# Difference between revisions of "Asymptotics of H t"

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## The Gamma function

The Gamma function is defined for $\mathrm{Re}(s) \gt 0$ by the formula

$\Gamma(s) = \int_0^\infty x^s e^{-x} \frac{dx}{x}$

and hence by change of variables

$\Gamma(s) = \int_{-\infty}^\infty \exp( s u - e^u )\ du. \quad (1.1)$

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

$\Gamma(s) = \int_C \exp( s u - e^u )\ du \quad (1.1')$

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

The Gamma function obeys the Euler reflection formula

$\Gamma(s) \Gamma(1-s) = \frac{\pi}{\sin(\pi s)} \quad (1.2)$

and the duplication formula

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

In particular one has

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

and thus on combining (3) and (4)

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

Since $s \Gamma(s) = \Gamma(s+1)$, we have

$\frac{s(s-1)}{2} \Gamma(\frac{s}{2}) = 2 \Gamma(\frac{s+4}{2}) - 3 \Gamma(\frac{s+2}{2}). \quad (1.6)$

We have the Stirling approximation

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

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

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

Another crude but useful approximation is

$\Gamma(s+h) \approx \Gamma(s) s^h (1.8)$

for $s$ as above and $h=O(1)$.

## The Riemann-Siegel formula for $t=0$

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

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

where $w^{s-1} := \exp((s-1) \log w)$ and we use the branch of the logarithm with imaginary part in $[0,2\pi)$, and $C_M$ is any contour from $+\infty$ to $+\infty$ going once anticlockwise around the zeroes $2\pi i m$ of $e^w-1$ with $|m| \leq M$, 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 $s$, so it will suffice to establish that

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

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

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

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

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

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

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

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

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

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

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

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

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

and the claim follows from the dominated convergence theorem. $\Box$

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

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

where

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

and

$R_{0,N,M}(s) := \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \frac{e^{-i\pi s} \Gamma(1-s)}{2\pi i} \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1}\ dw. \quad(2.3)$

## A contour integral

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

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

can be given explicitly by the formula

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

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

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

The right-hand side expands as

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

which we can write as

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

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

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

and the claim then follows after some algebra. $\Box$

We conclude from (2.3) that

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

## Heuristic approximation at $t=0$

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

$R_{0,N,M}(s) = \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \frac{e^{-i\pi s} \Gamma(1-s)}{2\pi i} \int_{C_M - 2\pi i M} \frac{(z+2\pi i M)^{s-1} e^{-Nz}}{e^z-1}\ dz$

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

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

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

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

and hence by Lemma 2

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

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

If we drop the $R_{0,N,M}$ term, we have

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

From (2.2) and (1.7) we have

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

when $s = (1+ix-y)/2$ and

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

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

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

## Extending the Riemann-Siegel formula to positive $t$

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

$H_t(z) = \frac{1}{8} \xi_t(1+\frac{iz}{2}).$

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

In view of (2.1), we will have

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

where $F_{t,N}, R_{t,N,M}$ are the heat flow evolutions of $F_{0,N}, R_{0,N,M}$ respectively.

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

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

and hence by (1.1')

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

We can now evolve to obtain

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

By integrating on $C$ rather than the real axis, the integrals remain absolutely convergent here.

Evolving $R_{0,N,M}$ is a bit trickier. From (1.5) one has

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

which can be rewritten using (1.6) as

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

For $\mathrm{Im}(s) \gt 0$, we have the geometric series formula

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

and from this and (1.1') we can rewrite $R_{0,N,M}(s)$ as

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

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

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

## Approximation for $t\gt0$

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

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

Next, we shift $u$ by $\log \frac{s+4}{2}$ to obtain

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

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

$\frac{t}{16} (u + \log \frac{s+4}{2\pi n^2})^2 ) \approx \frac{t}{16} \log^2 \frac{s+4}{2\pi n^2}$

and then we undo the shift to obtain

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

which by (1) becomes

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

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

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

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

$\log \frac{w}{2\sqrt{\pi}} - \frac{u}{2} \approx i \pi/4$

and so we may heuristically obtain

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

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

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

The $u$ integral can be evaluated by (1.1) to obtain

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

and so by comparison with (2.3) we have

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

In particular, from (4.1) we have

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

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

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

$x+iy = T + \frac{a+ib}{\log T}; \quad t = \frac{\tau}{\log T}$

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

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

leading to

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

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

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

one thus has

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

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