Effective bounds on H t - second approach

Elementary asymptotics

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

Lemma 1 Let [math]\displaystyle{ x \gt 0 }[/math].

1. If [math]\displaystyle{ a,b \gt 0 }[/math] are such that [math]\displaystyle{ x \gt b/a }[/math], then

[math]\displaystyle{ O_{\leq}(\frac{a}{x}) + O_{\leq}( \frac{b}{x^2} ) = O_{\leq}( \frac{a}{x-b/a} ). }[/math]

2. If [math]\displaystyle{ x \gt 1 }[/math], then

[math]\displaystyle{ \log(1 + O_{\leq}(\frac{1}{x}) ) = O_{\leq}(\frac{1}{x-1}) }[/math]

or equivalently

[math]\displaystyle{ 1 + O_{\leq}(\frac{1}{x}) = \exp( O_{\leq}(\frac{1}{x-1}) ). }[/math]

3. If [math]\displaystyle{ x \gt 1/2 }[/math], then

[math]\displaystyle{ \exp( O_{\leq}(\frac{1}{x}) ) = 1 + O_{\leq}( \frac{1}{x-0.5} ). }[/math]

4. We have [math]\displaystyle{ \exp(O_{\leq}(x)) = 1 + O_{\leq}(e^x-1) }[/math].

5. If [math]\displaystyle{ z }[/math] is a complex number with [math]\displaystyle{ |\mathrm{Im}(z)|\gt 1 }[/math] or [math]\displaystyle{ \mathrm{Re} z \gt 1 }[/math], then

[math]\displaystyle{ \Gamma(z) = \sqrt{2\pi} \exp( (z-\frac{1}{2}) \log z - z + O_{\leq}( \frac{1}{12(|z| - 0.33)} )). }[/math]

6. We have

[math]\displaystyle{ \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}. }[/math]

7. If [math]\displaystyle{ a,b \gt 0 }[/math] and [math]\displaystyle{ x \geq x_0 \geq \exp(b/a) }[/math], then

[math]\displaystyle{ \log^a x \leq \frac{\log^a x_0}{x_0^b} x^b. }[/math]

Proof Claim 1 follows from the geometric series formula

[math]\displaystyle{ \frac{a}{x-b/a} = \frac{a}{x} + \frac{b}{x^2} + \frac{b^2/a}{x^3} + \dots. }[/math]

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

[math]\displaystyle{ \log( 1 + O_{\leq}(\frac{1}{x}) ) = O_{\leq}(\frac{1}{x} + \frac{1}{2x^2} + \frac{1}{3x^3} + \dots) }[/math]

which on comparison with the geometric series formula

[math]\displaystyle{ \frac{1}{x-1} = \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + \dots }[/math]

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

[math]\displaystyle{ \exp( O_{\leq}(\frac{1}{x}) ) = 1 + O_{\leq}(\frac{1}{x} + \frac{1}{2! x^2} + \frac{1}{3! x^3} + \dots) }[/math]

with the geometric series formula

[math]\displaystyle{ \frac{1}{x-0.5} = \frac{1}{x} + \frac{1}{2x^2} + \frac{1}{2^2 x^3} + \dots }[/math]

and note that [math]\displaystyle{ k! \geq 2^k }[/math] for all [math]\displaystyle{ k \geq 2 }[/math].

