Difference between revisions of "Asymptotics of H t"

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== Asymptotics for <math>t=0</math> ==
+
== The Gamma function ==
  
The approximate functional equation (see e.g. [T1986, (4.12.4)]) asserts that
+
The Gamma function is defined for <math>\mathrm{Re}(s) > 0</math> by the formula
  
:<math>\displaystyle \zeta(s) = \sum_{n \leq N} \frac{1}{n^s} + \pi^{s-1/2} \frac{\Gamma((1-s)/2)}{\Gamma(s/2)} \sum_{n \leq N} \frac{1}{n^{1-s}} + O( t^{-\sigma/2} )</math>
+
:<math>\Gamma(s) = \int_0^\infty x^s e^{-x} \frac{dx}{x} </math>  
  
for <math>s = \sigma +it</math> with <math>t</math> large, <math>0 < \sigma < 1</math>, and <math>N := \sqrt{t/2\pi}</math>.  This implies that
+
and hence by change of variables
  
:<math>\displaystyle \xi(s) = F(s) + F(1-s) + O( \Gamma(\frac{s+4}{2}) t^{-\sigma/2} )</math>
+
:<math>\Gamma(s) = \int_{-\infty}^\infty \exp( s u - e^u )\ du. \quad (1.1)</math>
  
where
+
It can be extended to other values of <math>s</math> by analytic continuation or by contour shifting; for instance, if <math>Im(s)>0</math>, one can write
  
:<math>\displaystyle F(s) := \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \sum_{n=1}^N \frac{1}{n^s}.</math>
+
:<math>\Gamma(s) = \int_C \exp( s u - e^u )\ du \quad (1.1')</math>
  
Writing
+
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.
  
:<math>\displaystyle \frac{s(s-1)}{2} \Gamma(s/2) = 2 \Gamma(\frac{s+4}{2}) - 3 \Gamma(\frac{s+2}{2})</math>
+
The Gamma function obeys the Euler reflection formula
  
we have <math>F(s) = 2 F_0(s) - 3 F_{-1}(s)</math>, where
+
:<math>\Gamma(s) \Gamma(1-s) = \frac{\pi}{\sin(\pi s)} \quad (1.2)</math>
  
:<math>\displaystyle F_j(s) := \pi^{-s/2} \Gamma(\frac{s+4}{2} + j)  \sum_{n=1}^N \frac{1}{n^s}.</math>
+
and the duplication formula
  
The <math>F_{-1}</math> term sums to <math>O( \Gamma(\frac{s+4}{2}) t^{-\sigma/2} )</math>, hence
+
:<math>\Gamma(1-s) = \frac{\Gamma(\frac{1-s}{2}) \Gamma(1-\frac{s}{2})}{\pi^{1/2} 2^s}. \quad (1.3)</math>  
  
:<math>\displaystyle \xi(s) = 2F_0(s) + 2F_0(1-s) + O( \Gamma(\frac{s+4}{2}) t^{-\sigma/2} )</math>
+
In particular one has
  
and thus
+
:<math>\Gamma(\frac{s}{2}) \Gamma(1-\frac{s}{2}) = \frac{\pi}{\sin(\pi s/2)} \quad (1.4)</math>
  
:<math>\displaystyle H(x+iy) = \frac{1}{4} F_0( \frac{1+ix-y}{2} ) + \frac{1}{4} \overline{F_0( \frac{1+ix+y}{2} )} + O( \Gamma(\frac{9+ix+y}{2}) x^{-(1+y)/2} ).</math>
+
and thus on combining (3) and (4)
  
One would expect the <math>\sum_{n=1}^N \frac{1}{n^s}</math> term to remain more or less bounded (this is basically the Lindelof hypothesis), leading to the heuristics
+
:<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>
  
:<math>\displaystyle |F_0(\frac{1+ix \pm y}{2})| \asymp \Gamma(\frac{9+ix \pm y}{2}).</math>
+
Since <math>s \Gamma(s) = \Gamma(s+1)</math>, we have
  
Since <math>\Gamma(\frac{9+ix - y}{2}) \approx \Gamma(\frac{9+ix+y}{2}) (ix)^{-y}</math>, we expect the <math>F_0( \frac{1+ix+y}{2} )</math> term to dominate once <math>y \gg \frac{1}{\log x}</math>.
+
:<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>
  
