Effective bounds on H t - second approach

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 (1.1) }[/math]

we conclude that

[math]\displaystyle{ H_t(z) = \frac{1}{8} \xi_t(\frac{1+iz}{2}) \quad (1.2) }[/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]

The Riemann-Siegel formula gives, for any [math]\displaystyle{ N,M }[/math],

[math]\displaystyle{ \xi(s) = \sum_{n=1}^N F_{0,n}(s) + \sum_{m=1}^M F_{0,m}(1-s) + R_{0,N,M}(s) }[/math]

where

[math]\displaystyle{ F_{0,n}(s) := \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \frac{1}{n^s} \quad (1.3) }[/math]

and

[math]\displaystyle{ R_{0,N,M}(s) := \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \frac{e^{-i\pi s} \Gamma(1-s)}{2\pi i} \int_{C_M} \frac{w^s e^{-Nw}}{e^w-1}\ dw. \quad (1.4) }[/math]

By linearity of the heat equation, we thus have

[math]\displaystyle{ \xi_t(s) = \sum_{n=1}^N F_{t,n}(s) + \sum_{m=1}^M F_{t,m}(1-s) + R_{t,N,M}(s) \quad (1.5) }[/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 \quad (1.6) }[/math]

and

[math]\displaystyle{ R_{t,N,M}(s) = \int_{-\infty}^\infty R_{0,N,M}( s + \sqrt{t} u ) \frac{1}{\sqrt{\pi}} e^{-u^2}\ du. \quad (1.7) }[/math]

We will control (1.7) using the triangle inequality:

[math]\displaystyle{ |R_{t,N,M}(s)| \leq \int_{-\infty}^\infty |R_{0,N,M}( s + \sqrt{t} u )| \frac{1}{\sqrt{\pi}} e^{-u^2}\ du. \quad (1.8) }[/math]

For (1.6) we will be a little bit sneakier. Let [math]\displaystyle{ \alpha = alpha_n(s) }[/math] be a complex number to be chosen later, then by shifting [math]\displaystyle{ u }[/math] by [math]\displaystyle{ \sqrt{t} \alpha/2 }[/math] we see that

[math]\displaystyle{ F_{t,n}(s) = \exp( -\frac{t}{4} \alpha^2 ) \int_{-\infty}^\infty \exp(- \sqrt{t} \alpha u) F_{0,n}( s + \sqrt{t} u + \frac{t}{2} \alpha ) \frac{1}{\sqrt{\pi}} e^{-u^2}\ du. \quad (1.9) }[/math]

We will choose [math]\displaystyle{ \alpha }[/math] in order to remove most of the oscillation in [math]\displaystyle{ u }[/math] from the [math]\displaystyle{ F_{0,n} }[/math] term, so that the triangle inequality may be efficiently applied.

Explicit error terms in the Gamma function

From equations (1.13), (3.1), (3.14) and (3.15) of [B1994], we see that for [math]\displaystyle{ s }[/math] in the upper half-plane, we have the Stirling approximation

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

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

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

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

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

for [math]\displaystyle{ \mathrm{Re}(s) \leq 0 }[/math], where [math]\displaystyle{ C_2 := \frac{1}{2}(1 + \zeta(2)) = \frac{1}{2}(1 + \frac{\pi^2}{6}) }[/math]. Assuming for instance that

[math]\displaystyle{ \mathrm{Im}(s) \geq 1\quad (2.1) }[/math]

we have [math]\displaystyle{ \frac{1}{|1-e^{-2\pi i s}|} \leq \frac{1}{1-e^{-2\pi}} }[/math], and hence we have

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

whenever holds. In particular we have

[math]\displaystyle{ \Gamma(s) = \sqrt{2\pi} \exp( (s-\frac{1}{2}) \log s - s ) (1 + O_{\leq}( \frac{1}{12|s|} + \frac{0.0205}{|s|^2} ) ) }[/math]

[math]\displaystyle{ \Gamma(s) = \sqrt{2\pi} \exp( (s-\frac{1}{2}) \log s - s ) (1 + O_{\leq}( \frac{1}{12 (|s| - 0.246)} ). \quad (2.2) }[/math]

