# Effective bounds on H t - second approach

## Elementary asymptotics

We use $O_{\leq}(X)$ to denote any quantity bounded in magnitude by at most $X$. An equality $A=B$ using this notation means that any quantity of the form $A$ is also of the form $B$, though we do not require the converse to be true (thus we break the symmetry of the equality relation). Thus for instance $O_{\leq}(1) + O_{\leq}(1) = O_{\leq}(3)$. We have the following elementary estimates:

Lemma 1 Let $x \gt 0$.

1. If $a,b \gt 0$ are such that $x \gt b/a$, then

$O_{\leq}(\frac{a}{x}) + O_{\leq}( \frac{b}{x^2} ) = O_{\leq}( \frac{a}{x-b/a} ).$

2. If $x \gt 1$, then

$\log(1 + O_{\leq}(\frac{1}{x}) ) = O_{\leq}(\frac{1}{x-1})$

or equivalently

$1 + O_{\leq}(\frac{1}{x}) = \exp( O_{\leq}(\frac{1}{x-1}) ).$

3. If $x \gt 1/2$, then

$\exp( O_{\leq}(\frac{1}{x}) ) = 1 + O_{\leq}( \frac{1}{x-0.5} ).$

4. We have $\exp(O_{\leq}(x)) = 1 + O_{\leq}(e^x-1)$.

5. If $z$ is a complex number with $|\mathrm{Im}(z)|\gt1$ or $\mathrm{Re} z \gt 1$, then

$\Gamma(z) = \sqrt{2\pi} \exp( (z-\frac{1}{2}) \log z - z + O_{\leq}( \frac{1}{12(|z| - 0.33)} )).$

6. We have

$\frac{e^{x^2}-1}{2x} \leq \int_0^x e^{u^2}\ du \leq \frac{e^{x^2}-1}{2x} + \frac{e^{x^2}-x^2-1}{x^3}.$

7. If $a,b \gt 0$ and $x \geq x_0 \geq \exp(b/a)$, then

$\log^a x \leq \frac{\log^a x_0}{x_0^b} x^b.$

Proof Claim 1 follows from the geometric series formula

$\frac{a}{x-b/a} = \frac{a}{x} + \frac{b}{x^2} + \frac{b^2/a}{x^3} + \dots.$

For Claim 2, we use the Taylor expansion of the logarithm to note that

$\log( 1 + O_{\leq}(\frac{1}{x}) ) = O_{\leq}(\frac{1}{x} + \frac{1}{2x^2} + \frac{1}{3x^3} + \dots)$

which on comparison with the geometric series formula

$\frac{1}{x-1} = \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + \dots$

gives the claim. Similarly for Claim 3 we may compare the Taylor expansion

$\exp( O_{\leq}(\frac{1}{x}) ) = 1 + O_{\leq}(\frac{1}{x} + \frac{1}{2! x^2} + \frac{1}{3! x^3} + \dots)$

with the geometric series formula

$\frac{1}{x-0.5} = \frac{1}{x} + \frac{1}{2x^2} + \frac{1}{2^2 x^3} + \dots$

and note that $k! \geq 2^k$ for all $k \geq 2$.

Claim 4 follows from the trivial identity $e^x = 1 + (e^x-1)$ and the elementary inequality $e^{-x} \geq 1 - (e^x-1)$. For Claim 5, we may use the functional equation $\Gamma(\overline{z}) = \overline{\Gamma(z)}$ to assume that $\mathrm{Im}(z) \geq 0$. We use equations (1.13), (3.1), (3.14) and (3.15) of [B1994] to obtain the Stirling approximation

$\Gamma(z) = \sqrt{2\pi} \exp( (z-\frac{1}{2}) \log z - z ) (1 + \frac{1}{12 z} + R_2(z) )$

where the remainder $R_2(z)$ obeys the bound

$|R_2(z)| \leq (2 \sqrt{2}+1) \frac{C_2 \Gamma(2)}{(2\pi)^3 |z|^2}$

for $\mathrm{Re}(z) \geq 0$ and

$|R_2(z)| \leq (2 \sqrt{2}+1) \frac{C_2 \Gamma(2)}{(2\pi)^3 |z|^2 |1 - e^{2\pi i z}|}$

for $\mathrm{Re}(z) \leq 0$, where

$C_2 := \frac{1}{2} (1 + \zeta(2)) = \frac{1}{2} (1 + \frac{\pi^2}{6}).$

In the latter case, we have $\mathrm{Im}(z) \geq 1$ by hypothesis, and hence $|1 - e^{2\pi i z}| \geq 1 - e^{-2\pi}$. We conclude that in all ranges of $z$ of interest, we have

$|R_2(z)| \leq \frac{0.0205}{|z|^2}$

and hence by Claim 1

$\Gamma(z) = \sqrt{2\pi} \exp( (z-\frac{1}{2}) \log z - z ) (1 + O_{\leq}( \frac{1}{12(|z| - 0.246)} ))$

and the claim then follows by Claim 2.