== Asymptotics for <math>t > 0</math> ==
+
We have the Stirling approximation
  
Let <math>z=x+iy</math> for large <math>x</math> and positive bounded <math>y</math>.  We have
+
:<math>\Gamma(s) = \sqrt{2\pi/s} \exp( s \log s - s + O(1/|s|) )</math>
  
:<math>\displaystyle H_t(z) = \frac{1}{2} \int_{-\infty}^\infty e^{tu^2} \Phi(u) \exp(izu)\ du</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
  
where
+
:<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>
  
:<math>\displaystyle \Phi(u) = \sum_{n=1}^\infty (2\pi^2 n^4 e^{9u} - 3\pi n^2 e^{5u}) \exp(-\pi n^2 e^{4u}).</math>
+
Another crude but useful approximation is
  
We can shift contours to
+
:<math>\Gamma(s+h) \approx \Gamma(s) s^h (1.8) </math>
  
:<math>\displaystyle H_t(z) = \frac{1}{2} \int_{i\theta-\infty}^{i\theta+\infty} e^{tu^2} \Phi(u) \exp(izu)\ du</math>
+
for <math>s</math> as above and <math>h=O(1)</math>.
  
to any <math>-\pi/8 < \theta < \pi/8</math> that we please; it seems that a good choice will be <math>\theta = \mathrm{arg} (ix+y+9) \approx \frac{\pi}{8} - \frac{y+9}{x}</math>.  By symmetry, we thus have
+
== The Riemann-Siegel formula for <math>t=0</math> ==
  
:<math>\displaystyle H_t(z) = G_t(x+iy) + \overline{G_t(x-iy)}</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
 
+
where
+
 
+
:<math>\displaystyle G_t(z) := \int_{i\theta}^{i\theta+\infty} e^{tu^2} \Phi(u) \exp(izu)\ du.</math>
+
 
+
By Fubini's theorem we have
+
 
+
:<math>\displaystyle G_{t}(x \pm i y) = \sum_{n=1}^\infty \pi^2 n^4 \int_{i\theta}^{i\theta+\infty} \exp( tu^2 - \pi n^2 e^{4u} + (ix \mp y + 9) u)\ du</math>
+
:<math> \displaystyle - \sum_{n=1}^\infty \frac{3}{2} \pi n^2 \int_{i\theta}^{i\theta+\infty} \exp( tu^2 - \pi n^2 e^{4u} + (ix \mp y + 5) u)\ du.</math>
+
 
+
The second terms end up being about <math>O(1/x)</math> the size of the first terms and we will ignore them for now. Making the change of variables <math>u = \frac{1}{4} \log \frac{ix \pm y + 9}{4\pi n^2} + v</math>, we basically have
+
 
+
:<math> \displaystyle G_t(x \pm iy) \approx \sum_{n=1}^\infty \pi^2 n^4 (\frac{ix \pm y+9}{4\pi n^2})^{\frac{ix \mp y+9}{4}}
+
\int_{-\frac{1}{4} \log \frac{|ix\pm y+9|}{4\pi n^2}}^\infty \exp( \frac{t}{16} (\log \frac{ix \pm y+9}{4\pi n^2} + v)^2 + (ix \mp y + 9) (v - \frac{1}{4} e^{4v}) )\ dv.</math>
+
The function <math>\exp( (ix \mp y + 9) (v - \frac{1}{4} e^{4v}) )</math> decays rapidly away from <math>v=0</math>.  This suggests firstly that this integral is going to be very small when <math>n \gg N := \sqrt{x/4\pi}</math> (since the left limit of integration will then be to the right of the origin), so we will assume heuristically that <math>n</math> is now restricted to the range <math>n \leq N</math>.  Next, we approximate <math>\exp( \frac{t}{16} (\log \frac{ix \pm y+9}{4\pi n^2} + v)^2)</math> by <math>\exp( \frac{t}{16} \log^2 \frac{ix \pm y+9}{4\pi n^2} )</math>, and then send the left limit off to infinity to obtain (heuristically)
+
 
+
:<math> \displaystyle G_t(x \pm iy) \approx \sum_{n \leq N} \pi^2 n^4 (\frac{ix \pm y+9}{4\pi n^2})^{\frac{ix \mp y+9}{4}}
+
\exp( \frac{t}{16} \log^2 \frac{ix \pm y+9}{4\pi n^2} ) \int_{-\infty}^\infty \exp( (ix \mp y + 9) (v - \frac{1}{4} e^{4v}) )\ dv.</math>
+
 
