Asymptotics of H t

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The Gamma function

The Gamma function is defined for [math]\mathrm{Re}(s) \gt 0[/math] by the formula

[math]\Gamma(s) = \int_0^\infty x^s e^{-x} \frac{dx}{x} [/math]

and hence by change of variables

[math]\Gamma(s) = \int_{-\infty}^\infty \exp( s u - e^u )\ du. \quad (1.1)[/math]

It can be extended to other values of [math]s[/math] by analytic continuation or by contour shifting; for instance, if [math]Im(s)\gt0[/math], one can write

[math]\Gamma(s) = \int_C \exp( s u - e^u )\ du \quad (1.1')[/math]

where [math]C[/math] is a contour from [math]+i\infty[/math] to [math]\infty[/math] that stays within a bounded distance of the upper imaginary and right real axes.

The Gamma function obeys the Euler reflection formula

[math]\Gamma(s) \Gamma(1-s) = \frac{\pi}{\sin(\pi s)} \quad (1.2)[/math]

and the duplication formula

[math]\Gamma(1-s) = \frac{\Gamma(\frac{1-s}{2}) \Gamma(1-\frac{s}{2})}{\pi^{1/2} 2^s}. \quad (1.3)[/math]

In particular one has

[math]\Gamma(\frac{s}{2}) \Gamma(1-\frac{s}{2}) = \frac{\pi}{\sin(\pi s/2)} \quad (1.4)[/math]

and thus on combining (3) and (4)

[math] \Gamma(s/2) \Gamma(1-s) = \frac{\pi^{1/2}}{2^s \sin(\pi s/2)} \Gamma(\frac{1-s}{2}) \quad(1.5)[/math]

Since [math]s \Gamma(s) = \Gamma(s+1)[/math], we have

[math]\frac{s(s-1)}{2} \Gamma(\frac{s}{2}) = 2 \Gamma(\frac{s+4}{2}) - 3 \Gamma(\frac{s+2}{2}). \quad (1.6)[/math]

We have the Stirling approximation

[math]\Gamma(s) = \sqrt{2\pi/s} \exp( s \log s - s + O(1/|s|) )[/math]

whenever [math]\mathrm{Re}(s) \gg 1 [/math]. If we have [math]s = \sigma+iT[/math] for some large [math]T[/math] and bounded [math] \sigma \gg 1 [/math], this gives

[math]\Gamma(s) \approx \sqrt{2\pi} T^{\sigma -1/2} e^{-\pi T/2} \exp(i (T \log T - T + \pi \sigma/2 - \pi/4)). (1.7)[/math]

Another crude but useful approximation is

[math]\Gamma(s+h) \approx \Gamma(s) s^h (1.8) [/math]

for [math]s[/math] as above and [math]h=O(1)[/math].

The Riemann-Siegel formula for [math]t=0[/math]

Proposition 1 (Riemann-Siegel formula for [math]t=0[/math]) For any natural numbers [math]N,M[/math] and complex number [math]s[/math] that is not an integer, we have

[math]\zeta(s) = \sum_{n=1}^N \frac{1}{n^s} + \pi^{s-\frac{1}{2}} \frac{\Gamma((1-s)/2)}{\Gamma(s/2)} \sum_{m=1}^M \frac{1}{m^{1-s}} + \frac{e^{-i\pi s} \Gamma(1-s)}{2\pi i} \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1}\ dw[/math]

where [math]w^{s-1} := \exp((s-1) \log w)[/math] and we use the branch of the logarithm with imaginary part in [math][0,2\pi)[/math], and [math]C_M[/math] is any contour from [math]+\infty[/math] to [math]+\infty[/math] going once anticlockwise around the zeroes [math]2\pi i m[/math] of [math]e^w-1[/math] with [math]|m| \leq M[/math], but does not go around any other zeroes.