Controlling [math]\displaystyle{ F_{0,n}(s) }[/math]

Let [math]\displaystyle{ s = \sigma+iT }[/math] with [math]\displaystyle{ T \geq 10 }[/math] (say). From (1.3), (2.2), we have

[math]\displaystyle{ F_{0,n}(s') = G(s') (1 + O_{\leq}( \frac{1}{12(|s'|-0.246)} ) ) \quad (3.1) }[/math]

whenever [math]\displaystyle{ \mathrm{Im}(s') \geq 1 }[/math], where [math]\displaystyle{ G = G_n }[/math] is the function

[math]\displaystyle{ G(s') := \frac{s' (s'-1)}{2} \frac{1}{(\pi n^2)^{s'/2}} \sqrt{2\pi} \exp( (s' - \frac{1}{2}) \log s' - s' ). \quad (3.2) }[/math]

We compute the log-derivative of [math]\displaystyle{ G }[/math] as

[math]\displaystyle{ \frac{G'}{G}(s') = \frac{1}{2s'} + \frac{1}{s'-1} + \frac{1}{2} \log \frac{s'}{2\pi n^2}. \quad (3.3) }[/math]

We apply (1.9) with

[math]\displaystyle{ \alpha := \frac{G'}{G}(s). }[/math]

We observe that the imaginary part of [math]\displaystyle{ \alpha }[/math] is at most [math]\displaystyle{ -0.15 }[/math], so that [math]\displaystyle{ s + \sqrt{t} u + \frac{t}{2} \alpha }[/math] has imaginary part at least [math]\displaystyle{ T-0.08 }[/math] if [math]\displaystyle{ t \leq 1 }[/math], which we now assume. Thus by (3.1) we have

[math]\displaystyle{ F_{t,n}(s) = \exp( -\frac{t}{4} \alpha^2 ) \int_{-\infty}^\infty \exp(- \sqrt{t} \alpha u) G( s + \sqrt{t} u + \frac{t}{2} \alpha ) (1 + O_{\leq}(\frac{1}{12(T - 0.254)})) \frac{1}{\sqrt{\pi}} e^{-u^2}\ du. \quad (3.4) }[/math]

From (3.3) we have

[math]\displaystyle{ (\frac{G'}{G})'(s') = -\frac{1}{2(s')^2} + \frac{1}{(s'-1)^2} + \frac{1}{2 s'} }[/math]

and in particular whenever [math]\displaystyle{ s' }[/math] lies between [math]\displaystyle{ s }[/math] and [math]\displaystyle{ s + \sqrt{t} u + \frac{t}{2} \alpha }[/math] (so has imaginary part at least [math]\displaystyle{ T - 0.08 }[/math]) one has

[math]\displaystyle{ |(\frac{G'}{G})'(s')| \leq \frac{1}{2 (T - 0.08)^2} + \frac{1}{(T - 0.08)^2} + \frac{1}{2 (T - 0.08)} }[/math]

[math]\displaystyle{ = \frac{1}{2 (T - 0.08)} (1 + \frac{3}{T-0.08}) }[/math]

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

Thus by Taylor's theorem with remainder we have

[math]\displaystyle{ \log G( s + \sqrt{t} u + \frac{t}{2} \alpha ) = \log G(s) + (\sqrt{t} u + \frac{t}{2} \alpha) \alpha + O_{\leq}( \frac{1}{4 (T-3.08)} |\sqrt{t} u + \frac{t}{2} \alpha|^2 ) }[/math]

and thus

[math]\displaystyle{ F_{t,n}(s) = \exp( \frac{t}{4} \alpha^2 ) G(s) \int_{-\infty}^\infty \exp( O_{\leq}( \frac{1}{4 (T-3.08)} |\sqrt{t} u + \frac{t}{2} \alpha|^2 ) ) (1 + O_{\leq}(\frac{1}{12(T - 0.254)})) \frac{1}{\sqrt{\pi}} e^{-u^2}\ du. \quad (3.5) }[/math]