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-1} 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}} = 1.001869\dots }[/math], and hence we have

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

whenever (2.1) 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_n(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_n }[/math] is the function

[math]\displaystyle{ G_n(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'_n}{G_n}(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 = \alpha_n = \alpha_n(s) := \frac{G'_n}{G_n}(s) }[/math]

[math]\displaystyle{ = \frac{1}{2s} + \frac{1}{s-1} + \frac{1}{2} \log \frac{s}{2\pi n^2}. }[/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_n }[/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_n^2 ) \int_{-\infty}^\infty \exp(- \sqrt{t} \alpha_n u) G_n( s + \sqrt{t} u + \frac{t}{2} \alpha_n) (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'_n}{G_n})'(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_n }[/math] (so has imaginary part at least [math]\displaystyle{ T - 0.08 }[/math]) one has

[math]\displaystyle{ |(\frac{G'_n}{G_n})'(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_n( s + \sqrt{t} u + \frac{t}{2} \alpha_n ) = \log G_n(s) + (\sqrt{t} u + \frac{t}{2} \alpha_n) \alpha_n + O_{\leq}( \frac{1}{4 (T-3.08)} |\sqrt{t} u + \frac{t}{2} \alpha_n|^2 ) }[/math]

and thus

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

Since [math]\displaystyle{ \frac{1}{\sqrt{\pi}} e^{-u^2} }[/math] integrates to one, this implies that

[math]\displaystyle{ F_{t,n}(s) = \exp( \frac{t}{4} \alpha_n^2 ) G_n(s) (1 + O_{\leq}(\epsilon_n)) \quad (3.6) }[/math]

where

[math]\displaystyle{ \epsilon_n := \frac{1}{12(T - 0.254)} \int_{-\infty}^\infty \exp( \frac{1}{4 (T-3.08)} (\sqrt{t} |u| + \frac{t}{2} |\alpha_n|)^2 ) \frac{1}{\sqrt{\pi}} e^{-u^2}\ du }[/math]

[math]\displaystyle{ + \int_{-\infty}^\infty (\exp( \frac{1}{4 (T-3.08)} (\sqrt{t} |u| + \frac{t}{2} |\alpha_n|)^2 )-1) \frac{1}{\sqrt{\pi}} e^{-u^2}\ du. }[/math]

By the mean value theorem we have [math]\displaystyle{ \exp(x) - 1 \leq x \exp(x ) }[/math] for non-negative [math]\displaystyle{ x }[/math], thus

[math]\displaystyle{ \epsilon_n \leq \frac{1}{4 \sqrt{\pi} (T-3.08)} \int_{-\infty}^\infty \exp( \frac{1}{4 (T-3.08)} (\sqrt{t} |u| + \frac{t}{2} |\alpha_n|)^2 - u^2) ( (\sqrt{t} |u| + \frac{t}{2} |\alpha_n|)^2 + \frac{1}{3} )\ du. }[/math]

The integrand is even, so we may write the integral as twice the integral on the positive axis and then bound the resulting integral by the entire real axis again, hence

[math]\displaystyle{ \epsilon_n \leq \frac{1}{2 \sqrt{\pi} (T-3.08)} \int_{-\infty}^\infty \exp( \frac{1}{4 (T-3.08)} (\sqrt{t} u + \frac{t}{2} |\alpha|)^2 - u^2) ( (\sqrt{t} u + \frac{t}{2} |\alpha_n|)^2 + \frac{1}{3} )\ du. }[/math]

At this point we do some slightly crude calculations to simplify the expression. We may bound [math]\displaystyle{ (\sqrt{t} u + \frac{t}{2} |\alpha_n|)^2 \leq 2t u^2 + \frac{t^2}{2} |\alpha_n|^2 }[/math] hence

[math]\displaystyle{ \epsilon_n \leq \frac{\exp( \frac{t^2}{8 (T-3.08)} |\alpha_n|^2) }{2 \sqrt{\pi} (T-3.08)} \int_{-\infty}^\infty \exp( - (1 - \frac{t}{2 (T-3.08)}) u^2 ) ( 2tu^2 + \frac{t^2}{2} |\alpha_n|^2 + \frac{1}{3} )\ du. }[/math]

We can evaluate this exactly to obtain

[math]\displaystyle{ \epsilon_n \leq \frac{\exp( \frac{t^2}{8 (T-3.08)} |\alpha_n|^2) }{2(T-3.08)} ( \frac{\frac{t^2}{2} |\alpha_n|^2 + \frac{1}{3}}{(1 - \frac{t}{2 (T-3.08)})^{1/2}} + \frac{2t}{(1 - \frac{t}{2 (T-3.08)})^{3/2}} ). }[/math]

We raise the 1/2 exponent to 3/2:

[math]\displaystyle{ \epsilon_n \leq \frac{\exp( \frac{t^2}{8 (T-3.08)} |\alpha_n|^2) }{2(T-3.08)} \frac{\frac{t^2}{2} |\alpha_n|^2 + 2t + \frac{1}{3}}{(1 - \frac{t}{2 (T-3.08)})^{3/2}}. }[/math]

