi+\eta} \alpha^{-v} \zeta_1(u,\alpha) e^{-2\pi\ensuremath{\mathrm{i}} n\alpha}\, \ensuremath{\mathrm{d}} \alpha \\ = \mathcal{O}\left( \frac{N^{\sigma}}{|n-t/2\pi|}\right) + \mathcal{O}\left( \frac{N^{1-\sigma}}{t|n-t/2\pi|}\right) + \mathcal{O}\left( \frac{N^{-\sigma}}{|n-t/2\pi|}\right) + \mathcal{O}\left( \frac{tN^{-\sigma}}{|n-t/2\pi|}\right).\end{multline*}
By choosing $N= \mathcal{O}(t^{1/2\sigma})$ we find
\[ \left|\int_1^{t/2\pi+\eta} \alpha^{-v} \zeta_1(u,\alpha) e^{-2\pi\ensuremath{\mathrm{i}} n\alpha}\, \ensuremath{\mathrm{d}} \alpha \right| \lesssim \frac{ t^{1/2}}{|n-t/2\pi|}. \]
Similarly
\[ \left| \int_1^{t/2\pi+\eta} \alpha^{-u} \zeta_1(v,\alpha) e^{-2\pi\ensuremath{\mathrm{i}} n\alpha}\, \ensuremath{\mathrm{d}} \alpha \right| \lesssim \frac{ t^{1/2}}{|n-t/2\pi|}. \]
In particular, for $u=\sigma+\ensuremath{\mathrm{i}} t$ and $v=\bar u$
\[ \sum_{n>t/\pi} |q_n(u,v)|^2 \lesssim_\eta \sum_{|n|>t/\pi} \left[ \frac{t}{|n-t/2\pi|^2} + \frac{t}{|n+t/2\pi|^2} + \frac{t^{2-4\sigma}}{n^2} \right] = \mathcal{O}\left( t^{2-4\sigma}\right). \]
For $\sigma \geq \tfrac{1}{2}$ this term is bounded. Now by Parseval's theorem
\begin{align*} \int_0^1 |\zeta_1(u,\alpha)|^4\, \ensuremath{\mathrm{d}} \alpha &= \sum_n |q_n(u,\bar u)|^2 \\
&\lesssim_\eta  \sum_{|n|\leq t/\pi} \left| \int_1^{t/2\pi+\eta} \alpha^{-\bar{u}} \zeta_1(u,\alpha) e^{-2\pi\ensuremath{\mathrm{i}} n \alpha}\, \ensuremath{\mathrm{d}} \alpha + \frac{t^{1-2\sigma}}{n}\right|^2 + \mathcal{O}\left( t^{2-4\sigma}\right) \\
&\lesssim_\eta  \sum_{|n|\leq t/\pi} \left| \int_1^{t/2\pi+\eta} \alpha^{-\bar{u}} \zeta_1(u,\alpha) e^{-2\pi\ensuremath{\mathrm{i}} n \alpha}\, \ensuremath{\mathrm{d}} \alpha \right|^2 + \mathcal{O}\left( t^{2-4\sigma}\right)
\end{align*}
Taking $\sigma=\tfrac{1}{2}$ and using this in Theorem 1 gives rise to the estimate in Theorem 2.

\subsubsection*{Acknowledgements}
The second author acknowledges support from EPSRC via a Senior Fellowship award.

\small


\section{Introduction}\label{intro}
A chaotic system is one whose motion is sensitive to initial conditions \cite{strogatz1994nonlinear}.  Since different initial conditions lead to different trajectories for the same dynamical system, it is expected that the trajectories cannot coincide. The possibility of two chaotic systems with different trajectories to follow the same trajectory by the introduction of a control function, as proposed by \cite{Pecora1990}, has been an interesting research area for scientists in nonlinear dynamics.  This is partly due to the applicability in different fields such as communication technology, security, neuroscience, atmospheric physics and electronics.