For Claim 6, observe that the three expressions here have the power series $\sum_{k=0}^\infty \frac{x^{2k+1}}{2(k+1)!}$, $\sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1) k!}$, and $\sum_{k=0}^\infty \frac{x^{2k+1}}{2(k+1)!} + \frac{x^{2k+1}}{(k+2)!}$. The claim now follows from the elementary inequalities

$\frac{1}{2(k+1)} \leq \frac{1}{2k+1} = \frac{1}{2(k+1)} + \frac{1}{(k+1)(4k+2)} \leq \frac{1}{2(k+1)} + \frac{1}{(k+1) (k+2)}.$

For Claim 7, it suffices to show that the function $x \mapsto \frac{\log^a x}{x^b}$ is nonincreasing for $x \geq \exp(b/a)$. Taking logarithms and writing $y = \log x$, it suffices to show that $a \log y - by$ is non-increasing for $y \geq b/a$, but this is clear from taking a derivative. $\Box$

## Heat flow

The functions $H_t(z)$ obey the backwards heat equation $\partial_t H_t(z)=-\partial_{zz} H_t(z)$ with initial condition $H_0(z)=\frac{1}{8} \xi(\frac{1+iz}{2})$. Making the change of variables

$s := \frac{1+iz}{2} \quad (2.1)$

we conclude that

$H_t(z) = \frac{1}{8} \xi_t(\frac{1+iz}{2}) \quad (2.2)$

where $\xi_t$ solves the forward heat equation $\partial_t \xi_t(s) = \frac{1}{4} \partial_{ss} \xi_t(s)$ with initial condition $\xi_0(s) = \xi(s)$. By the fundamental solution to the heat equation, we thus have

$\xi_t(s) = \int_{-\infty}^\infty \xi(s+\sqrt{t} u) \frac{1}{\sqrt{\pi}} e^{-u^2}\ du.$

Now suppose that $s = \sigma+iT$ for some $T \geq T_0 \geq 10$ (say). From equations (1.1), (3.1) of [A2011] we have

$\xi(s) = \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) {\mathcal R}(s) + \frac{s(s-1)}{2} \pi^{-(1-s)/2} \Gamma((1-s)/2) \overline{{\mathcal R}(\overline{1-s})}$

where

${\mathcal R}(s) = \sum_{n=1}^N \frac{1}{n^s} + E_{0,N}(s)$
$a := \sqrt{\frac{\mathrm{Im}(s)}{2\pi}}$
$N := \lfloor a \rfloor$

and $E_{0,N}(s)$ is an error term, holomorphic in the upper half-plane, to be estimated in more detail later. We therefore have

$\xi(s) = (\sum_{n=1}^N F_{0,n}( \sigma + iT) + \overline{F_{0,n}( 1-\sigma + iT)}) + G_{0,N}(\sigma+iT) + \overline{G_{0,N}(1-\sigma+iT)}$

where

$F_{0,n}(s) := \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \frac{1}{n^s} \quad (2.3)$

and

$G_{0,N}(s) := \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) E_{0,N}(s). \quad (2.4)$

By the fundamental solution, we thus have

$\xi_t(s) = (\sum_{n=1}^N F_{t,n}( \sigma + iT) + \overline{F_{t,n}( 1-\sigma + iT)}) + G_{t,N}(\sigma+iT) + \overline{G_{t,N}(1-\sigma+iT)} \quad (2.5)$

where

$F_{t,n}(s) := \int_{-\infty}^\infty F_{0,n}(s +\sqrt{t} u) \frac{1}{\sqrt{\pi}} e^{-u^2}\ du$
$G_{t,N}(s) := \int_{-\infty}^\infty G_{0,N}(s +\sqrt{t} u) \frac{1}{\sqrt{\pi}} e^{-u^2}\ du.$

