Second attempt at computing H t(x) for negative t

We are interested in approximating

$H_t(x) = \int_{\bf R} \frac{1}{8} \xi(\frac{1+ix}{2} + i |t|^{1/2} v) \frac{1}{\sqrt{\pi}} e^{-v^2}\ dv \quad (1.1)$

in the regime when $x$ is large and $t$ is large and negative.

To cancel off an exponential decay factor in the $\xi$ function, it is convenient to shift the v variable by $\pi |t|^{1/2}/8$, thus

$H_t(x) = \frac{1}{8\sqrt{\pi}} \int_{\bf R} \xi(\frac{1+ix}{2} + i |t|^{1/2} v - \pi i |t|/8) e^{-(v - \pi |t|^{1/2}/8)^2}\ dv \quad (1.2)$
$= \frac{\exp( \pi^2 t / 64)}{8\sqrt{\pi}} \int_{\bf R} \xi(\frac{1+i\tilde x}{2} + i |t|^{1/2} v) e^{-v^2 + \pi |t|^{1/2} v / 4}\ dv \quad (1.3)$

where

$\tilde x := x - \pi |t|/4 = x + \frac{\pi t}{4}. \quad (1.4)$

Now from the definition of $\xi$ and the Stirling approximation we have

$\frac{1}{8} \xi(s) \approx M_0(s) \zeta(s)\quad (1.5)$

where $M_0$ is defined in (6) of the writeup. Thus

$H_t(x) \approx \frac{\exp( \pi^2 t / 64)}{\sqrt{\pi}} \int_{\bf R} M_0(\frac{1+i\tilde x}{2} + i |t|^{1/2} v) \zeta(\frac{1+i\tilde x}{2} + i |t|^{1/2} v) e^{-v^2 + \pi |t|^{1/2} v / 4}\ dv.\quad (1.6)$

By Taylor expansion we have

$M_0(\frac{1+i\tilde x}{2} + i |t|^{1/2} v) \approx M_0(\frac{1+i\tilde x}{2}) \exp( \alpha( \frac{1+i\tilde x}{2} ) i |t|^{1/2} v + \alpha'(\frac{1+i \tilde x}{2}) \frac{-|t| v^2}{2} )\quad (1.7)$

where $\alpha$ is defined in equation (8) of the writeup. We have the approximations

$\alpha(\frac{1+i\tilde x}{2} ) \approx \frac{1}{2} \log \frac{\tilde x}{4\pi} + \frac{i\pi}{4} \quad (1.8)$

and

$\alpha'(\frac{1+i\tilde x}{2} ) \approx \frac{-i}{\tilde x} \quad (1.9)$

and hence

$H_t(x) \approx \frac{\exp( \pi^2 t / 64)}{\sqrt{\pi}} M_0(\frac{1+i\tilde x}{2}) \int_{\bf R} \exp( \frac{i |t|^{1/2} v}{2} \log \frac{\tilde x}{4\pi} - \pi |t|^{1/2} v/4 + i |t| v^2 / 2\tilde x) \zeta(\frac{1+i\tilde x}{2} + i |t|^{1/2} v) e^{-v^2 + \pi |t|^{1/2} v / 4}\ dv.\quad (1.10)$

The two factors of $\exp( \pi |t|^{1/2} v/4 )$ cancel. If we now write

$N := \sqrt{\frac{\tilde x}{4\pi}}\quad (1.11)$

and

$u := |t|/N^2 = 4\pi |t|/\tilde x,\quad (1.12)$

we conclude that

$H_t(x) \approx \frac{\exp( \pi^2 t / 64)}{\sqrt{\pi}} M_0(\frac{1+i\tilde x}{2}) \int_{\bf R} \exp( i |t|^{1/2} v \log N + i u v^2 / 8 \pi) \zeta(\frac{1+i\tilde x}{2} + i |t|^{1/2} v) e^{-v^2}\ dv.\quad (1.13)$

If we formally write $\zeta(s) = \sum_n \frac{1}{n^s}$ (ignoring convergence issues) we obtain