+
Making the change of variables <math>w := \frac{ix \mp y + 9}{4} e^{4v}</math> we see that
+
 
+
:<math>\int_{-\infty}^\infty \exp( (ix \mp y + 9) (v - \frac{1}{4} e^{4v}) )\ dv = \frac{1}{4} \Gamma(\frac{ix \mp y + 9}{4}) (\frac{4}{ix \mp y + 9})^{\frac{ix \mp y+9}{4}}</math>
+
 
+
and thus
+
:<math> \displaystyle G_t(x \pm iy) \approx \Gamma(\frac{ix \mp y + 9}{4}) \sum_{n \leq N} \frac{\pi^2}{4} n^4 (\frac{1}{\pi n^2})^{\frac{ix \mp y+9}{4}}
+
\exp( \frac{t}{16} \log^2 \frac{ix \pm y+9}{4\pi n^2} ) </math>
+
which simplifies a bit to
+
:<math> \displaystyle G_t(x \pm iy) \approx \frac{1}{4} \pi^{-\frac{ix \mp y + 1}{4}} \Gamma(\frac{ix \mp y + 9}{4}) \sum_{n \leq N} \frac{\exp( \frac{t}{16} \log^2 \frac{ix \pm y+9}{4\pi n^2} )}{n^{\frac{1 \mp y + ix}{2}}} </math>
+
and thus we heuristically have
+
:<math> H_t(x+iy) \approx \frac{1}{4} F_t( \frac{1+ix-y}{2} ) + \frac{1}{4} \overline{F_t( \frac{1+ix+y}{2} )} </math>
+
where
+
:<math>F_t( s ) := \pi^{-s/2} \Gamma(\frac{s+4}{2}) \sum_{n \leq N} \frac{\exp( \frac{t}{16} \log^2 \frac{s+4}{2\pi n^2} )}{n^{s}}.</math>
+
Here we can view <math>N</math> as a function of <math>s</math> by the formula <math>N = \mathrm{Im}(s)/2\pi</math>.
+
 
+
To understand these asymptotics better, let us inspect <math>H_t(x+iy)</math> for <math>t>0</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 > \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(\frac{1+ix-y}{2}) \approx \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 \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(\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].
+
 
+
== Riemann-Siegel formula ==
+
 
+
'''Proposition 1'''  (Riemann-Siegel formula) 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>
 
:<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.
 
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.
Line 139: Line 78:
  
 
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
 
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)</math>
+
:<math>\xi(s) = F_{0,N}(s) + \overline{F_{0,M}(\overline{1-s})} + R_{0,N,M}(s) \quad(2.1)</math>
 
where
 
where
:<math>F_{0,N}(s) := \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \sum_{n=1}^N \frac{1}{n^s}</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>
 