To estimate these integrals for $s$ equal to $\sigma$ or $1-\sigma$, we perform a contour shift on each integral (translating $u$ by $\sqrt{t} \alpha_n/2$ or $\sqrt{t} \beta_N/2$) to obtain

$F_{t,n}(s) := \exp( - \frac{t}{4} \alpha_n^2) \int_{-\infty}^\infty \exp( - \sqrt{t} \alpha_n u) F_{0,n}(s +\sqrt{t} u + \frac{t}{2} \alpha_n) \frac{1}{\sqrt{\pi}} e^{-u^2}\ du \quad (2.6)$
$G_{t,N}(s) := \exp( - \frac{t}{4} \beta_N^2) \int_{-\infty}^\infty \exp( - \sqrt{t} \beta_N u) G_{0,N}(s +\sqrt{t} u + \frac{t}{2} \beta_N) \frac{1}{\sqrt{\pi}} e^{-u^2}\ du. \quad (2.7)$

where $\alpha_n = \alpha_n(s), \beta_N = \beta_N(s)$ are complex parameters with imaginary part greater than $-T$ to be chosen later (basically one chooses these parameters to kill off as much oscillation or exponential growth in the integrands as possible).

## Approximating $F_{t,n}(s)$

Let $s = \sigma+iT$ with $T \geq T_0 \geq 10$ and let $t \leq 1/2$. From (2.3) and Lemma 1.5 we have

$F_{0,n}(s') = H_{0,n}(s') \exp( O_{\leq}( \frac{1}{6(T - 0.66)} ) )$

where

$H_{0,n}(s') := \frac{s' (s'-1)}{2} \pi^{-s'/2} \frac{1}{n^{s'}} \sqrt{2\pi} \exp( (\frac{s'}{2} - \frac{1}{2}) \log \frac{s'}{2} - \frac{s'}{2} ). \quad (3.1)$

We compute the log-derivative of $H_{0,n}(s')$ as

$\partial_{s'} \log H_{0,n}(s') = \frac{1}{2s'} + \frac{1}{s'-1} + \frac{1}{2} \log \frac{s'}{2\pi n^2}$

and the second derivative as

$\partial_{s's'} \log H_{0,n}(s') = -\frac{1}{2(s')^2} - \frac{1}{(s'-1)^2} - \frac{1}{2 s'}. \quad (3.2)$

We set

$\alpha = \alpha_n(s) := \partial_{s} \log H_{0,n}(s) = \frac{1}{2s} + \frac{1}{s-1} + \frac{1}{2} \log \frac{s}{2\pi n^2} \quad (3.3);$

observe that the imaginary part of $\alpha_n(s)$ is at least $-\frac{1}{2T_0} - \frac{1}{T_0} \geq -0.15$. In particular, every point $s'$ on the line segment connecting $s$ to $s + \sqrt{t} u + \frac{t}{2} \alpha_n(s)$ for any real $u$ has real part at least $T_0 - 0.08$. By (3.2) followed by Lemma 1.1 we have

$\partial_{s's'} \log H_{0,n}(s') = O_{\leq}( \frac{0.5}{(T - 0.08)^2} + \frac{1}{(T-0.08)^2} + \frac{0.5}{(T_0 - 0.08)} )$
$= O_{\leq}( \frac{1}{2 (T - 3.08)} )$

and hence by Taylor's theorem with remainder

$\log H_{0,n}(s + \sqrt{t} u + \frac{t}{2} \alpha_n(s)) = \log H_{0,n}(s) + (\sqrt{t} u + \frac{t}{2} \alpha_n(s)) \alpha_n(s) + O_{\leq}( \frac{1}{4 (T - 3.08)} |\sqrt{t} u + \frac{t}{2} \alpha_n(s)|^2 ).$

We conclude (estimating $\frac{1}{T_0-0.66}$ by $\frac{1}{T_0 - 3.08}$) that

$F_{0,n}(s') = H_{0,n}(s) \exp( (\sqrt{t} u + \frac{t}{2} \alpha_n(s)) \alpha_n(s) + O_{\leq}( \frac{1}{4 (T - 3.08)} ( |\sqrt{t} u + \frac{t}{2} \alpha_n(s)|^2 + \frac{2}{3} ) ) )$

and hence by (2.6)

$F_{t,n}(s) = \exp( \frac{t}{4} \alpha_n^2 ) H_{0,n}(s) \int_{-\infty}^\infty \exp( O_{\leq}( \frac{1}{4 (T - 3.08)} ( |\sqrt{t} u + \frac{t}{2} \alpha_n(s)|^2 + \frac{2}{3} ) ) ) \frac{1}{\sqrt{\pi}} e^{-u^2}\ du.$