There are several methods for the synchronization of chaotic systems.  These methods include active control, Open Plus Closed Loop (OPCL), backstepping, feedback control, adaptive control, sliding mode and others.  A comparison of performance of a modified active control method and backstepping control  on synchronization of integer order system has been investigated \cite{ojo2013comparison}.  The active control method was found ``to be simpler with more stable synchronization time and hence more suitable for practical implementation".  The active control method was also found to have the best stability and convergence when compared with the direct method and OPCL method for fractional order systems \cite{ogunjo2017comparison}.

Generally, complete synchronization between a drive system $y_i$ and response system $x_i$ is said to occur if $\lim_{t\rightarrow+\infty}||y_i - x_i|| = 0$ and anti-synchronization if $\lim_{t\rightarrow+\infty}||y_i + x_i|| = 0$. If the error term is such that $\lim_{t\rightarrow+\infty}||y_i - \alpha x_i|| = 0$, where $\alpha$ is a positive integer, we have projective synchronization.  According to \cite{Zhou2017}, $\delta$ synchronization is defined by the error given as $\lim_{t\rightarrow+\infty}||y_i \pm x_i|| \leq \delta$, where $\delta$ has small value.  Other forms of synchronization include phase synchronization, anticipated synchronization, lag synchronization etc.  The possibility of one or more of these synchronization scheme in a single synchronization has not been explored.

Different synchronization methods and techniques have been used to study synchronization between two similar integer order systems \cite{ojo2012synchronization}, two dissimilar integer order systems of same dimension \cite{MOTALLEBZADEH20123643,femat2002}, two similar or dissimilar systems with different dimensions \cite{ojo2013mixed,ojo2014increased,ogunjo2013increased}, three or more integer order system (compound, combination-combination synchronization) \cite{ojo2014reduced1,ojo2014reduced2,Mahmoud2016}, discrete systems \cite{liu2008,kloeden2004,ma2007}, fractional order system of similar dimension \cite{LU20051125}, fractional order synchronization of different dimension \cite{Bhalekar2014,ayub2017},  circuit implementation of synchronization \cite{adelakun2017dynamics} and synchronization between integer order and fractional order systems \cite{chen2012}.

The study of chaotic systems has evolved over time from integer order dynamical systems to cover partial differential equations, time delayed differential equations, fractional order differential equations and even time series data.  The prevalence of integer order system was the lack of solution methods for fractional differential equations \cite{chen2009} and its inherent complexity \cite{gutierrez2010fractional}.  The Gr$\ddot{u}$nwald-Letnikov definition of fractional order systems, the fractional order derivative of order $\alpha$ can be written as \cite{petras2011fractional}
\begin{equation}\label{gl}
    D_t^\alpha f(t) = \lim_{h\rightarrow 0}\frac{1}{h^\alpha} \sum_{j=0}^\infty (-1)^j\begin{pmatrix}
                                                   \alpha \\
                                                   j\\
                                                 \end{pmatrix} f(t-jh)
\end{equation}
where the binomial coefficients can be written in terms of the Gamma function as
\begin{equation*}
    \begin{pmatrix}
    \alpha \\
    j\\
    \end{pmatrix}
    =
    \frac{\Gamma(\alpha + 1)}{\Gamma(j+1)\Gamma(\alpha - j +1)}
\end{equation*}
The Riemann-Liouville definition of fractional derivative is given as
\begin{equation}\label{rl}
    D_t^{-\alpha}f(t) = \frac{1}{\Gamma(n-\alpha)}  \frac{d^n}{dt^n} \int_a^t \frac{f(\tau)}{(t-\tau)^{\alpha +1}}d\tau
\end{equation}
The Caputo fractiional derivatives can be written as
\begin{equation}\label{caputo}
    D_t^{\alpha} f(t) = \frac{1}{\Gamma(n-\alpha)}\int_a^t \frac{f^{(n)}(\tau)}{(t-\tau)^{\alpha - n + 1}}d\tau,\,\, n-1<\alpha<n
\end{equation}
Fractional order systems have been found as a useful model in many engineering, physical and biological systems.