and
 
and
:<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.</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>
In preparation for applying the heat equation, we now write <math>F_{0,N}(s)</math> and <math>R_{0,N,M}(s)</math> as Fourier-Laplace transforms for <math>s</math> in the first quadrant.  Firstly we have
+
:<math>\frac{s(s-1)}{2} \Gamma(s/2) = 2 \Gamma(\frac{s+4}{2}) - 3 \Gamma(\frac{s+2}{2})</math>
+
so that
+
:<math>F_{0,N}(s) = \sum_{n=1}^N 2 \Gamma(\frac{s+4}{2}) (\pi n^2)^{-s/2} - 3 \Gamma(\frac{s+2}{2}) (\pi n^2)^{-s/2}.</math>
+
Since (by a change of variables <math>x=e^u</math>) we have
+
:<math>\Gamma(z) = \int_0^\infty x^{s} e^{-x}\ \frac{dx}{x} = \int_{-\infty}^\infty \exp( su - e^u )\ du \quad (1)</math>
+
we thus have
+
:<math>F_{0,N}(s) = \sum_{n=1}^N 2 \int_{-\infty}^\infty \exp( \frac{s+4}{2} u - e^u - \frac{s}{2} \log \pi n^2 )\ du
+
- 3 \int_{-\infty}^\infty \exp( \frac{s+2}{2} u - e^u - \frac{s}{2} \log \pi n^2 )\ du.</math>
+
We can shift the <math>u</math> contour to a contour <math>\Gamma</math> that starts at <math>i\infty</math> and ends at <math>+\infty</math>, staying within a bounded distance of the lower imaginary and right real axes.    One should think of the first summand as the main term and the second one as a lower order term (about <math>O(1/|s|)</math> smaller in practice).
+
In <math>s</math> coordinates, the backwards heat equation becomes
+
:<math>\partial_t F_{0,N}(s) = \frac{1}{4} \partial_{ss} F_{0,N}(s)</math>
+
and so the evolution becomes
+
:<math>F_{t,N}(s) = \sum_{n=1}^N 2 \int_{\Gamma} \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_{\Gamma} \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.</math>
+
Note that because we shifted contours to <math>\Gamma</math>, the integrand here remains absolutely integrable.
+
If we shift the first variable of integration by <math>\log \frac{s+4}{2}</math> and the second by <math>\log \frac{s+2}{2}</math>, we obtain
+
:<math>F_{t,N}(s) = 2 \sum_{n=1}^N \frac{\exp( \frac{s+4}{2} \log \frac{s+4}{2} - \frac{s+4}{2})}{(\pi n^2)^{s/2}} \int_\Gamma \exp( \frac{s+4}{2} (1 + u - e^u) + \frac{t}{16} (u + \log \frac{s+4}{2\pi n^2})^2 )\ du</math>
+
:<math> - 3 \sum_{n=1}^N \frac{\exp( \frac{s+2}{2} \log \frac{s+2}{2} - \frac{s+2}{2})}{(\pi n^2)^{s/2}} \int_\Gamma \exp( \frac{s+2}{2} (1 + u - e^u) + \frac{t}{16} (u + \log \frac{s+2}{2\pi n^2})^2 )\ du.</math>
+
Now we manipulate <math>R_{0,N,M}(s)</math>.  Firstly, 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> we have
+
:<math> \Gamma(s/2) \Gamma(1-s) = \frac{\pi^{1/2}}{2^s \sin(\pi s/2)} \Gamma(\frac{1-s}{2})</math>
+
and thus
+
:<math>R_{0,N,M}(s) = \frac{s(s-1)}{2} \frac{\pi^{(1-s)/2}}{2^s \sin(\pi s/2)} \Gamma(\frac{1-s}{2}) \frac{e^{-i\pi s}}{2\pi i} \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1}\ dw.</math>
+
Next, we have
+
:<math>\frac{s(s-1)}{2} \Gamma(\frac{1-s}{2}) = 2 \Gamma(\frac{5-s}{2}) - 3 \Gamma(\frac{3-s}{2})</math>
+
and hence by (1)
+
:<math>R_{0,N,M}(s) = 2 \frac{\pi^{(1-s)/2}}{2^s \sin(\pi s/2)} \frac{e^{-i\pi s}}{2\pi i} \int_{-\infty}^\infty \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 \frac{\pi^{(1-s)/2}}{2^s \sin(\pi s/2)} \frac{e^{-i\pi s}}{2\pi i} \int_{-\infty}^\infty \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1} \exp( \frac{3-s}{2} u - e^u ) \ dw du.</math>
+
We can shift the <math>u</math> contour to a contour <math>\overline{\Gamma}</math> that starts at <math>-i\infty</math> and ends at <math>+\infty</math>, staying within a bounded distance of the lower imaginary and right real axes.  Next, from the geometric series formula one has
+
:<math> \frac{1}{\sin(\pi s/2)} = -2i e^{i\pi s/2} \sum_{n=0}^\infty e^{\pi i n s}</math>
+
and hence
+
:<math>R_{0,N,M}(s) = - 2 \frac{\pi^{(-s-1)/2}}{2^s} \sum_{n=0}^\infty e^{\pi i (n-\frac{1}{2})s} \int_{\overline{\Gamma}} \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 \frac{\pi^{(-s-1)/2}}{2^s} \sum_{n=0}^\infty e^{\pi i (n-\frac{1}{2})s} \int_{\overline{\Gamma}} \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1} \exp( \frac{3-s}{2} u - e^u ) \ dw du.</math>
+
We can then evolve by the heat flow to obtain
+
:<math>R_{t,N,M}(s) = - 2 \frac{\pi^{(-s-1)/2}}{2^s} \sum_{n=0}^\infty e^{\pi i (n-\frac{1}{2})s} \int_\Gamma \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1} \exp( \frac{5-s}{2} u - e^u + \frac{t}{16} (-u + \pi i (2n-1) + \log \frac{w^2}{4\pi})^2 ) \ dw du</math>
+
:<math> + 3 \frac{\pi^{(-s-1)/2}}{2^s} \sum_{n=0}^\infty e^{\pi i (n-\frac{1}{2})s} \int_{\overline{\Gamma}} \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1} \exp( \frac{3-s}{2} u - e^u + \frac{t}{16} (-u + \pi i (2n-1) + \log \frac{w^2}{4\pi})^2 ) ) \ dw du,</math>
+
noting that the integrals here are absolutely convergent for <math>t > 0</math>.  We now have the exact formula
+
  