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

[math]\displaystyle{ \Gamma(z) = \sqrt{2\pi} \exp( (z-\frac{1}{2}) \log z - z ) (1 + \frac{1}{12 z} + R_2(z) ) }[/math]

where the remainder [math]\displaystyle{ R_2(z) }[/math] obeys the bound

[math]\displaystyle{ |R_2(z)| \leq (2 \sqrt{2}+1) \frac{C_2 \Gamma(2)}{(2\pi)^3 |z|^2} }[/math]

for [math]\displaystyle{ \mathrm{Re}(z) \geq 0 }[/math] and

[math]\displaystyle{ |R_2(z)| \leq (2 \sqrt{2}+1) \frac{C_2 \Gamma(2)}{(2\pi)^3 |z|^2 |1 - e^{2\pi i z}|} }[/math]

for [math]\displaystyle{ \mathrm{Re}(z) \leq 0 }[/math], where

[math]\displaystyle{ C_2 := \frac{1}{2} (1 + \zeta(2)) = \frac{1}{2} (1 + \frac{\pi^2}{6}). }[/math]

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

[math]\displaystyle{ |R_2(z)| \leq \frac{0.0205}{|z|^2} }[/math]

and hence by Claim 1

[math]\displaystyle{ \Gamma(z) = \sqrt{2\pi} \exp( (z-\frac{1}{2}) \log z - z ) (1 + O_{\leq}( \frac{1}{12(|z| - 0.246)} )) }[/math]

and the claim then follows by Claim 2.

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

[math]\displaystyle{ \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)}. }[/math]

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

Heat flow

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

[math]\displaystyle{ s := \frac{1+iz}{2} \quad (2.1) }[/math]

we conclude that

[math]\displaystyle{ H_t(z) = \frac{1}{8} \xi_t(\frac{1+iz}{2}) \quad (2.2) }[/math]

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

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

Now suppose that [math]\displaystyle{ s = \sigma+iT }[/math] for some [math]\displaystyle{ T \geq T_0 \geq 10 }[/math] (say). From equations (1.1), (3.1) of [A2011] (see also Riemann-Siegel formula) we have

[math]\displaystyle{ \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})} }[/math]

where

[math]\displaystyle{ {\mathcal R}(s) = \sum_{n=1}^N \frac{1}{n^s} + E_{0,N}(s) }[/math]

[math]\displaystyle{ a := \sqrt{\frac{\mathrm{Im}(s)}{2\pi}} }[/math]

[math]\displaystyle{ N := \lfloor a \rfloor }[/math]

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

[math]\displaystyle{ \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)} }[/math]

where

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

and

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

By the fundamental solution, we thus have

[math]\displaystyle{ \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) }[/math]

where

[math]\displaystyle{ F_{t,n}(s) := \int_{-\infty}^\infty F_{0,n}(s +\sqrt{t} u) \frac{1}{\sqrt{\pi}} e^{-u^2}\ du }[/math]

[math]\displaystyle{ G_{t,N}(s) := \int_{-\infty}^\infty G_{0,N}(s +\sqrt{t} u) \frac{1}{\sqrt{\pi}} e^{-u^2}\ du. }[/math]

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

[math]\displaystyle{ 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) }[/math]

[math]\displaystyle{ 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) }[/math]

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

Approximating [math]\displaystyle{ F_{t,n}(s) }[/math]

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

[math]\displaystyle{ F_{0,n}(s') = H_{0,n}(s') \exp( O_{\leq}( \frac{1}{6(T - 0.66)} ) ) }[/math]

where

[math]\displaystyle{ 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) }[/math]

We compute the log-derivative of [math]\displaystyle{ H_{0,n}(s') }[/math] as

[math]\displaystyle{ \partial_{s'} \log H_{0,n}(s') = \frac{1}{2s'} + \frac{1}{s'-1} + \frac{1}{2} \log \frac{s'}{2\pi n^2} }[/math]

and the second derivative as

[math]\displaystyle{ \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) }[/math]

We set

[math]\displaystyle{ \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); }[/math]

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

[math]\displaystyle{ \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)} ) }[/math]

[math]\displaystyle{ = O_{\leq}( \frac{1}{2 (T - 3.08)} ) }[/math]

and hence by Taylor's theorem with remainder

[math]\displaystyle{ \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 ). }[/math]

We conclude (estimating [math]\displaystyle{ \frac{1}{T_0-0.66} }[/math] by [math]\displaystyle{ \frac{1}{T_0 - 3.08} }[/math]) that