In this present work, we aim to investigate the possibility of coexistence of different synchronization scheme in the synchronization of four chaotic systems (two drives and two response systems).  Specifically, we aim to implement synchronization, anti-synchronization and projective synchronization on different dimensions in a fractional order combination synchronization using the method of active control.  We believe, if implemented, it will will enhance faster, robust and more secure information transmission.  To the best of our knowledge, this has not been reported in literature.
\section{System Description}\label{system}
The integer order Chen system was introduced by \cite{GuanrongChen1999} as
\begin{equation}\label{chen1}
    \begin{split}
      \dot{x} &= a(y - x) \\
      \dot{y} &= (c-a)x + cy - xz\\
      \dot{z} &= -bz + xy
    \end{split}
\end{equation}

The fractional order chaotic Chen system was introduced by \cite{Li2004} as
\begin{equation}\label{eqn_d1}
    \begin{split}
      D^{\alpha} x_1 &= \sigma (x_2 - x_1) \\
      D^{\alpha} x_2 &= (c-a) x_1 - x_1x_3 + cx_2\\
      D^{\alpha} x_3 &= x_1 x_2 - b x_3
    \end{split}
\end{equation}
The system was found to be chaotic when $(a,b,c)=(35,3,28)$ and $0.7 \leq \alpha \leq 0.9$.  However, by varying parameter $a$ rather than parameter $c$ as in \cite{Li2004}, the system was found to be chaotic in the region $0.1\leq \alpha \leq 0.1$ \cite{LU2006685}. The phase space of the fractional order Chen system is shown in figure \ref{fig1}.   Various successful attempts have been made at synchronization of the integer, hyperchaotic, and fractional order Chen system \cite{hegazi2011,long2010,deng2005,Chen2008}.

\section{Design and implementation of synchronization scheme}\label{sec_des_imp}
The co-existence of different synchronization scheme within commensurate fractional order Chen system will be studied.  Suitable controllers are designed (Section \ref{design}) and numerical simulations presented in Section \ref{simulation} to verify the proposed controllers. \cite{Bhalekar2014}

\subsection{Design of controllers}\label{design}

Let the two drive system be defined as
\begin{equation}\label{eqn_d1}
    \begin{split}
      D^{p_1} x_1 &= \sigma (x_2 - x_1) \\
      D^{p_2} x_2 &= (c-a) x_1 - x_1x_3 + cx_2\\
      D^{p_3} x_3 &= x_1 x_2 - b x_3
    \end{split}
\end{equation}
and
\begin{equation}\label{eqn_d2}
    \begin{split}
      D^{q_1} y_1 &= \sigma (y_2 - y_1) \\
      D^{q_2} y_2 &= (c-a) y_1 - y_1y_3 + cy_2\\
      D^{q_3} y_3 &= y_1 y_2 - b y_3
    \end{split}
\end{equation}
Defining the two response systems as
\begin{equation}\label{eqn_d3}
    \begin{split}
      D^{r_1} z_1 &= \sigma (z_2 - z_1) + u_1 \\
      D^{r_2} z_2 &= (c-a) z_1 - z_1z_3 + cz_2 + u_2\\
      D^{r_3} z_3 &= z_1 z_2 - b z_3 + u_3
    \end{split}
\end{equation}
and
\begin{equation}\label{eqn_d4}
    \begin{split}
      D^{s_1} w_1 &= \sigma (w_2 - w_1) + u_4 \\
      D^{s_2} w_2 &= (c-a) w_1 - w_1w_3 + cw_2 + u_5\\
      D^{s_3} w_3 &= w_1 w_2 - b w_3 + u_6
    \end{split}
\end{equation}
where the six active control functions $u_1,\, u_2,\, u_3,\, u_4,\, u_5,\, u_6$ introduced in equations \ref{eqn_d3} and \ref{eqn_d4} are control functions to be determined.

