Polymath15 test problem

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We are initially focusing attention on the following

Test problem For [math]t=y=0.4[/math], can one prove that [math]H_t(x+iy) \neq 0[/math] for all [math]x \geq 0[/math]?

If we can show this, it is likely that (with the additional use of the argument principle, and some further information on the behaviour of [math]H_t(x+iy)[/math] at [math]y=0.4[/math]) that one can show that [math]H_t(x+iy) \neq 0[/math] for all [math]y \geq 0.4[/math] as well. This would give a new upper bound

[math]\Lambda \leq 0.4 + \frac{1}{2} (0.4)^2 = 0.48[/math]

for the de Bruijn-Newman constant.

For very small values of [math]x[/math] we expect to be able to establish this by direct calculation of [math]H_t(x+iy)[/math]. For medium or large values, the strategy is to use a suitable approximation

[math]H_t(x+iy) \approx A + B[/math]

for some relatively easily computable quantities [math]A = A_t(x+iy), B = B_t(x+iy)[/math] (it may possibly be necessary to use a refined approximation [math]A+B-C[/math] instead). The quantity [math]B[/math] contains a non-zero main term [math]B_0[/math] which is expected to roughly dominate. To show [math]H_t(x+iy)[/math] is non-zero, it would suffice to show that

[math] \frac{|H_t - A - B|}{|B_0|} \lt \frac{|A + B|}{|B_0|}.[/math]

Thus one will seek upper bounds on the error [math]\frac{|H_t - A - B|}{|B_0|}[/math] and lower bounds on [math]\frac{|A+B|}{|B_0|}[/math] for various ranges of [math]x[/math]. Numerically it seems that the RHS stays above 0.4 as soon as [math]x[/math] is moderately large, while the LHS stays below 0.1, which looks promising for the rigorous arguments.

Choices of approximation

There are a number of slightly different approximations we have used in previous discussion. The first approximation was [math]A+B[/math], where

[math]A := \frac{1}{8} \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}[/math]
[math]B := \frac{1}{8} \frac{s(s-1)}{2} \pi^{-(1-s)/2} \Gamma((1-s)/2) \sum_{n=1}^N \frac{\exp(\frac{t}{16} \log^2 \frac{5-s}{2\pi n^2})}{n^{1-s}}[/math]
[math]B_0 := \frac{1}{8} \frac{s(s-1)}{2} \pi^{-(1-s)/2} \Gamma((1-s)/2) \exp( \frac{t}{16} \log^2 \frac{5-s}{2\pi} )[/math]
[math]s := \frac{1-y+ix}{2}[/math]
[math]N := \lfloor \sqrt{\frac{\mathrm{Im} s}{2\pi}} \rfloor = \lfloor \sqrt{\frac{x}{4\pi}} \rfloor.[/math]

There is also the refinement [math]A+B-C[/math], where

[math] C:= \frac{1}{8} \exp(-\frac{t\pi^2}{64}) \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \frac{e^{-i\pi s} \Gamma(1-s)}{2\pi i} (2\pi i N)^{s-1} \Psi( \frac{s}{2\pi i N}-N )[/math]
[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 i}{8}).[/math]

The first approximation was modified slightly to [math]A'+B'[/math], where

[math]A' := \frac{2}{8} \pi^{-s/2} \sqrt{2\pi} \exp( (\frac{s+4}{2}-\frac{1}{2}) \log \frac{s+4}{2} - \frac{s+4}{2}) \sum_{n=1}^N \frac{\exp(\frac{t}{16} \log^2 \frac{s+4}{2\pi n^2})}{n^s}[/math]
[math]B' := \frac{2}{8} \pi^{-(1-s)/2} \sqrt{2\pi} \exp( (\frac{5-s}{2}-\frac{1}{2}) \log \frac{5-s}{2} - \frac{5-s}{2}) \sum_{n=1}^N \frac{\exp(\frac{t}{16} \log^2 \frac{5-s}{2\pi n^2})}{n^{1-s}}[/math]
[math]B'_0 := \frac{2}{8} \pi^{-(1-s)/2} \sqrt{2\pi} \exp( (\frac{5-s}{2}-\frac{1}{2}) \log \frac{5-s}{2} - \frac{5-s}{2}) \exp( \frac{t}{16} \log^2 \frac{5-s}{2\pi} ) [/math]
[math]s := \frac{1-y+ix}{2}[/math]
[math]N := \lfloor \sqrt{\frac{\mathrm{Im} s}{2\pi}} \rfloor = \lfloor \sqrt{\frac{x}{4\pi}} \rfloor.[/math]

