# Difference between revisions of "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:

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

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. We thus have

$H_t(x) \approx M ( \exp( iNc_0 + ibc_0 a^{-1} ) \mathrm{Ai}( - (c_0 a^{-1/3} + b^2 a^{-4/3}) ) + \exp( iNc_1 + ibc_1 a^{-1} ) \mathrm{Ai}( - (c_1 a^{-1/3} + b^2 a^{-4/3}) ) ) \quad (1.56)$

where M is the multiplier

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

\sum_{m \in {\bf Z}} g_m(N,u) \quad (1.41)[/itex]