Proof This equation is in [T1986, p. 82], but we give a proof here. The right-hand side is meromorphic in [math]s[/math], so it will suffice to establish that

  1. The right-hand side is independent of [math]N[/math];
  2. The right-hand side is independent of [math]M[/math];
  3. Whenever [math]\mathrm{Re}(s)\gt1[/math] and [math]s[/math] is not an integer, the right-hand side converges to [math]\zeta(s)[/math] if [math]M=0[/math] and [math]N \to \infty[/math].

We begin with the first claim. It suffices to show that the right-hand sides for [math]N[/math] and [math]N-1[/math] agree for every [math]N \gt 1[/math]. Subtracting, it suffices to show that

[math]0 = \frac{1}{N^s} + \frac{e^{-i\pi s} \Gamma(1-s)}{2\pi i} \int_{C_M} \frac{w^{s-1} (e^{-Nw} - e^{-(N-1)w})}{e^w-1}\ dw.[/math]

The integrand here simplifies to [math]- w^{s-1} e^{-Nw}[/math], which on shrinking [math]C_M[/math] to wrap around the positive real axis becomes [math]N^{-s} \Gamma(s) (1 - e^{2\pi i(s-1)})[/math]. The claim then follows from the Euler reflection formula [math]\Gamma(s) \Gamma(1-s) = \frac{\pi}{\sin(\pi s)}[/math].

Now we verify the second claim. It suffices to show that the right-hand sides for [math]M[/math] and [math]M-1[/math] agree for every [math]M \gt 1[/math]. Subtracting, it suffices to show that

[math]0 = \pi^{s-\frac{1}{2}} \frac{\Gamma((1-s)/2)}{\Gamma(s/2)} \frac{1}{M^{1-s}} + \frac{e^{-i\pi s} \Gamma(1-s)}{2\pi i} \int_{C_M - C_{M-1}} \frac{w^{s-1} e^{-Nw}}{e^w-1}\ dw.[/math]

The contour [math]C_M - C_{M-1}[/math] encloses the simple poles at [math]+2\pi i M[/math] and [math]-2\pi i M[/math], which have residues of [math](2\pi i M)^{s-1} = - i (2\pi M)^{s-1} e^{\pi i s/2}[/math] and [math](-2\pi i M)^{s-1} = i (2\pi M)^{s-1} e^{3\pi i s/2}[/math] respectively. So, on canceling the factor of [math]M^{s-1}[/math] it suffices to show that

[math]0 = \pi^{s-\frac{1}{2}} \frac{\Gamma((1-s)/2)}{\Gamma(s/2)} + e^{-i\pi s} \Gamma(1-s) (2\pi)^{s-1} i (e^{3\pi i s/2} - e^{\pi i s/2}).[/math]

But this follows from the duplication formula [math]\Gamma(1-s) = \frac{\Gamma(\frac{1-s}{2}) \Gamma(1-\frac{s}{2})}{\pi^{1/2} 2^s}[/math] and the Euler reflection formula [math]\Gamma(\frac{s}{2}) \Gamma(1-\frac{s}{2}) = \frac{\pi}{\sin(\pi s/2)}[/math].

Finally we verify the third claim. Since [math]\zeta(s) = \lim_{N \to \infty} \sum_{n=1}^\infty \frac{1}{n^s}[/math], it suffices to show that

[math]\lim_{N \to \infty} \int_{C_0} \frac{w^{s-1} e^{-Nw}}{e^w-1}\ dw = 0.[/math]

We take [math]C_0[/math] to be a contour that traverses a [math]1/N[/math]-neighbourhood of the real axis. Writing [math]C_0 = \frac{1}{N} C'_0[/math], with [math]C'_0[/math] independent of [math]N[/math], we can thus write the left-hand side as

[math]\lim_{N \to \infty} N^{-s} \int_{C'_0} \frac{w^{s-1} e^{-w}}{e^{w/N}-1}\ dw,[/math]

and the claim follows from the dominated convergence theorem. [math]\Box[/math]

Applying the Riemann-Siegel formula to the Riemann xi function [math]\xi(s) = \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \zeta(s)[/math], we have