In Effective bounds on H_t - second approach, a more refined approximation [math]A^{eff} + B^{eff}[/math] was introduced:

[math] A^{eff} := \frac{1}{8} \exp( \frac{t}{4} \alpha_1(\frac{1-y+ix}{2})^2 ) H_{0,1}(\frac{1-y+ix}{2}) \sum_{n=1}^N \frac{1}{n^{\frac{1-y+ix}{2} + \frac{t \alpha_1(\frac{1-y+ix}{2})}{2} - \frac{t}{4} \log n}}[/math]
[math] B^{eff} := \frac{1}{8} \exp( \frac{t}{4} \overline{\alpha_1(\frac{1+y+ix}{2})}^2 ) \overline{H_{0,1}(\frac{1+y+ix}{2})} \sum_{n=1}^N \frac{1}{n^{\frac{1+y-ix}{2} + \frac{t \overline{\alpha_1(\frac{1+y+ix}{2})}}{2} - \frac{t}{4} \log n}}[/math]
[math] B^{eff}_0 := \frac{1}{8} \exp( \frac{t}{4} \overline{\alpha_1(\frac{1+y+ix}{2})}^2 ) \overline{H_{0,1}(\frac{1+y+ix}{2})} [/math]
[math]H_{0,1}(s) := \frac{s (s-1)}{2} \pi^{-s/2} \sqrt{2\pi} \exp( (\frac{s}{2} - \frac{1}{2}) \log \frac{s}{2} - \frac{s}{2} )[/math]
[math] \alpha_1(s) := \frac{1}{2s} + \frac{1}{s-1} + \frac{1}{2} \log \frac{s}{2\pi} [/math]
[math] N := \lfloor \sqrt{ \frac{T'}{2\pi}} \rfloor[/math]
[math] T' := \frac{x}{2} + \frac{\pi t}{8}.[/math]

There is a refinement [math]A^{eff}+B^{eff}-C^{eff}[/math], where

[math]C^{eff} := \frac{1}{8} \exp( \frac{t\pi^2}{64}) \frac{s'(s'-1)}{2} (-1)^N ( \pi^{-s'/2} \Gamma(s'/2) a^{-\sigma} C_0(p) U + \pi^{-(1-s')/2} \Gamma((1-s')/2) a^{-(1-\sigma)} \overline{C_0(p)} \overline{U})[/math]
[math]s' := \frac{1-y}{2} + iT' = \frac{1-y+ix}{2} + \frac{\pi i t}{8} [/math]
[math]a := \sqrt{\frac{T'}{2\pi}}[/math]
[math]p := 1 - 2(a-N)[/math]
[math]\sigma := \mathrm{Re} s' = \frac{1-y}{2}[/math]
[math]U := \exp( -i (\frac{T'}{2} \log \frac{T'}{2\pi} - \frac{T'}{2} - \frac{\pi}{8} ))[/math]
[math]C_0(p) := \frac{ \exp( \pi i (p^2/2 + 3/8) )- i \sqrt{2} \cos(\pi p/2)}{2 \cos(\pi p)}.[/math]

One can also replace [math]C^{eff}[/math] by the very slightly different quantity

[math]\tilde C^{eff} :=\frac{2 e^{-\pi i y/8}}{8} \exp( \frac{t\pi^2}{64}) (-1)^N \mathrm{Re}( H_{0,1}(iT') C_0(p) U e^{\pi i/8} ).[/math]

Finally, a simplified approximation is [math]A^{toy} + B^{toy}[/math], where

[math] A^{toy} := B^{toy}_0 \exp(i ((\frac{x}{2} + \frac{\pi t}{8}) \log \frac{x}{4\pi} - \frac{x}{2} - \frac{\pi}{4} )) N^{-y} \sum_{n=1}^N \frac{1}{n^{\frac{1-y+ix}{2} + \frac{t}{4} \log \frac{N^2}{n} + \pi i t/8}}[/math]
[math] B^{toy} := B^{toy}_0 \sum_{n=1}^N \frac{1}{n^{\frac{1+y-ix}{2} + \frac{t}{4} \log \frac{N^2}{n} - \pi i t/8}}[/math]
[math] B^{toy}_0 := \frac{\sqrt{2}}{4} \pi^2 N^{\frac{7+y}{2}} \exp( i (-\frac{x}{4} \log \frac{x}{4\pi} + \frac{x}{4} + \frac{9-y}{8} \pi) + \frac{t}{16} (\log \frac{x}{4\pi} - \frac{\pi i}{2})^2 ) e^{-\pi x/8} [/math]
[math] N := \lfloor \sqrt{\frac{x}{4\pi}} \rfloor.[/math]