Since $\frac{1}{\sqrt{\pi}} e^{-u^2}\ du$ integrates to $1$, we thus have from Lemma 1.4 that

$F_{t,n}(s) = \exp( \frac{t}{4} \alpha_n^2 ) H_{0,n}(s) (1 + O_{\leq}(\epsilon_n(s)) ) \quad (3.4)$

where

$\epsilon_n(s) := \int_{-\infty}^\infty (\exp( \frac{1}{4 (T - 3.08)} ( |\sqrt{t} u + \frac{t}{2} \alpha_n(s)|^2 + \frac{2}{3} ) ) - 1 ) \frac{1}{\sqrt{\pi}} e^{-u^2}\ du.$

To simplify this expression, we use the elementary inequality

$|\sqrt{t} u + \frac{t}{2} \alpha_n|^2 \leq 2 |\sqrt{t} u|^2 + 2 |\frac{t}{2} \alpha_n|^2$

to bound

$\frac{1}{4 (T - 3.08)} ( |\sqrt{t} u + \frac{t}{2} \alpha_n(s)|^2 + \frac{2}{3} ) \leq \frac{1}{2(T-3.08)} ( t u^2 + \frac{t^2}{4} |\alpha_n(s)|^2 + \frac{1}{3} ).$

Applying the mean value theorem estimate $e^x - 1 \leq x e^x$, we conclude that

$\epsilon_n(s) \leq \frac{1}{2\sqrt{\pi} (T - 3.08)} \int_{-\infty}^\infty ( tu^2 + \frac{t^2}{4} |\alpha_n|^2 + \frac{1}{3} ) \exp( - (1 - \frac{t}{2(T-3.08)}) u^2 + \frac{1}{2(T-3.08)} (\frac{t^2}{4} |\alpha_n(s)|^2 + \frac{1}{3}) )\ du$

the right-hand side may be evaluated exactly as

$\epsilon_n(s) \leq \frac{1}{2 (T - 3.08)} \exp(\frac{1}{2(T-3.08)} (\frac{t^2}{4} |\alpha_n(s)|^2 + \frac{1}{3})) (1 - \frac{t}{2(T-3.08)})^{-1/2} (\frac{t^2}{4} |\alpha_n(s)|^2 + \frac{1}{3} + \frac{t}{1 - \frac{t}{2(T-3.08)}} ).$

Roughly speaking, this bound is of the form $O( \frac{\log^2 \frac{T}{2\pi n^2}}{T} )$. We can clean it up a bit as follows. Firstly, from Lemma 1.2 and $t \leq 0.5$ we have

$1 - \frac{t}{2(T-3.08)} = \exp( O_{\leq}(\frac{t}{2(T - 3.33)} ) )$

and similarly

$\frac{1}{2 (T - 3.08)} \frac{1}{1 - \frac{t}{2(T-3.08)}} = \frac{1}{2(T-3.08-2t)} \leq \frac{1}{2(T-3.33)}$

and so

$\epsilon_n(s) \leq \frac{1}{2 (T - 3.33)} \exp(\frac{1}{2(T-3.33)} (\frac{t^2}{4} |\alpha_n(s)|^2 + \frac{1}{3} + t)) (\frac{t^2}{4} |\alpha_n(s)|^2 + \frac{1}{3} + t ).$

From (3.1), (3.3) we have $H_{0,n}(s) = H_{0,1}(s) \frac{1}{n^s}$ and $\alpha_n(s) = \alpha_1(s) - \log n$. We conclude that

$\sum_{n=1}^N F_{t,n}(s) = \exp( \frac{t}{4} \alpha_1(s)^2 ) H_{0,1}(s) ( \sum_{n=1}^N \frac{1}{n^{s + \frac{t \alpha_1(s)}{2} - \frac{t}{4} \log n}} + O_{\leq}( \frac{1}{T-3.33} \epsilon'(s) )) \quad (3.5)$

where

$\epsilon'(s) := \frac{1}{2} \sum_{n=1}^N \frac{1}{n^{\sigma + \frac{t \mathrm{Re} \alpha_1(s)}{2} - \frac{t}{4} \log n}} \exp(\frac{1}{2(T-3.33)} (\frac{t^2}{4} |\alpha_1(s) - \log n|^2 + \frac{1}{3} + t)) (\frac{t^2}{4} |\alpha_1(s) - \log n|^2 + \frac{1}{3} + t ). \quad (3.6)$