[math]\xi(s) = F_{0,N}(s) + \overline{F_{0,M}(\overline{1-s})} + R_{0,N,M}(s) \quad(2.1)[/math]


[math]F_{0,N}(s) := \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \sum_{n=1}^N \frac{1}{n^s} \quad(2.2)[/math]


[math]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(2.3)[/math]

A contour integral

Lemma 2 Let [math]L[/math] be a line in the direction [math]\mathrm{arg} w = \pi/4[/math] passing between [math]0[/math] and [math]2\pi i [/math]. Then for any complex [math]\alpha[/math], the contour integral

[math]\Psi(\alpha) := \int_L \frac{\exp( \frac{i}{4\pi} z^2 + \alpha z)}{e^z - 1}\ dz[/math]

can be given explicitly by the formula

[math]\Psi(\alpha) = 2\pi \frac{\cos \pi(\frac{1}{2} \alpha^2 - \alpha - \frac{\pi}{8})}{\cos(\pi \alpha)} \exp( \frac{i \pi}{2} \alpha^2 - \frac{5 \pi}{8} )[/math].

Proof The integrand has a residue of [math]1[/math] at [math]0[/math], hence on shifting the contour downward by [math]2\pi i[/math] we have

[math]\Psi(\alpha) = -2\pi i + \int_L \frac{\exp( \frac{i}{4\pi} (z-2\pi i)^2 + \alpha (z-2\pi i) )}{e^z-1}\ dz.[/math]

The right-hand side expands as

[math]-2\pi i - e^{-2\pi i \alpha} \int_L \frac{\exp( \frac{i}{4\pi} z^2 + (\alpha+1) z)}{e^z-1}\ dz[/math]

which we can write as

[math]-2\pi i - e^{-2\pi i \alpha} (\Psi(\alpha) + \int_L \exp( \frac{i}{4\pi} z^2 + \alpha z\ dz).[/math]

The last integral is a standard gaussian integral, which can be evaluated as [math]-\sqrt{\frac{\pi}{i/4\pi}} \exp( \pi i \alpha^2)[/math]. Hence

[math]\Psi(\alpha) = -2\pi i - e^{-2\pi i \alpha} (\Psi(\alpha) - \sqrt{\frac{\pi}{i/4\pi}} \exp( \pi i \alpha^2)),[/math]

and the claim then follows after some algebra. [math]\Box[/math]

We conclude from (2.3) that

[math]R_{0,N,M}(s) \approx - 2 \Gamma(\frac{5-s}{2}) \frac{\pi^{(-s-1)/2}}{2^s} e^{-\pi i s/2} \exp( -\frac{t \pi^2}{64} ) (2\pi i M)^{s-1} \Psi(\frac{s-2\pi i MN}{2\pi i M})[/math]
[math] = i \Gamma(\frac{5-s}{2}) \pi^{-(s+1)/2} \exp( -\frac{t \pi^2}{64} ) (\pi M)^{s-1} \Psi(\frac{s}{2\pi i M} - N).[/math]

Heuristic approximation at [math]t=0[/math]

To estimate the remainder term [math]R_{0,N,M}(s)[/math] in (2.3) with [math]M,N = \sqrt{\mathrm{Im}(s) / 2\pi} + O(1)[/math], we make the change of variables [math]w = z + 2\pi i M[/math] to obtain

[math] 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 - 2\pi i M} \frac{(z+2\pi i M)^{s-1} e^{-Nz}}{e^z-1}\ dz[/math]

Steepest descent heuristics suggest that the dominant portion of this integral comes when [math]z=O(1)[/math]. In this regime we may Taylor expand

[math] (z+2\pi i M)^{s-1} = (2\pi i M)^{s-1} \exp( (s-1) \log(1 + \frac{z}{2\pi i M}) ) [/math]
[math] \approx (2\pi i M)^{s-1} \exp( (s-1) \frac{z}{2\pi i M} -\frac{s-1}{2} (\frac{z}{2\pi i M})^2 )[/math]
[math] \approx (2\pi i M)^{s-1} \exp( s \frac{z}{2\pi i M} + \frac{i}{4\pi} z^2 );[/math]

using this approximation and then shifting the contour to [math]-L[/math] (cf. [T1986, Section 4.16], we conclude that