Here is a table comparing the size of the various main terms:

[math]x[/math] [math]B_0[/math] [math]B'_0[/math] [math]B^{eff}_0[/math] [math]B^{toy}_0[/math]
[math]10^3[/math] [math](3.4405 + 3.5443 i) \times 10^{-167}[/math] [math](3.4204 + 3.5383 i) \times 10^{-167}[/math] [math](3.4426 + 3.5411 i) \times 10^{-167}[/math] [math](2.3040 + 2.3606 i) \times 10^{-167}[/math]
[math]10^4[/math] [math](-1.1843 - 7.7882 i) \times 10^{-1700}[/math] [math](-1.1180 - 7.7888 i) \times 10^{-1700}[/math] [math](-1.1185 - 7.7879 i) \times 10^{-1700}[/math] [math](-1.1155 - 7.5753 i) \times 10^{-1700}[/math]
[math]10^5[/math] [math](-7.6133 + 2.5065 i) * 10^{-17047} [/math] [math](-7.6134 + 2.5060 i) * 10^{-17047} [/math] [math](-7.6134 + 2.5059 i) * 10^{-17047} [/math] [math](-7.5483 + 2.4848 i) * 10^{-17047} [/math]
[math]10^6[/math] [math](-3.1615 - 7.7093 i) * 10^{-170537} [/math] [math](-3.1676 - 7.7063 i) * 10^{-170537} [/math] [math](-3.1646 - 7.7079 i) * 10^{-170537} [/math] [math](-3.1590 - 7.6898 i) * 10^{-170537} [/math]
[math]10^7[/math] [math](2.1676 - 9.6330 i) * 10^{-1705458} [/math] [math](2.1711 - 9.6236 i) * 10^{-1705458} [/math] [math](2.1571 - 9.6329 i) * 10^{-1705458} [/math] [math](2.2566 - 9.6000 i) * 10^{-1705458} [/math]

Here some typical values of [math]B/B_0[/math] (note that [math]B/B_0[/math] and [math]B'/B'_0[/math] are identical):

[math]x[/math] [math]B/B_0[/math] [math]B'/B'_0[/math] [math]B^{eff}/B^{eff}_0[/math] [math]B^{toy}/B^{toy}_0[/math]
[math]10^3[/math] [math]0.7722 + 0.6102 i[/math] [math]0.7722 + 0.6102 i[/math] [math]0.7733 + 0.6101 i[/math] [math]0.7626 + 0.6192 i[/math]
[math]10^4[/math] [math]0.7434 - 0.0126 i[/math] [math]0.7434 - 0.0126 i[/math] [math]0.7434 - 0.0126 i[/math] [math]0.7434 - 0.0124 i[/math]
[math]10^5[/math] [math]1.1218 - 0.3211 i[/math] [math]1.1218 - 0.3211 i [/math] [math]1.1218 - 0.3211 i [/math] [math]1.1219 - 0.3213 i [/math]
[math]10^6[/math] [math]1.3956 - 0.5682 i[/math] [math]1.3956 - 0.5682 i[/math] [math]1.3955 - 0.5682 i[/math] [math]1.3956 - 0.5683 i[/math]
[math]10^7[/math] [math]1.6400 + 0.0198 i[/math] [math]1.6400 + 0.0198 i[/math] [math]1.6401 + 0.0198 i[/math] [math]1.6400 - 0.0198 i[/math]

Here some typical values of [math]A/B_0[/math], which seems to be about an order of magnitude smaller than [math]B/B_0[/math] in many cases:

[math]x[/math] [math]A/B_0[/math] [math]A'/B'_0[/math] [math]A^{eff}/B^{eff}_0[/math] [math]A^{toy}/B^{toy}_0[/math]
[math]10^3[/math] [math]-0.3856 - 0.0997 i[/math] [math]-0.3857 - 0.0953 i[/math] [math]-0.3854 - 0.1002 i[/math] [math]-0.4036 - 0.0968 i[/math]
[math]10^4[/math] [math]-0.2199 - 0.0034 i[/math] [math]-0.2199 - 0.0036 i[/math] [math]-0.2199 - 0.0033 i[/math] [math]-0.2208 - 0.0033 i[/math]
[math]10^5[/math] [math]0.1543 + 0.1660 i[/math] [math]0.1543 + 0.1660 i [/math] [math]0.1543 + 0.1660 i [/math] [math]0.1544 + 0.1663 i [/math]
[math]10^6[/math] [math]-0.1013 - 0.1887 i[/math] [math]-0.1010 - 0.1889 i[/math] [math]-0.1011 - 0.1890 i[/math] [math]-0.1012 - 0.1888 i[/math]
[math]10^7[/math] [math]-0.1018 + 0.1135 i[/math] [math]-0.1022 + 0.1133 i[/math] [math]-0.1025 + 0.1128 i[/math] [math]-0.0986 + 0.1163 i[/math]

Here some typical values of [math]C/B_0[/math], which is significantly smaller than either [math]A/B_0[/math] or [math]B/B_0[/math]:

[math]x[/math] [math]C/B_0[/math] [math]C^{eff}/B^{eff}_0[/math]
[math]10^3[/math] [math]-0.1183 + 0.0697i[/math] [math]-0.0581 + 0.0823 i[/math]
[math]10^4[/math] [math]-0.0001 - 0.0184 i[/math] [math]-0.0001 - 0.0172 i[/math]
[math]10^5[/math] [math]-0.0033 - 0.0005i[/math] [math]-0.0031 - 0.0005i[/math]
[math]10^6[/math] [math]-0.0001 - 0.0006 i[/math] [math]-0.0001 - 0.0006 i[/math]
[math]10^7[/math] [math]-0.0000 - 0.0001 i[/math] [math]-0.0000 - 0.0001 i[/math]

Some values of [math]H_t[/math] and its approximations at small values of [math]x[/math] source source:

[math]x[/math] [math]H_t[/math] [math]A+B[/math] [math]A'+B'[/math] [math]A^{eff}+B^{eff}[/math] [math]A^{toy}+B^{toy}[/math] [math]A+B-C[/math] [math]A^{eff}+B^{eff}-C^{eff}[/math]
[math]10[/math] [math](3.442 - 0.168 i) \times 10^{-2}[/math] 0 0 0 N/A N/A [math](3.501 - 0.316 i) \times 10^{-2}[/math]
[math]30[/math] [math](-1.000 - 0.071 i) \times 10^{-4}[/math] [math](-0.650 - 0.188 i) \times 10^{-4}[/math] [math](-0.211 - 0.192 i) \times 10^{-4}[/math] [math](-0.670 - 0.114 i) \times 10^{-4}[/math] [math](-0.136 + 0.021 i) \times 10^{-4}[/math] [math](-1.227 - 0.058 i) \times 10^{-4}[/math] [math](-1.032 - 0.066 i) \times 10^{-4}[/math]
[math]100[/math] [math](6.702 + 3.134 i) \times 10^{-16}[/math] [math](2.890 + 3.667 i) \times 10^{-16}[/math] [math](2.338 + 3.742 i) \times 10^{-16}[/math] [math](2.955 + 3.650 i) \times 10^{-16}[/math] [math](0.959 + 0.871 i) \times 10^{-16}[/math] [math](6.158 + 12.226 i) \times 10^{-16}[/math] [math](6.763 + 3.074 i) \times 10^{-16} [/math]
[math]300[/math] [math](-4.016 - 1.401 i) \times 10^{-49}[/math] [math](-5.808 - 1.140 i) \times 10^{-49}[/math] [math](-5.586 - 1.228 i) \times 10^{-49}[/math] [math](-5.824 - 1.129 i) \times 10^{-49}[/math] [math](-2.677 - 0.327 i) \times 10^{-49}[/math] [math](-3.346 + 6.818 i) \times 10^{-49}[/math] [math](-4.032 - 1.408 i) \times 10^{-49}[/math]
[math]1000[/math] [math](0.015 + 3.051 i) \times 10^{-167}[/math] [math](-0.479 + 3.126 i) \times 10^{-167}[/math] [math](-0.516 + 3.135 i) \times 10^{-167}[/math] [math](-0.474 + 3.124 i) \times 10^{-167}[/math] [math](-0.406 + 2.051 i) \times 10^{-167}[/math] [math](0.175 + 3.306 i) \times 10^{-167}[/math] [math](0.017 + 3.047 i) \times 10^{-167}[/math]
[math]3000[/math] [math](-1.144+ 1.5702 i) 10^{-507}[/math] [math](-1.039+ 1.5534 i) 10^{-507}[/math] [math](-1.039+ 1.5552 i) 10^{-507}[/math] [math](-1.038+ 1.5535 i) 10^{-507}[/math] [math](-0.925+ 1.3933 i) 10^{-507}[/math] [math](-1.155+ 1.5686 i) 10^{-507}[/math] [math](-1.144+ 1.5701 i) 10^{-507}[/math]
[math]10000[/math] [math](-0.558 - 4.088 i) \times 10^{-1700}[/math] [math](-0.692 - 4.067 i) \times 10^{-1700}[/math] [math](-0.687 - 4.067 i) \times 10^{-1700}[/math] [math](-0.692 - 4.066 i) \times 10^{-1700}[/math] [math](-0.673 - 3.948 i) \times 10^{-1700}[/math] [math](-0.548 - 4.089 i) \times 10^{-1700}[/math] [math](-0.558 - 4.088 i) \times 10^{-1700}[/math]
[math]30000[/math] [math](3.160 - 6.737) \times 10^{-5110}[/math] [math](3.065 - 6.722) \times 10^{-5100}[/math] [math](3.066 - 6.722) \times 10^{-5100}[/math] [math](3.065 - 6.722) \times 10^{-5100}[/math] [math](2.853 - 6.286) \times 10^{-5100}[/math] [math](3.170 - 6.733) \times 10^{-5100}[/math] [math](3.160 - 6.737) \times 10^{-5100}[/math]