$H_t(x) \approx \frac{\exp( \pi^2 t / 64)}{\sqrt{\pi}} M_0(\frac{1+i\tilde x}{2}) \sum_n \int_{\bf R} \exp( i |t|^{1/2} v \log N + i u v^2 / 8 \pi) n^{-\frac{1+i\tilde x}{2} - i |t|^{1/2} v} e^{-v^2}\ dv\quad (1.14)$
$\approx \frac{\exp( \pi^2 t / 64)}{8\sqrt{\pi}} M_0(\frac{1+i\tilde x}{2}) N^{-\frac{1+i\tilde x}{2}} \sum_n \int_{\bf R} \exp( - i |t|^{1/2} v \log \frac{n}{N} + i u v^2 / 8 \pi -\frac{1+i\tilde x}{2} \log \frac{n}{N} ) e^{-v^2}\ dv\quad (1.15)$

We can compute the $v$ integral to obtain

$H_t(x) \approx \frac{\exp( \pi^2 t / 64)}{\sqrt{1-\frac{iu}{8\pi}}} M_0(\frac{1+i\tilde x}{2}) N^{-\frac{1+i\tilde x}{2}} \sum_n \exp( - \frac{|t| \log^2 \frac{n}{N}}{4 (1 - iu / 8 \pi)} -\frac{1+i\tilde x}{2} \log \frac{n}{N}).\quad (1.16)$

We approximate $\frac{1+i\tilde x}{2} \approx \frac{i\tilde x}{2}$ to obtain

$H_t(x) \approx \frac{\exp( \pi^2 t / 64)}{\sqrt{1-\frac{iu}{8\pi}}} M_0(\frac{1+i\tilde x}{2}) N^{-\frac{1+i\tilde x}{2}} \sum_n \exp( - \frac{|t| \log^2 \frac{n}{N}}{4 (1 - iu / 8 \pi)} -\frac{i\tilde x}{2} \log \frac{n}{N})\quad (1.17)$

and then we approximate $\log^2 \frac{n}{N} \approx \frac{(n-N)^2}{N^2}$ and $|t| = u N^2$ and $\tilde x= 4\pi N^2$ to obtain

$H_t(x) \approx \frac{\exp( \pi^2 t / 64)}{\sqrt{1-\frac{iu}{8\pi}}} M_0(\frac{1+i\tilde x}{2}) N^{-\frac{1+i\tilde x}{2}} \sum_n \exp( - \frac{u(n-N)^2}{4 (1 - iu / 8 \pi)} - 2\pi i N^2 \log \frac{n}{N}).\quad (1.18)$

Next, we use the Taylor approximation

$\log \frac{n}{N} \approx \frac{n-N}{N} - \frac{(n-N)^2}{2N^2} + \frac{(n-N)^3}{3N^3}\quad (1.19)$

to obtain

$H_t(x) \approx \frac{\exp( \pi^2 t / 64)}{\sqrt{1-\frac{iu}{8\pi}}} M_0(\frac{1+i\tilde x}{2}) N^{-\frac{1+i\tilde x}{2}} \sum_n \exp( - \frac{u(n-N)^2}{4 (1 - iu / 8 \pi)} - 2\pi i N(n-N) + \pi i (n-N)^2 - \frac{2\pi i}{3N} (n-N)^3).\quad (1.20)$

Writing

$N(n-N) = \frac{1}{2} n^2 - \frac{1}{2} N^2 - \frac{1}{2} (N-n)^2$

this becomes

$H_t(x) \approx \frac{\exp( \pi^2 t / 64)}{\sqrt{1-\frac{iu}{8\pi}}} M_0(\frac{1+i\tilde x}{2}) N^{-\frac{1+i\tilde x}{2}} e^{\pi i N^2} \sum_n \exp( - \frac{u(n-N)^2}{4 (1 - iu / 8 \pi)} - \pi i n^2 + 2 \pi i (n-N)^2 - \frac{2\pi i}{3N} (n-N)^3)\quad (1.23)$
$\approx \frac{\exp( \pi^2 t / 64)}{\sqrt{1-\frac{iu}{8\pi}}} M_0(\frac{1+i\tilde x}{2}) N^{-\frac{1+i\tilde x}{2}} e^{\pi i N^2} \sum_n \exp( \frac{2\pi i(n-N)^2}{(1 - iu / 8 \pi)} + \pi i n - \frac{2\pi i}{3N} (n-N)^3)\quad (1.24)$