[math] R_{0,N,M}(s) \approx - \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \frac{e^{-i\pi s} \Gamma(1-s)}{2\pi i} (2\pi i M)^{s-1}\int_L \frac{\exp( (\frac{s}{2\pi i M}-N)z + \frac{i}{4\pi} z^2 )}{e^z-1}\ dz[/math]

and hence by Lemma 2

[math] R_{0,N,M}(s) \approx - \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \frac{e^{-i\pi s} \Gamma(1-s)}{2\pi i} (2\pi i M)^{s-1}\Psi(\frac{s}{2\pi i M}-N). (4.1)[/math]

Using (1.7) one can calculate that this expression has magnitude [math]O( x^{6/4} e^{-\pi x/8} )[/math].

If we drop the [math]R_{0,N,M}[/math] term, we have

[math]H_0(x+iy) \approx \frac{1}{8} F_{0,N}(\frac{1+ix-y}{2}) + \frac{1}{8} \overline{F_{0,M}(\frac{1+ix+y}{2})}.[/math]

From (2.2) and (1.7) we have

[math]|\frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2)| \asymp x^{(7-y)/4} e^{-\pi x/8} [/math]

when [math]s = (1+ix-y)/2[/math] and

[math]|\frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2)| \asymp x^{(7+y)/4} e^{-\pi x/8} [/math]

when [math]s = (1+ix+y)/2[/math]. Thus we expect the second term to dominate, and typically we would expect

[math]|H_0(x+iy)| \asymp x^{(7+y)/4} e^{-\pi x/8}.[/math]

Extending the Riemann-Siegel formula to positive [math]t[/math]

Evolving [math]H_0(z) = \frac{1}{8} \xi(\frac{1+iz}{2})[/math] by the backwards heat equation [math]\partial_t H_t(z) = -\partial_{zz} H_t(z)[/math] is equivalent to evolving the Riemann [math]\xi[/math] function [math]\xi = \xi_0[/math] by the forwards heat equation [math]\partial_t \xi_t(s) = \frac{1}{4} \partial_{ss} \xi_t(s)[/math], and then setting

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

One way to do this is to expand [math]\xi_0(s)[/math] as a linear combination of exponentials [math]e^{\alpha s}[/math], and replace each such exponential by [math] \exp( \frac{t}{4} \alpha^2 ) e^{\alpha s}[/math] to obtain [math]\xi_t[/math]. Roughly speaking, this can be justified as long as everything is absolutely convergent.

In view of (2.1), we will have

[math]\xi_t(s) = F_{t,N}(s) + \overline{F_{t,M}(\overline{1-s})} + R_{t,N,M}(s) \quad(5.1)[/math]

where [math]F_{t,N}, R_{t,N,M}[/math] are the heat flow evolutions of [math]F_{0,N}, R_{0,N,M}[/math] respectively.

It is easy to evolve [math]F_{t,N}(s)[/math]. Firstly, from (1.6) one has

[math]F_{0,N}(s) = \sum_{n=1}^N 2 \frac{\Gamma(\frac{s+4}{2})}{(\pi n^2)^{s/2}} - 3 2 \frac{\Gamma(\frac{s+2}{2})}{(\pi n^2)^{s/2}}[/math]

and hence by (1.1')

[math]F_{0,N}(s) = \sum_{n=1}^N 2 \int_C \exp( \frac{s+4}{2} u - e^u - \frac{s}{2} \log(\pi n^2))\ du - 3 \int_C \exp( \frac{s+2}{2} u - e^u - \frac{s}{2} \log(\pi n^2) )\ du.[/math]

We can now evolve to obtain

[math]F_{t,N}(s) = \sum_{n=1}^N 2 \int_C \exp( \frac{s+4}{2} u - e^u - \frac{s}{2} \log(\pi n^2) + \frac{t}{16} (u - \log(\pi n^2))^2 )\ du - 3 \int_C \exp( \frac{s+2}{2} u - e^u - \frac{s}{2} \log(\pi n^2) + \frac{t}{16} (u - \log(\pi n^2))^2 )\ du (5.2).[/math]

By integrating on [math]C[/math] rather than the real axis, the integrals remain absolutely convergent here.