Controlling |A+B|/|B_0|

See Controlling A+B/B_0.

Mesh evaluations of [math]A^{eff}+B^{eff}/B^{eff}_0[/math] in the ranges

Here is a table of analytic lower bounds for [math]A^{eff}+B^{eff}/B^{eff}_0[/math] for [math]3 \leq N \leq 2000[/math].

Controlling |H_t-A-B|/|B_0|

See Controlling H_t-A-B/B_0.

Here is a table on bounds on error terms [math]E_1/B^{eff}_0, E_2/B^{eff}_0, E_3^*/B^{eff}_0[/math] for N=3 to 2000. Here is a table with some sharpened estimates from the PDF writeup.

Here is a graph depicting [math]|H_t-A^{eff}-B^{eff}/B_0^{eff}|[/math] and [math]E_1+E_2+E_3^*/|B_0^{eff}|[/math] for [math]x \leq 1600[/math].

Small values of x

Tables of [math]H_t(x+iy)[/math] for small values of [math]x[/math]:

Here are some snapshots of [math]H_t/B^{eff}_0[/math].

In this range we will need Bounding the derivative of H_t or Bounding the derivative of H_t - second approach or Bounding the derivative of H_t - third approach.

Tables of [math]H'_t(x+iy)[/math]:

Here is a table of [math]x[/math], pari/gp prec, [math]H_{t}, H^{'}_{t}, |H_{t}|, |H^{'}_{t}|, \frac{|H_{t}|}{|B_{0}^{eff}|}, \frac{|H^{'}_{t}|}{|B_{0}^{eff}|}[/math] for x=0 to x=30 with step size 0.01.

Here is a plot of [math]H_t/B_0[/math] for a rectangle [math] \{x+iy: 0 \leq x \leq 300; 0.4 \leq y \leq 0.45\}[/math]. Here is an adaptive mesh plot; here is a closeup near the origin.

Here is a script for verifying the absence of zeroes of [math]H_t[/math] in a rectangle. It can eliminate zeros in the rectangle [math]\{0 \leq x \leq 1000, 0.4 \leq y \leq 0.45\}[/math] when t = 0.4.

Large negative values of [math]t[/math]

See also Second attempt at computing H_t(x) for negative t.