By Poisson summation, this is

$H_t(x) \approx \frac{\exp( \pi^2 t / 64)}{\sqrt{1-\frac{iu}{8\pi}}} M_0(\frac{1+i\tilde x}{2}) N^{-\frac{1+i\tilde x}{2}} e^{\pi i N^2} \sum_{m \in {\bf Z}} \int_{\bf R} \exp( \frac{2\pi i(x-N)^2}{(1 - iu / 8 \pi)} + 2\pi i m x + \pi i x - \frac{2\pi i}{3N} (x-N)^3)\ dx\quad (1.25)$
$\approx \frac{\exp( \pi^2 t / 64)}{\sqrt{1-\frac{iu}{8\pi}}} M_0(\frac{1+i\tilde x}{2}) N^{-\frac{1+i\tilde x}{2}} e^{\pi i N^2} \sum_{m \in {\bf Z} + \frac{1}{2}} \int_{\bf R} \exp( \frac{2\pi i(x-N)^2}{1 - iu / 8 \pi} + 2\pi i m x - \frac{2\pi i}{3N} (x-N)^3)\ dx\quad (1.26)$
$\approx \frac{\exp( \pi^2 t / 64)}{\sqrt{1-\frac{iu}{8\pi}}} M_0(\frac{1+i\tilde x}{2}) N^{-\frac{1+i\tilde x}{2}} e^{\pi i N^2} \sum_{m \in {\bf Z} + \frac{1}{2}} e^{2\pi i m N} \int_{\bf R} \exp( - \frac{2\pi ix^2}{1 - iu / 8 \pi} + 2\pi i m x - \frac{2\pi i}{3N} x^3)\ dx\quad (1.27)$

The integral here can be evaluated as an Airy integral, but perhaps a Taylor expansion of the last term is a better approach?

We return to (1.18) and try to apply Poisson summation without using the Taylor expansion. We first observe that as $\exp( \pi i n^2 - \pi i n ) = 1$ for all integers $n$, we have

$H_t(x) \approx \frac{\exp( \pi^2 t / 64)}{\sqrt{1-\frac{iu}{8\pi}}} M_0(\frac{1+i\tilde x}{2}) N^{-\frac{1+i\tilde x}{2}} \sum_n \exp( - \frac{u(n-N)^2}{4 (1 - iu / 8 \pi)} + \pi i n^2 - \pi i n - 2\pi i N^2 \log \frac{n}{N}).\quad (1.28)$

(this is to try to reduce the oscillation of the phase near $n \sim N$). By Poisson summation (extending the range of n to the entire integers), this is

$H_t(x) \approx \frac{\exp( \pi^2 t / 64)}{\sqrt{1-\frac{iu}{8\pi}}} M_0(\frac{1+i\tilde x}{2}) N^{-\frac{1+i\tilde x}{2}} \sum_{m \in {\bf Z}} \int_{\bf R} \exp( - \frac{u(X-N)^2}{4 (1 - iu / 8 \pi)} + \pi i X^2 - \pi i X + 2\pi i m X - 2\pi i N^2 \log \frac{X}{N})\ dX.\quad (1.29)$

(using $X$ as the variable of integration to distinguish it from the initial variable $x$. Shifting $m$ by 1/2, this is

$H_t(x) \approx \frac{\exp( \pi^2 t / 64)}{\sqrt{1-\frac{iu}{8\pi}}} M_0(\frac{1+i\tilde x}{2}) N^{-\frac{1+i\tilde x}{2}}\exp(\pi i N^2) \sum_{m \in {\bf Z} + \frac{1}{2}} f_m( u, N ) \quad (1.30)$

where

$f_m(u,N) := \int_{\bf R} \exp( - \frac{u(X-N)^2}{4 (1 - iu / 8 \pi)} + \pi i X^2 - \pi i N^2 + 2\pi i m X - 2\pi i N^2 \log \frac{X}{N})\ dX.\quad (1.31)$

These functions should be roughly 1-periodic in N and decay quickly as $|m| \to \infty$; they are roughly like the m^th Fourier harmonics of $H_t$ in the N variable.