Evolving [math]R_{0,N,M}[/math] is a bit trickier. From (1.5) one has

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

which can be rewritten using (1.6) as

[math]2 \pi^{-s/2} \frac{e^{-i\pi s} \Gamma(\frac{5-s}{2})}{2^{s+1}\pi^{1/2} i \sin(\pi s/2)} \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1}\ dw[/math]
[math]-3 \pi^{-s/2} \frac{e^{-i\pi s} \Gamma(\frac{3-s}{2})}{2^{s+1}\pi^{1/2} i \sin(\pi s/2)} \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1}\ dw.[/math]

For [math]\mathrm{Im}(s) \gt 0[/math], we have the geometric series formula

[math]\frac{1}{\sin(\pi s/2)} = -2i e^{i\pi s/2} \sum_{n=0}^\infty e^{i \pi s n}[/math]

and from this and (1.1') we can rewrite [math]R_{0,N,M}(s)[/math] as

[math]2 \sum_{n=0}^\infty \pi^{-s/2} \frac{e^{-i\pi s/2} e^{i \pi s n}}{2^{s}\pi^{1/2} } \int_{\overline{C}} \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1} \exp( \frac{5-s}{2} u - e^u)\ dw\ du[/math]
[math]-3 \sum_{n=0}^\infty \pi^{-s/2} \frac{e^{-i\pi s/2} e^{i \pi s n}}{2^{s}\pi^{1/2}} \int_{\overline{C}} \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1} \exp( \frac{3-s}{2} u - e^u)\ dw\ du[/math]

where [math]\overline{C}[/math] is the complex conjugate of [math]C[/math]. Hence we can write [math]R_{t,N,M}(s)[/math] exactly as

[math]2 \sum_{n=0}^\infty \pi^{-s/2} \frac{e^{-i\pi s/2} e^{i \pi s n}}{2^{s}\pi^{1/2} \sin(\pi s/2)} \int_{\overline{C}}\int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1} \exp( \frac{5-s}{2} u - e^u + \frac{t}{4} (i \pi (n-1/2) + \log \frac{w}{2\sqrt{\pi}} - \frac{u}{2})^2 )\ dw\ du[/math]
[math]-3 \sum_{n=0}^\infty \pi^{-s/2} \frac{e^{-i\pi s/2} e^{i \pi s n}}{2^{s}\pi^{1/2} \sin(\pi s/2)} \int_{\overline{C}} \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1} \exp( \frac{3-s}{2} u - e^u + \frac{t}{4} (i \pi (n-1/2) + \log \frac{w}{2\sqrt{\pi}} - \frac{u}{2})^2 )\ dw\ du (5.3)[/math]

Approximation for [math]t\gt0[/math]

The above formulae are clearly unwieldy, so let us make a number of heuristic approximations to simplify them. We start with [math]F_{t,N}(s)[/math], assumig that the imaginary part of [math]s[/math] is large and positive and the real part is bounded. We first drop the second term of (5.2) as being lower order:

[math] F_{t,N}(s) \approx \sum_{n=1}^N 2 \int_C \exp( \frac{s+4}{2} u - e^u - \frac{s}{2} \log(\pi n^2) + \frac{t}{16} (u - \log(\pi n^2))^2 )\ du.[/math]