:<math>H_t(x+iy) = \frac{1}{8} ( F_{t,N}(\frac{1+ix-y}{2}) + \overline{F_{t,M}(\frac{1+ix+y}{2})} + R_{t,N,M}( \frac{1+ix+y}{2} ) ).</math>
+
== A contour integral ==
  
A good choice for <math>N,M</math> appears to be <math>N=M=\lfloor \sqrt{x/4\pi}\rfloor</math>.
+
'''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
Now for asymptotics.  For <math>\lambda</math> in the first quadrant, one has
+
:<math>\Psi(\alpha) := \int_L \frac{\exp( \frac{i}{4\pi} z^2 + \alpha z)}{e^z - 1}\ dz</math>
:<math> \int_\Gamma \exp( \lambda( 1+u-e^u) )\ du = \exp( \lambda - \lambda \log \lambda ) \Gamma(\lambda) </math>
+
can be given explicitly by the formula
and then on integrating by parts
+
:<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>.
:<math> \int_\Gamma \exp( \lambda( 1+u-e^u) ) (1 - e^u)\ du = 0</math>
+
and
+
:<math> \int_\Gamma \exp( \lambda( 1+u-e^u) ) u (1 - e^u)\ du = -\frac{1}{\lambda} \exp( \lambda - \lambda \log \lambda ) \Gamma(\lambda).</math>
+
This suggests the stationary phase asymptotic
+
:<math> \int_\Gamma \exp( \lambda( 1+u-e^u) ) f(u) = \exp( \lambda - \lambda \log \lambda ) \Gamma(\lambda) (f(0) + \frac{f''(0) - f'(0)}{2 \lambda}
+
+ O( \|f\|/\lambda^2) )</math>
+
for reasonable functions <math>f</math>, with a reasonable norm <math>\|f\|</math>.  This suggests
+
:<math>F_{t,N}(s) \approx 2 \sum_{n=1}^N \frac{\Gamma( \frac{s+4}{2} )}{(\pi n^2)^{s/2}} \exp( \frac{t}{16} \log^2 \frac{s+4}{2\pi n^2} )
+
(1 + (\frac{t}{16} + \frac{t^2}{128} \log^2 \frac{s+4}{2\pi n^2} - \frac{t}{16} \log \frac{s+4}{2\pi n^2} ) \frac{2}{s+4} )</math>
+
:<math>- 3  \sum_{n=1}^N \frac{\Gamma( \frac{s+2}{2} )}{(\pi n^2)^{s/2}} \exp( \frac{t}{16} \log^2 \frac{s+2}{2\pi n^2} ).</math>
+
Now we heuristically estimate <math>R_{t,N,M}(s) </math> term to top order.  We discard the second term as being lower order.  The critical point for <math>u</math> is <math>u = \log \frac{5-s}{2}</math>, so we approximate <math>\frac{t}{16} (-u + \pi i (2n-1) + \log \frac{w^2}{4\pi})^2 )</math> by <math>\frac{t}{16} (\pi i (2n-1) + \log \frac{w^2}{2\pi(5-s)})^2 )</math> and arrive at
+
:<math>R_{t,N,M}(s) \approx - 2 \Gamma(\frac{5-s}{2}) \frac{\pi^{(-s-1)/2}}{2^s} \sum_{n=0}^\infty e^{\pi i (n-\frac{1}{2})s}  \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1} \exp( \frac{t}{16} (\pi i (2n-1) + \log \frac{w}{2\pi(5-s)})^2 ) \ dw.</math>
+
The critical point of <math>w^s e^{-Nw}</math> occurs at <math>w = \frac{s}{N}</math>.  Approximating <math>\log \frac{w^2}{2\pi(5-s)}</math> by <math>\log \frac{s^2}{2\pi N^2 (5-s)} \approx \frac{\pi i}{2}</math>, we thus have
+
:<math>R_{t,N,M}(s) \approx - 2 \Gamma(\frac{5-s}{2}) \frac{\pi^{(-s-1)/2}}{2^s} \sum_{n=0}^\infty e^{\pi i (n-\frac{1}{2})s}  \exp( \frac{t}{16} (\pi i (2n-1/2))^2 ) \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1} \ dw.</math>
+
The <math>n=0</math> term should dominate quite strongly, thus
+
:<math>R_{t,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} ) \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1} \ dw.</math>
+
To compute the integral, we make the change of variables <math>w = 2\pi i M + z</math> and note that
+
:<math> w^{s-1} e^{-Nw} = (2\pi i M)^{s-1} \exp( - 2\pi i MN ) \exp( (s-1) \log(1 + \frac{z}{2\pi i M}) - N z)</math>
+
:<math> \approx (2\pi i M)^{s-1}\exp(  \frac{i}{4\pi} z^2 + \frac{s-2\pi i MN}{2\pi i M} z)</math>
+
so we expect
+
:<math> \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1} \ dw \approx (2\pi i M)^{s-1} \Phi(\frac{s-2\pi i MN}{2\pi i M})</math>
+
where
+
:<math>\Phi(\alpha) := \int_C \frac{\exp( \frac{i}{4\pi} z^2 + \alpha z)}{e^z - 1}\ dz</math>
+
where <math>C</math> is a contour that goes counter-clockwise around the origin (as well as all the lower imaginary zeroes of <math>e^{z-1}</math>).
+
 