We define the error states $e_1,\, e_2\, e_3$ as

\begin{equation}\label{eqn_d5}
    \begin{split}
      e_1 &= (x_1 + y_1) - (z_1 + w_1) \\
     e_2 &= (x_2 + y_2) + (z_2 + w_2) \\
     e_3 &= (x_3 + y_3) - \alpha (z_3 + w_3)
    \end{split}
\end{equation}
Substituting the drive systems (equations \ref{eqn_d1} and \ref{eqn_d2}) and response systems (equations \ref{eqn_d3} and \ref{eqn_d4}) into equation \ref{eqn_d5} and assuming a commensurate system, the error system is obtained as

\begin{equation}\label{eqn_d6}
    \begin{split}
      D^\mu e_1 &= -[ ae_1 + ae_2 - 2a(x_2 + y_2) + u_1 + u_4]\\
      D^\mu e_2 &= (a+c)e_1 + ce_2 + 2c(z_1 + w_1) - 2a(x_1 + y_1) - x_1x_3 - y_1y_3 -z_1z_3 - w_1w_3 + u_2 + u_5 \\
      D^\mu e_3 &= -[be_3 - x_1x_2 - y_1y_2 + \alpha z_1z_2 + \alpha w_1w_2 + \alpha u_3 + \alpha u_6
    \end{split}
\end{equation}
Active control inputs $u_i (i=1,2,3,4,5,6)$ are then defined as
\begin{equation}\label{eqn_d7}
    \begin{split}
      u_1 + u_4 &=  -[V_1 + 2a(x_2 + y_2) \\
      u_2 + u_5 &= V_2 - 2c(z_1 + w_1) + 2a(x_1 + y_1) + x_1x_3 + y_1y_3 + z_1z_3 + w_1w_3\\
      u_3 + u_6 &= \frac{1}{\alpha}[-V_3 + x_1x_2 + y_1y_2 - \alpha z_1z_2 - \alpha w_1w_2 ]
    \end{split}
\end{equation}
where the functions $V_i$ are to be obtained.  Substituting equation \ref{eqn_d7} into equation \ref{eqn_d6} yields
\begin{equation}\label{eqn_d8}
    \begin{split}
      D^\mu e_1 &= -ae_1 - ae_2 + V_1\\
      D^\mu e_2 &= (a+c)e_1 + ce_2 + V_2 \\
      D^\mu e_3 &= -be_3 + V_3
    \end{split}
\end{equation}
The synchronization error system (equation \ref{eqn_d8}) is a linear system with active control inputs $V_i$.  We design  an appropriate feedback control which stabilizes the system so that $e_i (i = 1,2,3) \rightarrow 0$ as $t\rightarrow\infty$, which implies that synchronization is achieved with the proposed feedback control.  There are many possible choices for the control inputs $V_i$, for simplicity, we chose
\begin{equation}\label{eqn_d9}
  \begin{bmatrix}
    V_1 \\
    V_2 \\
    V_3 \\
  \end{bmatrix}
  = C \begin{bmatrix}
    e_1 \\
    e_2 \\
    e_3 \\
  \end{bmatrix}
\end{equation}
where C is a $3\times3$ constant matrix.  In order to make the closed loop system stable, matrix C should be selected in such a way that the feedback system has eigenvalues $\lambda_i$ that satisfies the equation
\begin{equation}\label{eqn_d10}
    |\arg(\lambda_i)| > 0.5\pi\alpha, \,\, i = 1,2,\ldots.
\end{equation}
where $\lambda$ is the eigenvalue, I is an identity matrix and A is the coeffcient of the error state.  There are varieties of choices for choosing matrix C.  Matrix C is chosen as

\begin{equation}\label{eqn_d11}
    C = \begin{pmatrix}
          (c-\lambda) & a & 0 \\
          -(a+c) & -(c+\lambda) & 0 \\
          0 & 0 & (b-\lambda) \\
        \end{pmatrix}
\end{equation}
Using equation \ref{eqn_d11} in \ref{eqn_d9}, we obtain our control function as
\begin{equation}\label{eqn_d12}
    \begin{split}
      u_1 + u_4 &= -V_1 + 2a(x_2 + y_2) \\
      u_2 + u_5 &= V_2 - 2c(z_1 + w_1) + 2a(x_1 + y_1) + x_1x_3 + y_1y_3 + z_1z_3 + w_1w_3\\
      u_3 + u_6 &= \frac{1}{\alpha}[-V_3 + x_1x_2 + y_1y_2 - \alpha z_1z_2 - \alpha w_1w_2]
    \end{split}
\end{equation}

Based on the controllers obtained, two unique cases can be observed.

