Difference between revisions of "Second attempt at computing H t(x) for negative t"

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this becomes
 
this becomes
 
:<math> 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)</math>
 
:<math> 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)</math>
:<math> 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{2\pi i(n-N)^2}{(1 - iu / 8 \pi)} + \pi i n - \frac{2\pi i}{3N} (n-N)^3)\quad (1.24)</math>
+
:<math> \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)</math>
 
By Poisson summation, this is
 
By Poisson summation, this is
 
:<math> 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)</math>
 
:<math> 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)</math>

Revision as of 16:17, 2 January 2019

We are interested in approximating

[math]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)[/math]

in the regime when [math]x[/math] is large and [math]t[/math] is large and negative.

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

[math] 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)[/math]
[math] = \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)[/math]

where

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

Now from the definition of [math]\xi[/math] and the Stirling approximation we have

[math] \frac{1}{8} \xi(s) \approx M_0(s) \zeta(s)\quad (1.5)[/math]

where [math]M_0[/math] is defined in (6) of the writeup. Thus

[math] 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)[/math]

By Taylor expansion we have

[math] 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)[/math]

where [math]\alpha[/math] is defined in equation (8) of the writeup. We have the approximations

[math] \alpha(\frac{1+i\tilde x}{2} ) \approx \frac{1}{2} \log \frac{\tilde x}{4\pi} + \frac{i\pi}{4} \quad (1.8)[/math]

and

[math] \alpha'(\frac{1+i\tilde x}{2} ) \approx \frac{-i}{\tilde x} \quad (1.9)[/math]

and hence

[math] 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)[/math]

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

[math]N := \sqrt{\frac{\tilde x}{4\pi}}\quad (1.11)[/math]

and

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

we conclude that

[math] 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)[/math]

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

[math] 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)[/math]
[math] \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)[/math]

We can compute the [math]v[/math] integral to obtain

[math] 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)[/math]

We approximate [math]\frac{1+i\tilde x}{2} \approx \frac{i\tilde x}{2}[/math] to obtain

[math] 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)[/math]

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

[math] 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)[/math]

Next, we use the Taylor approximation

[math] \log \frac{n}{N} \approx \frac{n-N}{N} - \frac{(n-N)^2}{2N^2} + \frac{(n-N)^3}{3N^3}\quad (1.19)[/math]

to obtain

[math] 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)[/math]

Writing

[math]N(n-N) = \frac{1}{2} n^2 - \frac{1}{2} N^2 - \frac{1}{2} (N-n)^2[/math]

this becomes

[math] 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)[/math]
[math] \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)[/math]

By Poisson summation, this is

[math] 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)[/math]
[math] \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)[/math]
[math] \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)[/math]

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