Since

[math]\displaystyle{ (1 - \frac{t}{2 (T-3.08)})^{3/2} \geq 1 - \frac{3}{2} \frac{t}{2 (T-3.08)} }[/math]

we have

[math]\displaystyle{ 2(T-3.08) (1 - \frac{t}{2 (T-3.08)})^{3/2} \geq 2 (T - 3.83) }[/math]

and thus

[math]\displaystyle{ \epsilon \leq \frac{\exp( \frac{t^2}{8 (T-3.08)} |\alpha_n|^2)}{T-3.83} (\frac{t^2}{4} |\alpha_n|^2 + t + \frac{1}{6}). }[/math]

For instance, if [math]\displaystyle{ t \leq 0.4 }[/math], then

[math]\displaystyle{ \epsilon \leq \frac{\exp( \frac{|\alpha_n|^2}{20 (T-3.08)} )}{T-3.83} (\frac{|\alpha_n|^2}{25} + \frac{17}{30}). }[/math]

Roughly speaking this error has shape [math]\displaystyle{ O( |\alpha_n|^2 / T ) = O( \frac{\log^2 \frac{T}{n^2}}{T} ) }[/math] in practice.

Summing, and using [math]\displaystyle{ G_n(s) = G_1(s) / n^s }[/math], we have

[math]\displaystyle{ \sum_{n=1}^N F_{t,n}(s) = G_1(s) \sum_{n=1}^N \frac{\exp( \frac{t}{4} \alpha_n(s)^2 )}{n^s} + O_{\leq}( \frac{|G_1(s)|}{T-3.83} \sum_{n=1}^N \frac{\exp( \frac{t}{4} \mathrm{Re}(\alpha_n(s)^2) + \frac{|\alpha_n(s)|^2}{20(T-3.08)}) )}{n^\sigma} (\frac{|\alpha_n(s)|^2}{25} + \frac{17}{30}) ) ). \quad (3.7) }[/math]

We expect the main term to be comparable to [math]\displaystyle{ G_1(s) \exp( \frac{t}{4} \alpha_1(s)^2 ) }[/math], while the error term should have size about [math]\displaystyle{ O( |G_1(s)| T^{-1} N^{1-\sigma} ) = O( |G_1(s)| T^{-\frac{1 + \sigma}{2}} ) }[/math] (in the worst-case scenario in which there is no cancellation in the sum), which should be negligible.

Similar considerations hold for [math]\displaystyle{ \sum_{m=1}^M F_{t,m}(1-s) }[/math] (note one can write [math]\displaystyle{ F_{t,m}(1-s) }[/math] as [math]\displaystyle{ \overline{F_{t,m}\overline{1-s})} }[/math] to conjugate [math]\displaystyle{ 1-s }[/math] back into the upper half-plane).

Controlling [math]\displaystyle{ R_{t,N,M}(s) }[/math]

Suppose [math]\displaystyle{ s = \sigma+iT }[/math] with [math]\displaystyle{ T \gt 10 }[/math]. From equations of (1.1), (3.1), (3.3), (3.4) and (3.11) of [A2011] we have

[math]\displaystyle{ \xi(s) = \pi^{-s/2} \Gamma(s/2) {\mathcal R}(s) + \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} + (-1)^{N-1} U a^{-\sigma} (\sum_{k=0}^K \frac{C_k(p)}{a^k} + RS_K) }[/math]

[math]\displaystyle{ a := \sqrt{T/2\pi} }[/math]

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

[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]

[math]\displaystyle{ K }[/math] is any natural number, and [math]\displaystyle{ C_k(p), RS_k }[/math] are certain complex numbers. In particular

[math]\displaystyle{ {\mathcal R}(s) := \sum_{n=1}^N \frac{1}{n^s} + O_{\leq}( a^{-\sigma} (\sum_{k=0}^K \frac{|C_k(p)|}{a^k} + |RS_K|) ). \quad (3.1) }[/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 \frac{\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 (3.1), thus

[math]\displaystyle{ {\mathcal R}(s) := \sum_{n=1}^N \frac{1}{n^s} + O_{\leq}( a^{-\sigma} (\frac{1}{2} + \frac{|C_1(p)|}{a} + |RS_1|) ). }[/math]

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

[math]\displaystyle{ |C_1(p)| \leq \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{ {\mathcal R}(s) := \sum_{n=1}^N \frac{1}{n^s} + O_{\leq}( \frac{1}{2} a^{-\sigma} (1 + 0.400 \frac{9^\sigma}{a} + 0.346 \frac{2^{3\sigma/2}}{a^2} ) ) \quad (3.2). }[/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]

for [math]\displaystyle{ 1 \leq k \leq K }[/math], where [math]\displaystyle{ b := \sqrt{(3 - 2\log 2)\pi} }[/math]. One can check that this implies also that

[math]\displaystyle{ \frac{|C_k(p)|}{a^k} \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{ {\mathcal R}(s) := \sum_{n=1}^N \frac{1}{n^s} + O_{\leq}( \frac{1}{2} a^{-\sigma} (1 + (9/10)^{\lceil -\sigma\rceil} \sum_{1 \leq k \leq 4-\sigma} \frac{\Gamma(k/2)}{(\frac{10}{11} a)^{k}} ) ). }[/math]