We heuristically compute [math]H_t(x)[/math] in the regime where [math]x[/math] is large and [math]t[/math] is large and negative with [math]|t|/x \asymp 1[/math]. We shall only be interested in the zeroes and so we discard any multiplicative factor which is non-zero: we write [math]X \sim Y[/math] if X is equal (or approximately equal) to Y times something that is explicit and non-zero.

From equation (35) of the writeup we have

[math]H_t(x) = \int_{\bf R} \frac{1}{8} \xi(\frac{1+ix}{2} + i |t|^{1/2} v) \frac{1}{\pi} e^{-v^2}\ dv \quad (3.1)[/math]
[math] \sim \int_{\bf R} \xi(\frac{1+ix}{2} + i |t|^{1/2} v) e^{-v^2}\ dv. \quad (3.2)[/math]

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) \sim \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 (3.3)[/math]
[math] \sim \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 (3.4)[/math]


[math] \tilde x := x - \pi |t|/4 = x + \frac{\pi t}{4}. \quad (3.5)[/math]

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

[math] \xi(s) \sim M_0(s) \zeta(s)\quad (3.6)[/math]

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

[math] H_t(x) \sim \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 (3.7)[/math]

By Taylor expansion we have

[math] M_0(\frac{1+i\tilde x}{2} + i |t|^{1/2} v) \sim 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 (3.8)[/math]
[math] \sim \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 (3.9)[/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 (3.10)[/math]


[math] \alpha'(\frac{1+i\tilde x}{2} ) \approx \frac{-i}{\tilde x} \quad (3.11)[/math]

and hence

[math] H_t(x) \sim \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 (3.12)[/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 (3.13)[/math]


[math]u := |t|/N^2 = 4\pi |t|/\tilde x,\quad (3.14)[/math]

we conclude that

[math] H_t(x) \sim \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 (3.15)[/math]

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

[math] H_t(x) \sim \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 (3.16)[/math]
[math] \sim \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 (3.17)[/math]

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

[math] H_t(x) \sim \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 (3.18)[/math]

Using the Taylor approximation

[math] \log \frac{n}{N} \approx \frac{n-N}{N} - \frac{(n-N)^2}{2N^2} \quad (3.19)[/math]

and dropping some small terms, we obtain

[math] H_t(x) \sim \sum_n \exp( - \frac{|t| (n-N)^2}{4 N^2 (1 - iu/8\pi)} -\frac{i\tilde x}{2} \frac{n-N}{N} + \frac{i \tilde x (n-N)^2}{4N^2} ).\quad (3.20)[/math]

Writing [math]\tilde x = 4\pi N^2[/math] and [math]|t| = u N^2[/math], this becomes

[math] H_t(x) \sim \sum_n \exp( - \frac{2\pi u (n-N)^2}{8\pi - iu} -2 \pi i N(n-N) + \pi i (n-N)^2 ).\quad (3.21)[/math]


[math] N(n-N) = \frac{1}{2} n^2 - \frac{1}{2} (n-N)^2 - \frac{1}{2} N^2 \quad (3.22)[/math]

we thus have

[math] H_t(x) \sim \sum_n \exp( - \frac{2 \pi u (n-N)^2}{8 \pi - iu} - \pi i n^2 + 2 \pi i (n-N)^2 )\quad (3.23)[/math]
[math] \sim \sum_n \exp( \frac{16 \pi^2 i (n-N)^2}{8 \pi - iu} ) e^{\pi i n}\quad (3.24)[/math]
[math] \sim \theta_{01}( \frac{16 \pi N}{8\pi - iu}, \frac{16 \pi}{8\pi - iu} )\quad (3.25)[/math]

where [math]\theta_{01}[/math] is the theta function defined in this Wikipedia page. Using the Jacobi identity we then have

[math] H_t(x) \sim \theta_{10}(N, \frac{iu - 8\pi}{16 \pi} )\quad (3.26)[/math]
[math] \sim \theta( N + \frac{1}{2} \frac{iu - 8\pi}{16 \pi}, \frac{iu - 8\pi}{16 \pi})\quad (3.27)[/math]
[math] \sim \sum_n \exp( - \pi i n(n+1) / 2 ) e^{2\pi i (n+1/2) N} e^{-u n(n+1)/16}.\quad (3.28)[/math]

As a sanity check, one can verify that the RHS is real-valued, just as [math]H_t(x)[/math] is (by the functional equation).