[math]\displaystyle{ 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} ) ) ) }[/math]

and hence by (2.6)

[math]\displaystyle{ 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. }[/math]

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

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

where

[math]\displaystyle{ \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. }[/math]

To simplify this expression, we use the elementary inequality

[math]\displaystyle{ |\sqrt{t} u + \frac{t}{2} \alpha_n|^2 \leq 2 |\sqrt{t} u|^2 + 2 |\frac{t}{2} \alpha_n|^2 }[/math]

to bound

[math]\displaystyle{ \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} ). }[/math]

Applying the mean value theorem estimate [math]\displaystyle{ e^x - 1 \leq x e^x }[/math], we conclude that

[math]\displaystyle{ \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 }[/math]

the right-hand side may be evaluated exactly as

[math]\displaystyle{ \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)}} ). }[/math]

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

[math]\displaystyle{ 1 - \frac{t}{2(T-3.08)} = \exp( O_{\leq}(\frac{t}{2(T - 3.33)} ) ) }[/math]

and similarly

[math]\displaystyle{ \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)} }[/math]

and so

[math]\displaystyle{ \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 ). }[/math]

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

[math]\displaystyle{ \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) }[/math]

where

[math]\displaystyle{ \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) }[/math]

This quantity grows very slowly (like [math]\displaystyle{ O( \log^2 T ) }[/math]) and should be easy to control numerically. One can bound it somewhat crudely by

[math]\displaystyle{ \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 ) }[/math]

which assuming that

[math]\displaystyle{ \sigma + \frac{t \mathrm{Re} \alpha_1(s)}{2} - \frac{t}{4} \log N \gt 1 \quad (3.7) }[/math]

(which should be true for [math]\displaystyle{ T }[/math] moderately large) yields the bound

[math]\displaystyle{ \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) }[/math]

Bounding [math]\displaystyle{ G_{t,N}(s) }[/math]

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

[math]\displaystyle{ \beta_N := +\frac{\pi i}{4} }[/math]

to obtain

[math]\displaystyle{ 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 }[/math]

and hence

[math]\displaystyle{ |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 }[/math]

We change variables to write this as

[math]\displaystyle{ |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 }[/math]

where [math]\displaystyle{ T' := T + \frac{\pi t}{8} }[/math]. In particular,

[math]\displaystyle{ |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 }[/math]

where

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

is an average of two heat kernels.

From (2.4) and Lemma 1.5 we have

[math]\displaystyle{ |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')| }[/math]

[math]\displaystyle{ \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')| }[/math]

[math]\displaystyle{ =\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')| }[/math]

where

[math]\displaystyle{ a := \sqrt{T'/2\pi} \quad (4.2) }[/math]

and thus

[math]\displaystyle{ |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 }[/math]

where [math]\displaystyle{ w }[/math] is the weight

[math]\displaystyle{ 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) }[/math]

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

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

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

for any natural number [math]\displaystyle{ K }[/math], where

[math]\displaystyle{ p := 1-2(a-N) }[/math]

[math]\displaystyle{ U := \exp( -i(\frac{T'}{2} \log \frac{T'}{2\pi} - \frac{T'}{2} - \frac{\pi}{8})) }[/math]

and [math]\displaystyle{ C_k(p), RS_K }[/math] are certain contour integrals defined in [A2011]. In particular

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

where

[math]\displaystyle{ a_0 := \sqrt{\frac{T'_0}{2\pi}}. }[/math]

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

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

(removing the singularities at [math]\displaystyle{ p=\pm 1/2 }[/math]); since [math]\displaystyle{ p }[/math] ranges between -1 and 1, it is not difficult to establish the bound

[math]\displaystyle{ |C_0(p)| \leq \frac{1}{2}. }[/math]

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

For [math]\displaystyle{ \sigma \geq 0 }[/math], we shall take [math]\displaystyle{ K=1 }[/math] in (4.4), thus