This quantity grows very slowly (like $O( \log^2 T )$) and should be easy to control numerically. One can bound it somewhat crudely by

$\epsilon'(s) \leq \frac{1}{2} \sum_{n=1}^N \frac{1}{n^{\sigma + \frac{t \mathrm{Re} \alpha_1(s)}{2} - \frac{t}{4} \log N}} \exp(\frac{1}{2(T-3.33)} (\frac{t^2}{4} |\alpha_1(s)|^2 + \frac{1}{3} + t)) (\frac{t^2}{4} |\alpha_1(s)|^2 + \frac{1}{3} + t )$

which assuming that

$\sigma + \frac{t \mathrm{Re} \alpha_1(s)}{2} - \frac{t}{4} \log N \gt 1 \quad (3.7)$

(which should be true for $T$ moderately large) yields the bound

$\epsilon'(s) \leq \frac{1}{2} \zeta( \sigma + \frac{t \mathrm{Re} \alpha_1(s)}{2} - \frac{t}{4} \log N ) \exp(\frac{1}{2(T-3.33)} (\frac{t^2}{4} |\alpha_1(s)|^2 + \frac{1}{3} + t)) (\frac{t^2}{4} |\alpha_1(s)|^2 + \frac{1}{3} + t ) \quad (3.8)$

## Bounding $G_{t,N}(s)$

Let $s = \sigma_0+iT$ for some $T \geq T_0 \geq 10$, and let $0 \lt t \leq 1/2$. We will apply (2.7) with

$\beta_N := +\frac{\pi i}{4}$

to obtain

$G_{t,N}(s) = \exp( \frac{t \pi^2}{64} ) \int_{-\infty}^\infty \exp( \frac{\pi}{4} \sqrt{t} i u) G_{0,N}(s + \sqrt{t} u - \frac{\pi i t}{8} ) \frac{1}{\sqrt{\pi}} e^{-u^2}\ du$

and hence

$|G_{t,N}(s)| \leq \exp( \frac{t \pi^2}{64} ) \int_{-\infty}^\infty |G_{0,N}(s + \sqrt{t} u - \frac{\pi i t}{8} )| \frac{1}{\sqrt{\pi}} e^{-u^2}\ du$

We change variables to write this as

$|G_{t,N}(s)| \leq \exp( \frac{t \pi^2}{64} ) \int_{-\infty}^\infty |G_{0,N}(\sigma + i T')| \frac{1}{\sqrt{\pi t}} e^{-(\sigma-\sigma_0)^2/t}\ d\sigma$

where $T' := T + \frac{\pi t}{8}$. In particular,

$|G_{t,N}(\sigma_0+iT) + \overline{G_{t,N}(1-\sigma_0+iT)}| \leq 2 \exp( \frac{t \pi^2}{64} ) \int_{-\infty}^\infty |G_{0,N}(\sigma + i T' )| f(\sigma)\ d\sigma$

where

$f(\sigma) := \frac{1}{2\sqrt{\pi t}} (e^{-(\sigma-\sigma_0)^2/t} + e^{-((1-\sigma)-\sigma_0)^2/t}) \quad (4.1)$

is an average of two heat kernels.

From (2.4) and Lemma 1.5 we have

$|G_{0,N}(\sigma + i T' )| \leq \frac{(\sigma^2 + (T')^2)^{1/2} ((1-\sigma)^2 + (T')^2)^{1/2}}{2} \sqrt{2\pi} \pi^{-\sigma/2} |\exp( (\frac{\sigma-1}{2} + \frac{iT'}{2}) \log(\frac{\sigma}{2} + \frac{iT'}{2}) - (\frac{\sigma}{2} + \frac{iT'}{2}) + O_{\leq}( \frac{1}{12(T' - 0.33)} ) )| |E_{0,N}(\sigma+iT')|$
$\leq \frac{(1 + \frac{\sigma^2}{(T')^2})^{1/2} (1 + \frac{(1-\sigma)^2}{(T')^2})^{1/2} }{2} \sqrt{2\pi} \pi^{-\sigma/2} (T')^{(\sigma+3)/2} 2^{(1-\sigma)/2} e^{-\pi T'/4} \exp( \frac{\sigma-1}{4} \log (1 + \frac{\sigma^2}{(T')^2}) + \frac{T'}{2} \arctan \frac{\sigma}{T'} - \frac{\sigma}{2} + \frac{1}{12(T' - 0.33)}) |E_{0,N}(\sigma+iT')|$
$=\exp( - \frac{t\pi^2}{32}) (1 + \frac{\sigma^2}{(T')^2})^{1/2} (1 + \frac{(1-\sigma)^2}{(T')^2})^{1/2} \sqrt{\pi} (T')^{3/2} e^{-\pi T/4} a^\sigma \exp( \frac{\sigma-1}{4} \log (1 + \frac{\sigma^2}{(T')^2}) + \frac{T'}{2} \arctan \frac{\sigma}{T'} - \frac{\sigma}{2} + \frac{1}{12(T' - 0.33)}) |E_{0,N}(\sigma+iT')|$