Next, we shift [math]u[/math] by [math]\log \frac{s+4}{2}[/math] to obtain

[math] F_{t,N}(s) \approx \sum_{n=1}^N \frac{2 \exp( \frac{s+4}{2} \log \frac{s+4}{2} - \frac{s+4}{2})}{(\pi n^2)^{s/2}} \int_C \exp( \frac{s+4}{2} (1 + u - e^u) + \frac{t}{16} (u + \log \frac{s+4}{2\pi n^2})^2 )\ du.[/math]

Because the expression [math]\exp( \frac{s+4}{4} (1+u-e^u) )[/math] decays rapidly away from [math]u=0[/math], we can heuristically approximate

[math] \frac{t}{16} (u + \log \frac{s+4}{2\pi n^2})^2 ) \approx \frac{t}{16} \log^2 \frac{s+4}{2\pi n^2} [/math]

and then we undo the shift to obtain

[math] F_{t,N}(s) \approx \sum_{n=1}^N \frac{2}{(\pi n^2)^{s/2}} \int_{-\infty}^\infty \exp( \frac{s+4}{2} u - e^u + \frac{t}{16} \log^2\frac{s+4}{2\pi n^2} )\ du[/math]

which by (1) becomes

[math] F_{t,N}(s) \approx \sum_{n=1}^N \frac{2}{(\pi n^2)^{s/2}} \Gamma(\frac{s+4}{2}) \exp( \frac{t}{16} \log^2\frac{s+4}{2\pi n^2} ).\quad (6.1)[/math]

Reinstating the lower order term and applying (1.6), we have an alternate form

[math] F_{t,N}(s) \approx \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \sum_{n=1}^N \frac{\exp( \frac{t}{16} \log^2\frac{s+4}{2\pi n^2})}{n^s}.\quad (6.2)[/math]

We can perform a similar analysis for [math]R_{t,N,M}[/math]. Again, we drop the second term as being lower order. The [math]w[/math] integrand [math]w^{s-1} e^{-Nw}[/math] attains a maximum at [math]w = \frac{s}{N} \approx \sqrt{2\pi \mathrm{Im}(s)} i[/math] and the [math]u[/math] integrand [math]\exp( \frac{s+4}{2} u - e^u )[/math] attains a maximum at [math]u = \log \frac{s+4}{2} \approx \log \frac{\mathrm{Im}(s)}{2} + i \frac{\pi}{2}[/math], and hence

[math]\log \frac{w}{2\sqrt{\pi}} - \frac{u}{2} \approx i \pi/4[/math]

and so we may heuristically obtain

[math]2 \sum_{n=0}^\infty \pi^{-s/2} \frac{e^{-i\pi s/2} e^{i \pi s n}}{2^{s}\pi^{1/2} \sin(\pi s/2)} \int_{-\infty}^\infty \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1} \exp( \frac{5-s}{2} u - e^u + \frac{t\pi^2}{64} (4n-1) )\ dw\ du.[/math]

Because [math]e^{i \pi sn}[/math] decays incredibly rapidly in [math]n[/math], the [math]n=0[/math] term should dominate, thus giving

[math]2 \pi^{-s/2} \frac{e^{-i\pi s/2}}{2^{s}\pi^{1/2} \sin(\pi s/2)} \int_{-\infty}^\infty \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1} \exp( \frac{5-s}{2} u - e^u - \frac{t\pi^2}{64} )\ dw\ du.[/math]

The [math]u[/math] integral can be evaluated by (1.1) to obtain

[math]2 \pi^{-s/2} \frac{e^{-i\pi s/2} \Gamma(\frac{5-s}{2})}{2^{s}\pi^{1/2} \sin(\pi s/2)} \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1} \exp( - \frac{t\pi^2}{64} )\ dw[/math]

and so by comparison with (2.3) we have

[math]R_{t,N,M}(s) \approx \exp( - t \pi^2/64) R_{0,N,M}(s).[/math]

In particular, from (4.1) we have

[math] R_{t,N,M}(s) \approx - \exp( - t \pi^2/64) \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \frac{e^{-i\pi s} \Gamma(1-s)}{2\pi i} (2\pi i M)^{s-1} \Psi(\frac{s}{2\pi i M}-N). \quad(6.3)[/math]

Combining (6.2), (6.3), (5.1) we obtain an approximation to [math]\xi_t(s)[/math] and hence to [math]H_t(z) = \xi_t(\frac{1+iz}{2})[/math].