Shifting $X$ by $N$, we have

$f_m(u,N) = \int_{\bf R} \exp( - \frac{uX^2}{4 (1 - iu / 8 \pi)} + \pi i (X+N)^2 - \pi i N^2 + 2\pi i m (X+N) - 2\pi i N^2 \log(1+\frac{X}{N}) )\ dX\quad (1.32)$
$= \exp( 2\pi i m N) \int_{\bf R} \exp( \frac{2\pi i}{1-\frac{iu}{8\pi}} X^2 + 2\pi i m X - 2\pi i N^2 [\log(1+\frac{X}{N}) - \frac{X}{N} + \frac{X^2}{2N^2}] )\ dX. (1.33)$

If we ignore the term $2\pi i N^2 [\log(1+\frac{X}{N}) - \frac{X}{N} + \frac{X^2}{2N^2}] = O( X^3 / N )$, we would have

$f_m(u,N) \approx \exp( 2\pi i m N) \int_{\bf R} \exp( \frac{2\pi i}{1-\frac{iu}{8\pi}} X^2 + 2\pi i m X)\ dX (1.34)$
$= \sqrt{\pi i (1-\frac{iu}{8\pi})} \exp( 2\pi i m N - \frac{\pi i m^2}{2} (1 - \frac{iu}{8\pi}) )(1.35)$

hence

$H_t(x) \approx \exp( \pi^2 t / 64) \sqrt{\pi i} M_0(\frac{1+i\tilde x}{2}) N^{-\frac{1+i\tilde x}{2}}\exp(\pi i N^2) \sum_{m \in {\bf Z} + \frac{1}{2}} \exp( 2\pi i m N - \frac{\pi i m^2}{2} (1 - \frac{iu}{8\pi}) ) \quad (1.36)$
$= \exp( \pi^2 t / 64) \sqrt{\pi i} M_0(\frac{1+i\tilde x}{2}) N^{-\frac{1+i\tilde x}{2}}\exp(\pi i N^2) \sum_{m \in {\bf Z} + \frac{1}{2}} e^{-m^2 u / 16} \exp( 2\pi i m N - \frac{\pi i m^2}{2} ) \quad (1.37)$
$= \exp( \pi^2 t / 64) \sqrt{\pi i} M_0(\frac{1+i\tilde x}{2}) N^{-\frac{1+i\tilde x}{2}}\exp(\pi i N^2) \sum_{m \in {\bf Z} + \frac{1}{2}} e^{-m^2 u / 16} \cos( 2\pi m N ) \exp( - \frac{\pi i m^2}{2} ) \quad (1.38)$
$= 2\exp( \pi^2 t / 64 + \frac{\pi i}{8} - u/64) \sqrt{\pi} M_0(\frac{1+i\tilde x}{2}) N^{-\frac{1+i\tilde x}{2}}\exp(\pi i N^2) \sum_{n=0}^\infty e^{-n(n+1) u / 16} \cos( 2\pi (n+\frac{1}{2}) N ) \exp( - \frac{\pi i n(n+1)}{2} ) \quad (1.39)$

This seems to be an adequate approximation for large $u$, but not for small $u$.

We return now to (1.20) and write it slightly differently. We compute

$\exp( - 2\pi i N(n-N) + \pi i (n-N)^2 ) = \exp( \pi i n^2 - 4\pi i N n + 3 \pi i N^2 )$
$= \exp( -\pi i n - 4\pi i N n + 3 \pi i N^2 )$
$= \exp(- 2\pi i \{ 2N + \frac{1}{2} \} n + 3 \pi i N^2 )$

where $\{x\}$ is the fractional part. We thus have

$H_t(x) \approx \frac{\exp( \pi^2 t / 64)}{\sqrt{1-\frac{iu}{8\pi}}} M_0(\frac{1+i\tilde x}{2}) N^{-\frac{1+i\tilde x}{2}} e^{3\pi i N^2} \sum_n \exp( - \frac{u(n-N)^2}{4 (1 - iu / 8 \pi)} - 2\pi i n \{2N + \frac{1}{2}\} - \frac{2\pi i}{3N} (n-N)^3).\quad (1.40)$

When $u$ is small, the summands here are fairly slowly varying, which suggests the use of Poisson summation:

$H_t(x) \approx \frac{\exp( \pi^2 t / 64)}{\sqrt{1-\frac{iu}{8\pi}}} M_0(\frac{1+i\tilde x}{2}) N^{-\frac{1+i\tilde x}{2}} e^{3\pi i N^2} \sum_{m \in {\bf Z}} g_m(N,u) \quad (1.41)$

where

$g_m(N,u) = \int_{\bf R} \exp( - \frac{u(X-N)^2}{4 (1 - iu / 8 \pi)} + 2\pi i X (m - \{2N + \frac{1}{2}\}) - \frac{2\pi i}{3N} (X-N)^3)\ dX.\quad (1.42)$

The region of integration here should contain a neighbourhood of $N$ as this is where the integrand is largest. We can shift $X$ by $N$ to obtain

$g_m(N,u) = e^{2\pi i N (m - \{2N + \frac{1}{2}\})} \int_{\bf R} \exp( - \frac{uX^2}{4 (1 - iu / 8 \pi)} + 2\pi i X (m - \{2N + \frac{1}{2}\}) - \frac{2\pi i}{3N} X^3)\ dX.\quad (1.43)$

and now the region of integration should be centered at $X=0$.

Suppose we neglect the third order term $\frac{2\pi i}{3N} X^3$. Then

$g_m(N,u) \approx e^{2\pi i N (m - \{2N + \frac{1}{2}\})} \int_{\bf R} \exp( - \frac{uX^2}{4 (1 - iu / 8 \pi)} + 2\pi i X (m - \{2N + \frac{1}{2}\})\ dX\quad (1.44)$
$\approx \sqrt{\frac{4 \pi (1-iu/8\pi)}{u}} e^{2\pi i N (m - \{2N + \frac{1}{2}\})} \exp( -\frac{4 \pi^2 (m - \{2N + \frac{1}{2}\})^2 (1 - iu/8\pi) }{u} ) \quad (1.45)$

For small $u$, this decays quickly for $m$ far from $\{2N + \frac{1}{2}\}$. For $N$ close to an integer or half-integer, $\{2N+\frac{1}{2}\}$ is close to 1/2, and so the m=0 and m=1 terms should be dominant.

Let's compute the integral in (1.43). We have

$g_m(N,u) = \exp( i N c ) \int_{\bf R} \exp( -\frac{ia}{3} X^3 + i b X^2 + i c_m X )\ dX \quad (1.46)$

where

$a := \frac{2\pi}{N} \quad (1.47)$
$b := i \frac{u}{4(1-iu/8\pi)} \quad (1.48)$
$c_m := 2 \pi (m - \{ 2N + \frac{1}{2} \} ). \quad (1.49)$

Change variables $X = a^{-1/3} Y$:

$g_m(N,u) = a^{-1/3} \exp( i N c_m ) \int_{\bf R} \exp( -\frac{i}{3} Y^3 + i b a^{-2/3} Y^2 + i c_m a^{-1/3} Y )\ dY \quad (1.50)$

Change variables $Y = Z + b a^{-2/3}$ and shift the contour of integration:

$g_m(N,u) = a^{-1/3} \exp( i N c_m + 2 i b^3 a^{-2}/3 + i bc_m a^{-1}) \int_{\bf R} \exp( -\frac{i}{3} Z^3 + i(c_m a^{-1/3} + b^2 a^{-4/3}) Z )\ dZ.\quad (1.51)$

Since

$\mathrm{Ai}(x) = \frac{1}{\pi} \int_0^\infty \cos(\frac{t^3}{3} + xt)\ dt \quad (1.52)$
$= \frac{1}{2\pi} \int_{\bf R} \exp(i\frac{t^3}{3} + ixt)\ dt \quad (1.53)$
$= \frac{1}{2\pi} \int_{\bf R} \exp(-i\frac{Z^3}{3} - ixZ)\ dZ \quad (1.54)$

we thus conclude that

$g_m(N,u) = 2\pi a^{-1/3} \exp( i N c_m + 2 i b^3 a^{-2}/3 + i bc_m a^{-1}) \mathrm{Ai}( - (c_m a^{-1/3} + b^2 a^{-4/3}) ).\quad (1.55)$

Note: b is complex valued, so the argument of the Airy function is also complex. Hopefully this is not going to be an issue. Also as N is large, the argument is going to be fairly large. But this could be a good thing; it means that we could hope to use one of the known asymptotics of the Airy function to get a more tractable approximation.