where

$a := \sqrt{T'/2\pi} \quad (4.2)$

and thus

$|G_{t,N}(\sigma_0+iT) + \overline{G_{t,N}(1-\sigma_0+iT)}| \leq 2 \sqrt{\pi} \exp( -\frac{t \pi^2}{64} ) (T')^{3/2} e^{-\pi T/4} \int_{-\infty}^\infty a^\sigma |E_{0,N}(\sigma + i T' )| w(\sigma) f(\sigma)\ d\sigma$

where $w$ is the weight

$w(\sigma) := (1 + \frac{\sigma^2}{(T'_0)^2})^{1/2} (1 + \frac{(1-\sigma)^2}{(T'_0)^2})^{1/2} \exp( \frac{(\sigma-1)_+}{4} \log (1 + \frac{\sigma^2}{(T'_0)^2}) + (\frac{T'_0}{2} \arctan \frac{\sigma}{T'_0} - \frac{\sigma}{2}) 1_{\sigma \lt 0} + \frac{1}{12(T'_0 - 0.33)}) \quad (4.3)$

(which is close to 1 in practice) with $T'_0 := T_0 + \frac{\pi t}{8}$ and we have used the fact that $T' \arctan \frac{\sigma}{T'} - \sigma$ is negative for $\sigma \gt 0$ and decreasing in $T$ for negative $\sigma$.

We still have to estimate $a^\sigma |E_{0,N}(\sigma + i T' )|$. From equations (3.3), (3.4) and (3.11) of [A2011] we have

$a^\sigma E_{0,N}(\sigma+iT') = (-1)^{N-1} U (\sum_{k=0}^K \frac{C_k(p)}{a^k} + RS_K)$

for any natural number $K$, where

$p := 1-2(a-N)$
$U := \exp( -i(\frac{T'}{2} \log \frac{T'}{2\pi} - \frac{T'}{2} - \frac{\pi}{8}))$

and $C_k(p), RS_K$ are certain contour integrals defined in [A2011]. In particular

$|a^\sigma E_{0,N}(\sigma+iT')| \leq \sum_{k=0}^K \frac{|C_k(p)|}{a_0^k} + |RS_K| \quad (4.4)$

where

$a_0 := \sqrt{\frac{T'_0}{2\pi}}.$

From equation (5.2) of [A2011] we have the explicit form

$C_0(p) = \frac{e^{\pi i (p^2/2 + 3/8)} - i \sqrt{2} \cos(\pi p/2)}{2 \cos(\pi p)}$

(removing the singularities at $p=\pm 1/2$); since $p$ ranges between -1 and 1, it is not difficult to establish the bound

$|C_0(p)| \leq \frac{1}{2}.$

(This also follows from Theorem 6.1 of [A2011].)

For $\sigma \geq 0$, we shall take $K=1$ in (4.4), thus

$|a^\sigma E_{0,N}(\sigma+iT')| \leq \frac{1}{2} + \frac{|C_1(p)|}{a_0} + |RS_1|.$

From equation (4.1), (4.2), (4.7) of [A2011] we have the bounds

$|C_1(p)| = \frac{9^\sigma}{\sqrt{2} \pi} \frac{\Gamma(1/2)}{2} \leq 0.200 \times 9^\sigma$
$|RS_1(p)| \leq \frac{1}{7} 2^{3\sigma/2} \frac{\Gamma(1)}{(\frac{10}{11} a)^2} \leq 0.173 \frac{2^{3\sigma/2}}{a^2}$

and hence

$|a^\sigma E_{0,N}(\sigma+iT')| \leq \frac{1}{2} ( 1 + 0.400 \frac{9^\sigma}{a_0} + 0.346 \frac{2^{3\sigma/2}}{a_0^2} ) \quad (4.5).$