In the range [math]\displaystyle{ 0 \leq \sigma \leq 1/2 }[/math] we have the bound (3.2). Comparing terms we can combine the two estimates to obtain the slightly weaker bound

[math]\displaystyle{ {\mathcal R}(s) := \sum_{n=1}^N \frac{1}{n^s} + O_{\leq}( \frac{1}{2} a^{-\sigma} (1 + (9/10)^{\lceil -\sigma\rceil} \sum_{1 \leq k \leq 4-\sigma} (1.2)^k \frac{\Gamma(k/2)}{a^{k}} ) ) }[/math]

valid for [math]\displaystyle{ \sigma \leq 1/2 }[/math]. In particular, for [math]\displaystyle{ \sigma \geq 1/2 }[/math] we now have the bound

[math]\displaystyle{ |R_{0,N,N}(\sigma+iT)| \leq \frac{1}{2} \pi^{-\sigma/2} |\Gamma(\frac{\sigma+iT}{2})| a^{-\sigma} (1 + 0.400 \frac{9^\sigma}{a} + 0.346 \frac{2^{3\sigma/2}}{a^2} ) }[/math]

[math]\displaystyle{ + \frac{1}{2} \pi^{-(1-\sigma)/2} |\Gamma(\frac{1-\sigma+iT}{2})| a^{-(1-\sigma)} (1 + (9/10)^{\lceil \sigma\rceil - 1} \sum_{1 \leq k \leq 5+\sigma} (1.2)^k \frac{\Gamma(k/2)}{a^{k}} ) ) \quad (3.3). }[/math]

For [math]\displaystyle{ \sigma \leq 1/2 }[/math] we can use the functional equation [math]\displaystyle{ F_{0,N,N}(\sigma+iT) = \overline{F_{0,N,N}(1-\sigma+iT)} }[/math].

Now let us control [math]\displaystyle{ R_{t,N,N}(\sigma_0+iT) }[/math] where [math]\displaystyle{ \sigma_0 = \frac{1+y}{2} }[/math] (thus if [math]\displaystyle{ y=0.4 }[/math] then [math]\displaystyle{ \sigma_0 = 0.7 }[/math]). From (1.8) and a change of variables we have

[math]\displaystyle{ |R_{t,N,N}(\sigma_0+iT)| \leq \int_{-\infty}^\infty \frac{1}{\sqrt{\pi t}} e^{-(\sigma-\sigma_0)^2/t} |R_{0,N,N}(\sigma+iT)|\ d\sigma. }[/math]

From the functional equation we thus have

[math]\displaystyle{ |R_{t,N,N}(\sigma_0+iT)| \leq \int_{1/2}^\infty \frac{1}{\sqrt{\pi t}} (e^{-(\sigma-\sigma_0)^2/t}+e^{-(1-\sigma-\sigma_0)^2/t}) |R_{0,N,N}(\sigma+iT)|\ d\sigma. }[/math]

Since [math]\displaystyle{ \sigma\lt /math will be closer to \lt math\gt \sigma_0 }[/math] than [math]\displaystyle{ 1-\sigma }[/math], we conclude

[math]\displaystyle{ |R_{t,N,N}(\sigma_0+iT)| \leq 2 \int_{1/2}^\infty \frac{1}{\sqrt{\pi t}} e^{-(\sigma-\sigma_0)^2/t} |R_{0,N,N}(\sigma+iT)|\ d\sigma. }[/math]

Inserting (3.3), we conclude that

[math]\displaystyle{ |R_{t,N,N}(\sigma_0+iT)| \leq \int_{1/2}^\infty \frac{1}{\sqrt{\pi t}} e^{-(\sigma-\sigma_0)^2/t} \frac{1}{2} \pi^{-\sigma/2} |\Gamma(\frac{\sigma+iT}{2})| a^{-\sigma} (1 + 0.400 \frac{9^\sigma}{a} + 0.346 \frac{2^{3\sigma/2}}{a^2} )\ d\sigma }[/math]

[math]\displaystyle{ + \int_{1/2}^\infty \frac{1}{\sqrt{\pi t}} e^{-(\sigma-\sigma_0)^2/t} \pi^{-(1-\sigma)/2} |\Gamma(\frac{1-\sigma+iT}{2})| a^{-(1-\sigma)} (1 + (9/10)^{\lceil \sigma\rceil - 1} \sum_{1 \leq k \leq 5+\sigma} (1.2)^k \frac{\Gamma(k/2)}{a^{k}} ) ). }[/math]

From (2.2) one has

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

[math]\displaystyle{ \leq (1 + \frac{1}{12 (T-0.246)}) \sqrt{2\pi} e^{\pi T/4} \exp( \frac{\sigma-1}{4} \log \frac{\sigma^2+T^2}{4} + \frac{T}{2} \arctan \frac{\sigma}{T} - \frac{\sigma}{2} ) }[/math]

... need to use Stirling approximation to control the Gamma factors, so that one is integrating gaussians in sigma; then interchange sum and integral