Difference between revisions of "Effective bounds on H t - second approach"

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(Elementary asymptotics)
(Elementary asymptotics)
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:<math> \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>
:<math> \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>a,b > 0</math> and <math>x \geq x_0 \geq \exp(b/a)</math>, then
:<math> \log^a x \leq \frac{\log^a x_0}{x_0^b} x^b.</math>
'''Proof'''  Claim 1 follows from the geometric series formula
'''Proof'''  Claim 1 follows from the geometric series formula
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:<math> \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>
:<math> \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>x \mapsto \frac{\log^a x}{x^b}</math> is nonincreasing for <math>x \geq \exp(b/a)</math>.  Taking logarithms and writing <math>y = \log x</math>, it suffices to show that <math>a \log y - by</math> is non-increasing for <math>y \geq b/a</math>, but this is clear from taking a derivative.

Revision as of 20:38, 7 March 2018

Elementary asymptotics

We use [math]O_{\leq}(X)[/math] to denote any quantity bounded in magnitude by at most [math]X[/math]. An equality [math]A=B[/math] using this notation means that any quantity of the form [math]A[/math] is also of the form [math]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] O_{\leq}(1) + O_{\leq}(1) = O_{\leq}(3)[/math]. We have the following elementary estimates:

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

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

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

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

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

or equivalently

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

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

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

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

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

[math] \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] \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]a,b \gt 0[/math] and [math]x \geq x_0 \geq \exp(b/a)[/math], then

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

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

Claim 4 follows from the trivial identity [math]e^x = 1 + (e^x-1)[/math] and the elementary inequality [math]e^{-x} \geq 1 - (e^x-1)[/math]. For Claim 5, we may use the functional equation [math]\Gamma(\overline{z}) = \overline{\Gamma(z)}[/math] to assume that [math]\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] \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]R_2(z)[/math] obeys the bound

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

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

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

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

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

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

and hence by Claim 1

[math] \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]\sum_{k=0}^\infty \frac{x^{2k+1}}{2(k+1)!}[/math], [math]\sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1) k!}[/math], and [math]\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] \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]x \mapsto \frac{\log^a x}{x^b}[/math] is nonincreasing for [math]x \geq \exp(b/a)[/math]. Taking logarithms and writing [math]y = \log x[/math], it suffices to show that [math]a \log y - by[/math] is non-increasing for [math]y \geq b/a[/math], but this is clear from taking a derivative. [math]\Box[/math]

Heat flow

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

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

we conclude that

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

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

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

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

[math]\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]


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

and [math]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] \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]


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


[math] 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]\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]


[math] F_{t,n}(s) := \int_{-\infty}^\infty F_{0,n}(s +\sqrt{t} u) \frac{1}{\sqrt{\pi}} e^{-u^2}\ du[/math]
[math] 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]s[/math] equal to [math]\sigma[/math] or [math]1-\sigma[/math], we perform a contour shift on each integral (translating [math]u[/math] by [math]\sqrt{t} \alpha_n/2[/math] or [math]\sqrt{t} \beta_N/2[/math]) to obtain

[math] 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] 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]\alpha_n = \alpha_n(s), \beta_N = \beta_N(s)[/math] are complex parameters with imaginary part greater than [math]-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]F_{t,n}(s)[/math]

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

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


[math]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]H_{0,n}(s')[/math] as

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

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

and hence by Taylor's theorem with remainder

[math]\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]\frac{1}{T_0-0.66}[/math] by [math]\frac{1}{T_0 - 3.08}[/math]) that

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

[math]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]


[math]\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]|\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]\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]e^x - 1 \leq x e^x[/math], we conclude that

[math]\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]\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]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]t \leq 0.5[/math] we have

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

and similarly

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

[math] \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]


[math] \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]O( \log^2 T )[/math]) and should be easy to control numerically. One can bound it somewhat crudely by

[math] \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] \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]T[/math] moderately large) yields the bound

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

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

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

to obtain

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

[math]|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]


[math] 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] |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] \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] =\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]


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

and thus

[math]|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]w[/math] is the weight

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

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

[math]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]K[/math], where

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

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

[math]|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]


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

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

[math]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]p=\pm 1/2[/math]); since [math]p[/math] ranges between -1 and 1, it is not difficult to establish the bound

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

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

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

[math]|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] |C_1(p)| = \frac{9^\sigma}{\sqrt{2} \pi} \frac{\Gamma(1/2)}{2} \leq 0.200 \times 9^\sigma[/math]
[math] |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]|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]\sigma \lt 0[/math] the formulae are more complicated, mainly due to the condition [math]K+\sigma \geq 2[/math] required in Theorem 4.2 of [A2011]. We thus set [math] K:= \lfloor -\sigma \rfloor + 3[/math]. From equations (4.1), (4.2), (4.7) of [A2011] we now have the bounds

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



One can check that this implies also that

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

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

[math] |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]|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]


[math]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]\sigma \geq 0[/math] and

[math]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]\sigma \lt 0[/math].

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

Final bound

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

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

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

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

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

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

[math] 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] 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]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] 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] 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] 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] \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] 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]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]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]a_0 := \sqrt{\frac{T'_0}{2\pi}}[/math]
[math] \alpha_1(s) := \frac{1}{2s} + \frac{1}{s-1} + \frac{1}{2} \log \frac{s}{2\pi} [/math]
[math] N := \lfloor \sqrt{ \frac{T'}{2\pi}} \rfloor[/math]
[math] T := \frac{x}{2} [/math]
[math] T' := T + \frac{\pi t}{8} [/math]
[math] T'_0 := T_0 + \frac{\pi t}{8} [/math]