For $\sigma \lt 0$ the formulae are more complicated, mainly due to the condition $K+\sigma \geq 2$ required in Theorem 4.2 of [A2011]. We thus set $K:= \lfloor -\sigma \rfloor + 3$. From equations (4.1), (4.2), (4.7) of [A2011] we now have the bounds

$\frac{|C_k(p)|}{a^k} \leq \frac{2^{-\sigma}}{\sqrt{2} \pi} \frac{\Gamma(k/2)}{(ba)^k}$
$|RS_K(p)| \leq \frac{1}{2} (9/10)^{\lceil -\sigma \rceil} \frac{\Gamma((K+1)/2)}{(\frac{10}{11} a)^{K+1}},$

where

$b:=\sqrt{(3−2\log2)\pi}.$

One can check that this implies also that

$|C_k(p)| \leq \frac{1}{2} (9/10)^{\lceil -\sigma \rceil} \frac{\Gamma(k/2)}{(\frac{10}{11} a)^{k}}$

for $1 \leq k \leq K$. Since $K+1 \leq 4 - \sigma$, we thus have

$|a^\sigma E_{0,N}(\sigma+iT')| \leq \frac{1}{2} ( 1 + (9/10)^{\lceil -\sigma \rceil} \sum_{1 \leq k \leq 4-\sigma} (1.1)^k \frac{\Gamma(k/2)}{a_0^k}) ).$

We thus have

$|G_{t,N}(\sigma_0+iT) + \overline{G_{t,N}(1-\sigma_0+iT)}| \leq \sqrt{\pi} \exp( -\frac{t \pi^2}{64} ) (T')^{3/2} e^{\pi T/4} \int_{-\infty}^\infty v(\sigma) w(\sigma) f(\sigma)\ d\sigma \quad (4.6)$

where

$v(\sigma) := 1 + 0.400 \frac{9^\sigma}{a_0} + 0.346 \frac{2^{3\sigma/2}}{a_0^2} \quad (4.7)$

for $\sigma \geq 0$ and

$v(\sigma) := 1 + (9/10)^{\lceil -\sigma \rceil} \sum_{1 \leq k \leq 4-\sigma} (1.1)^k \frac{\Gamma(k/2)}{a_0^k} \quad (4.8)$

for $\sigma \lt 0$.

The integral $\int_{-\infty}^\infty v(\sigma) w(\sigma) f(\sigma)\ d\sigma$ should be numerically computable for any given value of $T_0, t, \sigma_0$ and should be close to 1 when $T_0$ is large.

## Final bound

From (2.5), (3.5), (4.6) we conclude that for $T \geq T_0 \geq 10$, one has

$\xi_t(\sigma_0+iT) = \exp( \frac{t}{4} \alpha_1(\sigma_0+iT)^2 ) H_{0,1}(\sigma+iT) ( \sum_{n=1}^N \frac{1}{n^{\sigma_0+iT + \frac{t \alpha_1(\sigma_0+iT)}{2} - \frac{t}{4} \log n}} + O_{\leq}( \frac{1}{T-3.33} \epsilon'(\sigma_0+iT) ))$
$+ \exp( \frac{t}{4} \overline{\alpha_1(1-\sigma_0+iT)}^2 ) \overline{H_{0,1}((1-\sigma_0)+iT)} ( \sum_{n=1}^N \frac{1}{n^{1-\sigma_0-iT + \frac{t \overline{\alpha_1(1-\sigma_0+iT)}}{2} - \frac{t}{4} \log n}} + O_{\leq}( \frac{1}{T-3.33} \epsilon'(1-\sigma_0+iT) ))$
$+ O_{\leq}( \sqrt{\pi} \exp( -\frac{t \pi^2}{64} ) (T')^{3/2} e^{-\pi T/4} \int_{-\infty}^\infty v(\sigma) w(\sigma) f(\sigma)\ d\sigma) \quad (5.1)$

where $\alpha_1$ is defined by (3.3), $H_{0,1}$ is defined by (3.1), $\epsilon'$ is defined by (3.6), $v$ is defined by (4.7), (4.8), $w$ is defined by (4.3), and $f$ is defined by (4.1).

If the inequality (3.7) holds, one can also bound $\epsilon'(s)$ using the estimate (3.8).