Numerically it looks like the m=0,1 terms dominate. Since $c_1 = c_0 + 2\pi$, we thus have from (1.41) and (1.55) that

$H_t(x) \approx M ( \mathrm{Ai}( - (c_0 a^{-1/3} + b^2 a^{-4/3}) ) + \exp( 2\pi iN + 2\pi ib a^{-1} ) \mathrm{Ai}( - (c_0 a^{-1/3} + b^2 a^{-4/3} + 2\pi a^{-1/3}) ) ) \quad (1.56)$

where M is the multiplier

$M := \frac{\exp( \pi^2 t / 64)}{\sqrt{1-\frac{iu}{8\pi}}} M_0(\frac{1+i\tilde x}{2}) N^{-\frac{1+i\tilde x}{2}} e^{3\pi i N^2} 2\pi a^{-1/3} \exp( i N c_0 + 2 i b^3 a^{-2}/3 + i bc_0 a^{-1}). \quad (1.57)$

If we now formally substitute the Airy approximation

$\mathrm{Ai}(z) \approx \frac{1}{2\sqrt{\pi}} z^{-1/4} e^{-2/3 z^{3/2}}$

we conclude that

$H_t(x) \approx \frac{1}{2\sqrt{\pi}} M ( ( - (c_0 a^{-1/3} + b^2 a^{-4/3}) )^{-1/4} e^{-2/3 (- (c_0 a^{-1/3} + b^2 a^{-4/3}))^{3/2}} + \exp( 2\pi iN + 2\pi ib a^{-1} ) ( - (c_0 a^{-1/3} + b^2 a^{-4/3} + 2\pi a^{-1/3}) )^{-1/4} e^{-2/3 (- (c_0 a^{-1/3} + b^2 a^{-4/3} + 2\pi a^{-1/3}))^{3/2}} ). \quad (1.58)$

It looks like the $b^2 a^{-4/3}$ term is large compared to $c_0 a^{-1/3}$ or $2\pi a^{-1/3}$ (because $a=2\pi/N$ is quite small), suggesting the use of the approximations

$(-(c_0 a^{-1/3} + b^2 a^{-4/3}))^{-1/4} = (- b^2 a^{-4/3})^{-1/4} ( 1 + \frac{a c_0}{b^2} )^{-1/4} \quad (1.59)$
$\approx (- b^2 a^{-4/3})^{-1/4} \quad (1.60)$

and

$-\frac{2}{3} (-(c_0 a^{-1/3} + b^2 a^{-4/3}))^{3/2} = -\frac{2}{3} (-b^2 a^{-4/3})^{3/2} ( 1 + \frac{a c_0}{b^2} )^{3/2} \quad (1.61)$
$\approx -\frac{2 i b^3}{3a^2} ( 1 + \frac{3 a c_0}{2 b^2} + \frac{3 a^2 c_0^2}{8 b^4}) \quad (1.62)$
$= -\frac{2 i b^3}{3a^2} - i \frac{c_0 b}{a} - i\frac{c_0^2}{4 b} \quad (1.63)$

and similarly with $c_0$ replaced by $c_1 = c_0 + 2\pi$ (here we have somewhat optimistically dropped all terms that have a positive power of $a$; if this proves inaccurate, we should go back and add more terms to the expansion). We conclude that

$H_t(x) \approx \frac{1}{2\sqrt{\pi}} M (- b^2 a^{-4/3})^{-1/4} e^{-\frac{ac_0}{4b^2} - \frac{2 i b^3}{3a^2} - i \frac{c_0}{ab} - i \frac{c_0^2}{4 b}} ( 1 + \exp( 2 \pi i N - \frac{\pi i(c_0+\pi)}{b} ) ). \quad (1.64)$

We return to (1.58). The ratio between the second summand and the first summand can be written as $\exp(S)$, where