To understand these asymptotics better, let us inspect [math]H_t(x+iy)[/math] for [math]t\gt0[/math] in the region

[math]x+iy = T + \frac{a+ib}{\log T}; \quad t = \frac{\tau}{\log T}[/math]

with [math]T[/math] large, [math]a,b = O(1)[/math], and [math]\tau \gt \frac{1}{2}[/math]. If [math]s = \frac{1+ix-y}{2}[/math], then we can approximate

[math] \pi^{-s/2} \approx \pi^{-\frac{1+iT}{4}}[/math]
[math] \Gamma(\frac{s+4}{2}) \approx \Gamma(\frac{9+iT}{2}) T^{\frac{ia-b}{4 \log T}} = \exp( \frac{ia-b}{4} ) \Gamma(\frac{9+iT}{2}) [/math]
[math] \frac{1}{n^s} \approx \frac{1}{n^{\frac{1+iT}{2}}}[/math]
[math] \exp( \frac{t}{16} \log^2 \frac{s+4}{2\pi n^2} ) \approx \exp( \frac{t}{16} \log^2 \frac{s+4}{2\pi} - \frac{t}{4} \log T \log n )[/math]
[math] \approx \exp( \frac{\tau}{16} \log T + \frac{i \pi \tau}{16} ) \frac{1}{n^{\frac{\tau}{4}}} [/math]

leading to

[math] F_{t,N}(\frac{1+ix-y}{2}) \approx 2\pi^{-\frac{1+iT}{4}} \Gamma(\frac{9+iT}{2}) \exp( \frac{ia-b}{4} ) \exp( \frac{\tau}{16} \log T + \frac{i \pi \tau}{16} ) \sum_n \frac{1}{n^{\frac{1+iT}{2} + \frac{\tau}{4}}}[/math]
[math] \approx 2\pi^{-\frac{1+iT}{4}} \Gamma(\frac{9+iT}{2}) \zeta(\frac{1+iT}{2} + \frac{\tau}{4}) \exp( \frac{ia-b}{4} ).[/math]

Similarly for [math]F_{t,N}(\frac{1+ix+y}{2}) [/math] (replacing [math]b[/math] by [math]-b[/math]). If we make a polar coordinate representation

[math] \frac{1}{2} \pi^{-\frac{1+iT}{4}} \Gamma(\frac{9+iT}{2}) \zeta(\frac{1+iT}{2} + \frac{\tau}{4}) = r_{T,\tau} e^{i \theta_{T,\tau}}[/math]

one thus has

[math] H_t(x+iy) \approx \frac{1}{2} ( r_{T,\tau} e^{i \theta_{T,\tau}} \exp( \frac{ia-b}{4} ) + r_{T,\tau} e^{-i \theta_{T,\tau}} \exp(\frac{-ia+b}{4}) ) [/math]
[math] = r_{T,\tau} \cos( \frac{a+ib}{4} + \theta_{T,\tau} ).[/math]

Thus locally [math]H_t(x+iy)[/math] behaves like a trigonometric function, with zeroes real and equally spaced with spacing [math]4\pi[/math] (in [math]a[/math]-coordinates) or [math]\frac{4\pi}{\log T}[/math] (in [math]x[/math] coordinates). Once [math]\tau[/math] becomes large, further increase of [math]\tau[/math] basically only increases [math]r_{T,\tau}[/math] and also shifts [math]\theta_{T,\tau}[/math] at rate [math]\pi/16[/math], causing the number of zeroes to the left of [math]T[/math] to increase at rate [math]1/4[/math] as claimed in [KKL2009].

Effective bounds

Moved to effective bounds on H_t and effective bounds on H_t - second approach.