+
We can calculate the integral explicitly:
+
 
+
'''Lemma 2''' For any complex <math>\alpha</math>, we have <math>\Phi(\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
 
'''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>\Phi(\alpha) = 2\pi i + \int_C \frac{\exp( \frac{i}{4\pi} (z-2\pi i)^2 + \alpha (z-2\pi i) )}{e^z-1}\ dz.</math>  
+
:<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
 
The right-hand side expands as
:<math>2\pi i - e^{-2\pi i \alpha} \int_C \frac{\exp( \frac{i}{4\pi} z^2 + (\alpha+1) z)}{e^z-1}\ dz</math>
+
:<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
 
which we can write as
:<math>2\pi i - e^{-2\pi i \alpha} (\Phi(\alpha) + \int_C \exp( \frac{i}{4\pi} z^2 + \alpha z\ dz).</math>
+
:<math>-2\pi i - e^{-2\pi i \alpha} (\Phi(\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
+
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>\Phi(\alpha) = 2\pi i - e^{-2\pi i \alpha} (\Phi(\alpha) + \sqrt{\frac{\pi}{i/4\pi}} \exp( \pi i \alpha^2)),</math>
+
:<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>
 
and the claim then follows after some algebra. <math>\Box</math>
  
We conclude that
+
We conclude from (2.3) that
:<math>R_{t,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} \Phi(\frac{s-2\pi i MN}{2\pi i M})</math>
+
:<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} \Phi(\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} \Phi(\frac{s}{2\pi i M} - N).</math>
 
:<math> = i \Gamma(\frac{5-s}{2}) \pi^{-(s+1)/2} \exp( -\frac{t \pi^2}{64} ) (\pi M)^{s-1} \Phi(\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} \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} \Psi(\frac{s}{2\pi i M}-N). (4.1)</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_{-\infty}^\infty \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) > 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) 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_{-\infty}^\infty \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_{-\infty}^\infty \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1} \exp( \frac{3-s}{2} u - e^u)\ dw\ du</math>
 +
and 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_{-\infty}^\infty \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_{-\infty}^\infty \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>0</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>, andhence
 +
:<math>\log \frac{w}{2\sqrt{\pi}} - \frac{u}{2} \approx i \pi/4</math>
 +
and so we may heuristically
 +
:<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} \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>0</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 > \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].

Revision as of 11:00, 7 February 2018

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]

where

[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]

and

[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} (\Phi(\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} \Phi(\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} \Phi(\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} \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} \Psi(\frac{s}{2\pi i M}-N). (4.1)[/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_{-\infty}^\infty \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) 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_{-\infty}^\infty \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_{-\infty}^\infty \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1} \exp( \frac{3-s}{2} u - e^u)\ dw\ du[/math]

and 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_{-\infty}^\infty \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_{-\infty}^\infty \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], andhence

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

and so we may heuristically

[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} \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].