$S := 2\pi i N + 2 \pi i b a^{-1} - \frac{2}{3} ( (- (c_0 a^{-1/3} + b^2 a^{-4/3} + 2\pi a^{-1/3}))^{3/2} - (- (c_0 a^{-1/3} + b^2 a^{-4/3} ))^{3/2} )- \frac{1}{4} \log( 1 + \frac{2\pi a^{-1/3}}{c_0 a^{-1/3} + b^2 a^{-4/3}} ) \hbox{ mod } \pi i/2 \quad (1.65)$

where the modulus is present due to possible ambiguities in the logarithm. The real part of $S$ (representing the magnitude of the ratios) is unambiguous though:

$\mathrm{Re} S := N \mathrm{Re}( i b ) - \frac{2}{3} \mathrm{Re} ( (- (c_0 a^{-1/3} + b^2 a^{-4/3} + 2\pi a^{-1/3}))^{3/2} - (- (c_0 a^{-1/3} + b^2 a^{-4/3} ))^{3/2} ) - \frac{1}{4} \log | 1 + \frac{2\pi a^{-1/3}}{c_0 a^{-1/3} + b^2 a^{-4/3}}|. \quad (1.66)$

The quantity $b^2$ can be computed as

$b^2 = - \frac{u^2}{16(1-iu/8\pi)^2} \quad (1.67)$
$= - \frac{u^2}{16(1+\frac{u^2}{64 \pi^2})^2} (1 + \frac{iu}{8\pi})^2 \quad (1.68)$
$= - \frac{u^2}{16(1+\frac{u^2}{64 \pi^2})^2} (1 - \frac{u^2}{64\pi^2} + \frac{iu}{4\pi}) \quad (1.69).$

In particular, $b^2$ has negative real part when $u \leq 8\pi$ (which seems to be the most interesting region), and for small $u$ the imaginary part of b is rather small compared to the real part. The quantity $c_0$ ranges between $-2\pi$ and 0, so we conclude that the argument

$-(c_0 a^{-1/3} + b^2 a^{-4/3}) \quad (1.70)$

has positive real part (and small imaginary part), and so the Airy function of this quantity should not oscillate but instead decay quite quickly. The real part of the quantity

$-(c_0 a^{-1/3} + b^2 a^{-4/3} + 2\pi a^{-1/3}) \quad (1.71)$

can change sign though. When it is negative, the Airy function should oscillate rapidly; as long as $\mathrm{Re}(S) \geq 0$, the amplitude of this oscillation exceeds the magnitude of the other term in (1.58) and so one should obtain a lot of zeroes of the real part of (1.58). But once $\mathrm{Re}(S) \lt 0$ then the oscillation is less than the magnitude of the non-oscillating term in (1.58) and one now expects basically no zeroes. When instead the real part of (1.71) is positive then both terms in (1.58) do not oscillate and one again expects few zeroes. So the oscillating zero pattern should be confined to the "sharkfin" region

$\mathrm{Re} S \gt 0; \quad \mathrm{Re} -(c_0 a^{-1/3} + b^2 a^{-4/3} + 2\pi a^{-1/3}) \lt 0. \quad (1.72)$

The second condition in (1.72) can be simplified to

$c_0 + 2\pi \gt -\mathrm{Re}(b^2) / a \quad (1.73)$

which by (1.69), (1.47), (1.49) is

$2\pi (1 - \{ 2N + \frac{1}{2} \} ) \gt \frac{N}{2\pi} \frac{u^2}{16(1+\frac{u^2}{64 \pi^2})^2} (1 - \frac{u^2}{64\pi^2}) \quad (1.74)$

If we neglect the $\frac{u^2}{64 \pi^2}$ terms, and approximates $N \approx \frac{1}{2} [2N + \frac{1}{2}]$ this approximates to

$\{ 2N + \frac{1}{2} \} \lt 1 - [2N + \frac{1}{2}] \frac{u^2}{128\pi^2} \quad (1.75)$

which is a parabolic region for each choice of integer part $[2N + \frac{1}{2}]$.