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

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

[math]\displaystyle{ |C_1(p)| = \frac{9^\sigma}{\sqrt{2} \pi} \frac{\Gamma(1/2)}{2} \leq 0.200 \times 9^\sigma }[/math]

[math]\displaystyle{ |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} }[/math]

and hence

[math]\displaystyle{ |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). }[/math]

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

[math]\displaystyle{ \frac{|C_k(p)|}{a^k} \leq \frac{2^{-\sigma}}{\sqrt{2} \pi} \frac{\Gamma(k/2)}{(ba)^k} }[/math]

[math]\displaystyle{ |RS_K(p)| \leq \frac{1}{2} (9/10)^{\lceil -\sigma \rceil} \frac{\Gamma((K+1)/2)}{(\frac{10}{11} a)^{K+1}}, }[/math]

where

[math]\displaystyle{ b:=\sqrt{(3−2\log2)\pi}. }[/math]

' One can check actually, there is an issue here: there appears to be a factor of [math]\displaystyle{ 2^{-\sigma} }[/math] rather than [math]\displaystyle{ (9/10)^{\lceil -\sigma \rceil} }[/math] that this implies also that

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

for [math]\displaystyle{ 1 \leq k \leq K }[/math]. Since [math]\displaystyle{ K+1 \leq 4 - \sigma }[/math], we thus have

[math]\displaystyle{ |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}) ). }[/math]

We thus have

[math]\displaystyle{ |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) }[/math]

where

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

for [math]\displaystyle{ \sigma \geq 0 }[/math] and

[math]\displaystyle{ 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) }[/math]

for [math]\displaystyle{ \sigma \lt 0 }[/math].

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

Final bound

From (2.5), (3.5), (4.6) we conclude that for [math]\displaystyle{ T \geq T_0 \geq 10 }[/math], one has

[math]\displaystyle{ \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) )) }[/math]

[math]\displaystyle{ + \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) )) }[/math]

[math]\displaystyle{ + 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) }[/math]

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

If the inequality (3.7) holds, one can also bound [math]\displaystyle{ \epsilon'(s) }[/math] using the estimate (3.8).

Bounds on [math]\displaystyle{ H_t(x+iy) }[/math] can be derived from those on [math]\displaystyle{ \xi_t(\sigma_0+iT) }[/math] using (2.2). In particular, one has

[math]\displaystyle{ |H_t(x+iy) - A^{eff} - B^{eff}| \leq E_1 + E_2 + E_3 }[/math]

whenever [math]\displaystyle{ x \geq 2T_0 \geq 20 }[/math], where

[math]\displaystyle{ 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}} }[/math]

[math]\displaystyle{ 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}} }[/math]

[math]\displaystyle{ 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} ) }[/math]

[math]\displaystyle{ 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}) }[/math]

[math]\displaystyle{ 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}) }[/math]

[math]\displaystyle{ 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 }[/math]

[math]\displaystyle{ \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 ) }[/math]

[math]\displaystyle{ 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) }[/math]

[math]\displaystyle{ 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)}) }[/math]

[math]\displaystyle{ 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} }[/math]

[math]\displaystyle{ a_0 := \sqrt{\frac{T'_0}{2\pi}} }[/math]

[math]\displaystyle{ \alpha_1(s) := \frac{1}{2s} + \frac{1}{s-1} + \frac{1}{2} \log \frac{s}{2\pi} }[/math]

[math]\displaystyle{ N := \lfloor \sqrt{ \frac{T'}{2\pi}} \rfloor }[/math]

[math]\displaystyle{ T := \frac{x}{2} }[/math]

[math]\displaystyle{ T' := T + \frac{\pi t}{8} }[/math]

[math]\displaystyle{ T'_0 := T_0 + \frac{\pi t}{8} }[/math]

Upper bounds on derivatives

See Bounding the derivative of H_t.