The control system can be defined as
\begin{equation}\label{eqn_d13}
    \begin{split}
      u_1  &= \frac{1}{2}[-V_1 + 2a(x_2 + y_2)] \\
      u_2  &= \frac{1}{2}[V_2 - 2c(z_1 + w_1) + 2a(x_1 + y_1) + x_1x_3 + y_1y_3 + z_1z_3 + w_1w_3]\\
      u_3  &= \frac{1}{2\alpha}[-V_3 + x_1x_2 + y_1y_2 - \alpha z_1z_2 - \alpha w_1w_2]\\
      u_4  &= u_1\\
      u_5  &= u_2\\
      u_6  &= u_3
    \end{split}
\end{equation}
It can also be defined as
The control system can be defined as
\begin{equation}\label{eqn_d14}
    \begin{split}
      u_1  &= -V_1 + 2a(x_2 + y_2)\\
      u_2  &= V_2 - 2c(z_1 + w_1) + 2a(x_1 + y_1) + x_1x_3 + y_1y_3 + z_1z_3 + w_1w_3\\
      u_3  &= \frac{1}{\alpha}[-V_3 + x_1x_2 + y_1y_2 - \alpha z_1z_2 - \alpha w_1w_2]\\
      u_4  &= 0\\
      u_5  &= 0\\
      u_6  &= 0
    \end{split}
\end{equation}

\subsection{Numerical simulation of Results}\label{simulation}
To verify the effectiveness of the synchronization scheme proposed in section \ref{design} using the method of active control, we used the initial conditions $x_i(-10,0.001,37)$, $y_i(37,-5,0)$, $w_i(-5,0.5,25)$ and $z_i(10,-5,15)$.  The order of the system was taken as 0.95.  A time step of 0.005 was used.  In the case of projective synchronization, the scaling parameter was taken to be 5. The parameters of the system are taken as $(a,b,c) = (35,3,28)$.  According to \cite{petras2011fractional}, the general numerical solution of the fractional differential equation
\begin{equation}\label{eqn_ff}
    _aD^q_t y(t) = f(y(t),t)
\end{equation}
can be expressed as
\begin{equation}\label{eqn_fg}
    y(t_k) = f(y(t_k),t_k)h^q - \sum_{j=v}^k c_j^{(q)} y(t_{t-j})
\end{equation}
where $c_i^{(q)}$ is given as
\begin{equation}\label{eqn_fh}
    \begin{split}
      c_0^{(q)} &= 1 \\
      c_j^{(q)} &= \left( 1 - \frac{1+q}{j}c_{j-1}^q \right)
    \end{split}
\end{equation}
The results for the two cases considered are shown in figures \ref{fig2} and \ref{fig3}.   From the results presented, the drives and responses were found to achieve synchronization as indicated by the convergence of the error terms to zero.  The effectiveness of the proposed scheme is hereby confirmed.

\section{Conclusion}
In this paper, a new synchronization scheme is proposed and implemented.  The modified hybrid synchronization that allows for the coexistence of different synchronization schemes was implemented in a compound synchronization of fractional order Chen system.  In particular, the controllers consists of complete synchronization, anti-synchronization, and projective synchronization.  We believe that this type of synchronization will offer better security and more robust.  There is the need to investigate the performance of this type of synchronization using different synchronization schemes.  Furthermore, it will be productive to study the behaviour of this scheme under different types and strength of noise.  Practical implementation of this scheme is also proposed.

\section*{Conflict of Interest}
The authors hereby declare that there is no conflict of interest.

\begin{figure}

  \includegraphics[width=\textwidth]{fig1}\\
  \caption{Phase space of the system. }\label{fig1}
\end{figure}

\begin{figure}

  \includegraphics[width=\textwidth]{fig2}\\
  \caption{Synchronization obtained from the realization of case 1. }\label{fig2}
\end{figure}

\begin{figure}

  \includegraphics[width=\textwidth]{fig3}\\
  \caption{Synchronization obtained from the realization of case 2. }\label{fig3}
\end{figure}

\clearpage
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