Bounds on $H_t(x+iy)$ can be derived from those on $\xi_t(\sigma_0+iT)$ using (2.2). In particular, one has

$|H_t(x+iy) - A^{eff} - B^{eff}| \leq E_1 + E_2 + E_3$

whenever $x \geq 2T_0 \geq 20$, where

$A^{eff} := \frac{1}{8} \exp( \frac{t}{4} \alpha_1(\frac{1-y+ix}{2})^2 ) H_{0,1}(\frac{1-y+ix}{2}) \sum_{n=1}^N \frac{1}{n^{\frac{1-y+ix}{2} + \frac{t \alpha_1(\frac{1-y+ix}{2})}{2} - \frac{t}{4} \log n}}$
$B^{eff} := \frac{1}{8} \exp( \frac{t}{4} \overline{\alpha_1(\frac{1+y+ix}{2})}^2 ) \overline{H_{0,1}(\frac{1+y+ix}{2})} \sum_{n=1}^N \frac{1}{n^{\frac{1+y-ix}{2} + \frac{t \overline{\alpha_1(\frac{1+y+ix}{2})}}{2} - \frac{t}{4} \log n}}$
$H_{0,1}(s) := \frac{s (s-1)}{2} \pi^{-s/2} \sqrt{2\pi} \exp( (\frac{s}{2} - \frac{1}{2}) \log \frac{s}{2} - \frac{s}{2} )$
$E_1 := \frac{1}{8 (T - 3.33)} \exp( \frac{t}{4} \mathrm{Re} \alpha_1(\frac{1-y+ix}{2})^2 ) |H_{0,1}(\frac{1-y+ix}{2})| \epsilon'(\frac{1-y+ix}{2})$
$E_2 := \frac{1}{8 (T - 3.33)} \exp( \frac{t}{4} \mathrm{Re} \alpha_1(\frac{1+y+ix}{2})^2 ) |H_{0,1}(\frac{1+y+ix}{2})| \epsilon'(\frac{1+y+ix}{2})$
$E_3 := \frac{1}{8} \sqrt{\pi} \exp( -\frac{t \pi^2}{64} ) (T')^{3/2} e^{-\pi T/4} \int_{-\infty}^\infty v(\sigma) w(\sigma) f(\sigma)\ d\sigma$
$\epsilon'(s) := \frac{1}{2} \sum_{n=1}^N \frac{1}{n^{\mathrm{Re}(s) + \frac{t \mathrm{Re} \alpha_1(s)}{2} - \frac{t}{4} \log n}} \exp(\frac{1}{2(T-3.33)} (\frac{t^2}{4} |\alpha_1(s) - \log n|^2 + \frac{1}{3} + t)) (\frac{t^2}{4} |\alpha_1(s) - \log n|^2 + \frac{1}{3} + t )$
$f(\sigma) := \frac{1}{2\sqrt{\pi t}} (e^{-(\sigma-(1-y)/2)^2/t} + e^{-(\sigma-(1+y)/2)^2/t}) \quad (4.1)$
$w(\sigma) := (1 + \frac{\sigma^2}{(T'_0)^2})^{1/2} (1 + \frac{(1-\sigma)^2}{(T'_0)^2})^{1/2} \exp( \frac{(\sigma-1)_+}{4} \log (1 + \frac{\sigma^2}{(T'_0)^2}) + (\frac{T'_0}{2} \arctan \frac{\sigma}{T'_0} - \frac{\sigma}{2}) 1_{\sigma \lt 0} + \frac{1}{12(T'_0 - 0.33)})$
$v(\sigma) := 1 + (0.400 \frac{9^\sigma}{a_0} + 0.346 \frac{2^{3\sigma/2}}{a_0^2}) 1_{\sigma \geq 0} + (9/10)^{\lceil -\sigma \rceil} \sum_{1 \leq k \leq 4-\sigma} (1.1)^k \frac{\Gamma(k/2)}{a_0^k} 1_{\sigma \lt 0}$
$a_0 := \sqrt{\frac{T'_0}{2\pi}}$
$\alpha_1(s) := \frac{1}{2s} + \frac{1}{s-1} + \frac{1}{2} \log \frac{s}{2\pi}$
$N := \lfloor \sqrt{ \frac{T'}{2\pi}} \rfloor$
$T := \frac{x}{2}$
$T' := T + \frac{\pi t}{8}$
$T'_0 := T_0 + \frac{\pi t}{8}$