Now we look at the other condition in (1.72). We have

$\mathrm{Re}(ib) = \frac{-u}{4(1+\frac{u^2}{64\pi^2})} \approx \frac{-u}{4} \quad (1.76)$

so by (1.66) the constraint is approximately

$-\frac{Nu}{4} - \frac{2}{3 a^{1/2}} \mathrm{Re} ( (- (c_0 + b^2 a^{-1} + 2\pi))^{3/2} - (- (c_0 + b^2 a^{-1} ))^{3/2} ) - \frac{1}{4} \log | 1 + \frac{2\pi}{c_0 + b^2 a^{-1}}| \gt 0. \quad (1.77)$

We use (1.69) to approximate

$b^2 \approx -u^2/16 \quad (1.78)$

and use (1.47) to then obtain

$-\frac{Nu}{4} - \frac{2 \sqrt{N}}{3 \sqrt{2\pi}} \mathrm{Re} ( (- (c_0 - \frac{Nu^2}{32 \pi} + 2\pi))^{3/2} - (- (c_0 - \frac{Nu^2}{32 \pi} ))^{3/2} ) - \frac{1}{4} \log | 1 + \frac{2\pi}{c_0 + \frac{Nu^2}{32 \pi}}| \gt 0. \quad (1.78)$

The last term looks lower order, let's drop it:

$-\frac{Nu}{4} - \frac{2 \sqrt{N}}{3 \sqrt{2\pi}} \mathrm{Re} ( (- (c_0 - \frac{Nu^2}{32 \pi} + 2\pi))^{3/2} - (- (c_0 - \frac{Nu^2}{32 \pi} ))^{3/2} ) \gt 0. \quad (1.79)$

If we are in the region (1.73), we'd expect $- (c_0 - \frac{Nu^2}{32 \pi} + 2\pi)$ negative, so $(- (c_0 - \frac{Nu^2}{32 \pi} + 2\pi))^{3/2}$ is purely imaginary. So we simplify to

$-\frac{Nu}{4} + \frac{2 \sqrt{N}}{3 \sqrt{2\pi}} (- (c_0 - \frac{Nu^2}{32 \pi} ))^{3/2} \gt 0. \quad (1.80)$

This rearranges to

$- (c_0 - \frac{Nu^2}{32 \pi} ) \gt ( \frac{3 \sqrt{2\pi}}{8} \sqrt{N} u)^{2/3} \quad (1.81)$

or

$\{ 2N + \frac{1}{2} \} \gt -\frac{Nu^2}{64 \pi^2} + (\frac{9}{256 \pi^2} N u^2)^{1/3} \quad (1.82)$

Setting $v = Nu^2$, the region where zeroes are expected to occur is now

$(\frac{9}{256\pi^2} v)^{1/3} - \frac{v}{64\pi^2} \lt \{ 2N + \frac{1}{2} \} \lt 1 - \frac{v}{64\pi^2} \quad (1.83)$

which reaches an apex when

$\{ 2N + \frac{1}{2} \} = \frac{5}{9}, Nu^2 = \frac{256 \pi^2}{9}. \quad \quad(1.84)$

Inside this "Sharkfin" region, we expect the second term of (1.56) to dominate, so the zeroes should occur roughly where

$\mathrm{Ai}( - (c_0 a^{-1/3} + b^2 a^{-4/3} + 2\pi a^{-1/3}) ) ) = 0. \quad (1.85)$

The zeroes of $\mathrm{Ai}(-z)$ occur when $\frac{2}{3} z^{3/2} +\frac{\pi}{4}$ is close to a multiple of $\pi$, so one should get something like

$\frac{2}{3} (c_0 a^{-1/3} + b^2 a^{-4/3} + 2\pi a^{-1/3}) )^{3/2} + \frac{\pi}{4} ) = 0 \hbox{ mod } \pi. \quad (1.86)$

Using (1.78) this becomes

$\frac{2}{3} a^{-1/2} (c_0 - \frac{u^2}{16 a} + 2\pi )^{3/2} + \frac{\pi}{4} = 0 \hbox{ mod } \pi \quad (1.87)$

hence by (1.47), (1.49)

$\frac{2\sqrt{N}}{3\sqrt{2\pi}} (- 2\pi \{ 2N + \frac{1}{2} \} - \frac{N u^2}{32 \pi} + 2\pi )^{3/2} + \frac{\pi}{4} = 0 \hbox{ mod } \pi \quad (1.88)$

or on dividing by $\pi$

$\frac{2\sqrt{N}}{3\sqrt{2}} (- 2 \{ 2N + \frac{1}{2} \} - \frac{N u^2}{32 \pi^2} + 2 )^{3/2} + \frac{1}{4} = 0 \hbox{ mod } 1 \quad (1.89)$