Polymath15 test problem: Difference between revisions

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:<math>U := \exp( -i (\frac{T'}{2} \log \frac{T'}{2\pi} - \frac{T'}{2} - \frac{\pi}{8} ))</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>
:<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
Finally, a simplified approximation is <math>A^{toy} + B^{toy}</math>, where
Line 221: Line 224:
|}
|}


== Controlling |A+B|/|B_0| ==
Some values of <math>H_t</math> and its approximations at small values of <math>x</math> [https://terrytao.wordpress.com/2018/03/02/polymath15-fifth-thread-finishing-off-the-test-problem/#comment-493456 source] [https://terrytao.wordpress.com/2018/03/02/polymath15-fifth-thread-finishing-off-the-test-problem/#comment-493715 source]:
 
See [[Controlling A+B/B_0]].
 
== Controlling |H_t-A-B|/|B_0| ==
 
As computed in [[Effective bounds on H_t - second approach]], there is an effective bound
:<math>|H_{eff} - A^{eff} - B^{eff}| \leq E_1 + E_2 + E_3</math>
where
:<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> E_1 := \frac{1}{8 (T - 3.33)} \exp( \frac{t}{4} \mathrm{Re} \alpha_1(\frac{1-y+ix}{2})^2 ) |H_{0,1}(\frac{1-y+ix}{2})| \epsilon'(\frac{1-y+ix}{2}) </math>
:<math> E_2 := \frac{1}{8 (T - 3.33)} \exp( \frac{t}{4} \mathrm{Re} \alpha_1(\frac{1+y+ix}{2})^2 ) |H_{0,1}(\frac{1+y+ix}{2})| \epsilon'(\frac{1+y+ix}{2}) </math>
:<math> E_3 := \frac{1}{8} \sqrt{\pi} \exp( -\frac{t \pi^2}{64} ) (T')^{3/2} e^{-\pi T/4}  \int_{-\infty}^\infty v(\sigma) w(\sigma) f(\sigma)\ d\sigma</math>
:<math> \epsilon'(s) := \frac{1}{2} \sum_{n=1}^N \frac{1}{n^{\mathrm{Re}(s) + \frac{t \mathrm{Re} \alpha_1(s)}{2} - \frac{t}{4} \log n}}
\exp(\frac{1}{2(T-3.33)} (\frac{t^2}{4} |\alpha_1(s) - \log n|^2 + \frac{1}{3} + t))
(\frac{t^2}{4} |\alpha_1(s) - \log n|^2 + \frac{1}{3} + t ) </math>
:<math> f(\sigma) := \frac{1}{2\sqrt{\pi t}} (e^{-(\sigma-(1-y)/2)^2/t} + e^{-(\sigma-(1+y)/2)^2/t}) \quad (4.1)</math>
:<math>w(\sigma) := (1 + \frac{\sigma^2}{(T'_0)^2})^{1/2} (1 + \frac{(1-\sigma)^2}{(T'_0)^2})^{1/2}
\exp( \frac{(\sigma-1)_+}{4} \log (1 + \frac{\sigma^2}{(T'_0)^2}) + (\frac{T'_0}{2} \arctan \frac{\sigma}{T'_0} - \frac{\sigma}{2}) 1_{\sigma < 0} + \frac{1}{12(T'_0 - 0.33)}) </math>
:<math>v(\sigma) := 1 + (0.400 \frac{9^\sigma}{a_0} + 0.346 \frac{2^{3\sigma/2}}{a_0^2}) 1_{\sigma \geq 0} + (9/10)^{\lceil -\sigma \rceil} \sum_{1 \leq k \leq 4-\sigma} (1.1)^k \frac{\Gamma(k/2)}{a_0^k} 1_{\sigma < 0} </math>
:<math>a_0 := \sqrt{\frac{T'_0}{2\pi}}</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>
:<math> T'_0 := T_0 + \frac{\pi t}{8} </math>
 
 
Comparison between <math>H^{eff} = A^{eff}+B^{eff}</math>, <math>A'+B'</math>, and the effective error bound <math>E_1+E_2+E_3</math> on <math>H - H^{eff}</math> at some points of <math>x</math> [https://terrytao.wordpress.com/2018/02/12/polymath15-third-thread-computing-and-approximating-h_t/#comment-492893 source]:


{| border=1
{| border=1
|-
|-
! style="text-align:left;"| <math>x</math>
! style="text-align:left;"| <math>x</math>
! <math>|H^{eff}/B'_0|</math>
! <math>H_t</math>
! <math>|(A'+B')/B'_0|</math>
! <math>A+B</math>
! <math>|(H^{eff}-(A'+B'))/B'_0|</math>
! <math>A'+B'</math>
! <math>|(H^{eff}-(A'+B'))/B'_0| + |(E_1+E_2+E_3)/B'_0|</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>
|-
|-
|10000
| <math>10</math>
|0.52
| <math>(3.442 - 0.168 i) \times 10^{-2}</math>
|0.52
| 0
|0.0006
| 0
|0.039
| 0
| N/A
| N/A
| <math>(3.501 - 0.316 i) \times 10^{-2}</math>
|-
|-
|12131
| <math>30</math>
|1.28
| <math>(-1.000 - 0.071 i) \times 10^{-4}</math>
|1.28
| <math>(-0.650 - 0.188 i) \times 10^{-4}</math>
|0.0004
| <math>(-0.211 - 0.192 i) \times 10^{-4}</math>
|0.033
| <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>
|-
|-
|15256
| <math>100</math>
|0.97
| <math>(6.702 + 3.134 i) \times 10^{-16}</math>
|0.97
| <math>(2.890 + 3.667 i) \times 10^{-16}</math>
|0.0003
| <math>(2.338 + 3.742 i) \times 10^{-16}</math>
|0.027
| <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>
|-
|-
|18432
| <math>300</math>
|0.68
| <math>(-4.016 - 1.401 i) \times 10^{-49}</math>
|0.68
| <math>(-5.808 - 1.140 i) \times 10^{-49}</math>
|0.0003
| <math>(-5.586 - 1.228 i) \times 10^{-49}</math>
|0.023
| <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>
|-
|-
|20567
| <math>1000</math>
|0.98
| <math>(0.015 + 3.051 i) \times 10^{-167}</math>
|0.98
| <math>(-0.479 + 3.126 i) \times 10^{-167}</math>
|0.0004
| <math>(-0.516 + 3.135 i) \times 10^{-167}</math>
|0.022
| <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>
|-
|-
|30654
| <math>3000</math>
|1.93
| <math>(-1.144+ 1.5702 i) 10^{-507}</math>
|1.93
| <math>(-1.039+ 1.5534 i) 10^{-507}</math>
|0.0004
| <math>(-1.039+ 1.5552 i) 10^{-507}</math>
|0.016
| <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>
|}
|}


The <math>E_3</math> error dominates the other two [https://terrytao.wordpress.com/2018/02/12/polymath15-third-thread-computing-and-approximating-h_t/#comment-492922 source]:
== Controlling |A+B|/|B_0| ==


{| border=1
See [[Controlling A+B/B_0]].
|-
 
! style="text-align:left;"| <math>x</math>
Mesh evaluations of <math>A^{eff}+B^{eff}/B^{eff}_0</math> in the ranges
! <math>\frac{E_3}{E_1+E_2}</math>
 
|-
* [https://drive.google.com/open?id=1qbkvCBIt_OHnrtcDJWQvN_FUt-1zPv4m N between 11 and 19]
|10000
* [https://drive.google.com/open?id=1YBIA5gRv2DUXX74MLwfn9F2J0QJBWUh_ N between 20 and 150] ([https://drive.google.com/open?id=1ZBX7jNGXhQZQ50t8UX4boTPW93dmRAun raw data]
|9.11
* [https://drive.google.com/file/d/1NvEv-1R4KTEchWMbJCpZA6Uf1xLkiYWM/view N between 151 and 300]
|-
* [https://drive.google.com/open?id=15Xf9GsaAzydl-39zyG9nei5aZyaHwsFe The (A+B)/B0 mesh data for N=300 to 20, y=0.45, t=0.4, c=0.065]
|15000
* [https://drive.google.com/open?id=1kK8tV2bRfACm1lUKRFIktca58ZV8L8U3  c=0.26 for N=7 to 19, y=0.4,t=0.4]
|14.97
* https://drive.google.com/open?id=13_mzqvtaZCghmj7oAZtDRXnkb2zbxQH3 c=0.26 for N=19 to 7, y=0.45,t=0.4]
|-
 
|20000
[https://github.com/km-git-acc/dbn_upper_bound/blob/master/dbn_upper_bound/python/research/mod_abbeff_lower_Nbounds.csv 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>].
|19.26
 
|-
== Controlling |H_t-A-B|/|B_0| ==
|50000
 
|32.39
See [[Controlling H_t-A-B/B_0]].
|-
 
|100000
[https://github.com/km-git-acc/dbn_upper_bound/blob/master/dbn_upper_bound/python/research/bounded_normalized_E1_and_E2_and_E3_and_overall_error.csv 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].  [https://github.com/km-git-acc/dbn_upper_bound/blob/master/dbn_upper_bound/python/research/e1_e2_e3_sharper_Nbound.csv Here is a table] with some sharpened estimates from the PDF writeup.
|42.99
 
|-
[https://ibb.co/b7baZc 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>.
|<math>10^7</math>
 
|87.23
== Small values of x ==
|}
 
Tables of <math>H_t(x+iy)</math> for small values of <math>x</math>:
 
* [https://github.com/km-git-acc/dbn_upper_bound/tree/master/dbn_upper_bound/python/research/H_t%20at%20small%20x x=0 to x=300 with step size 0.1]
* [https://pastebin.com/fim2swFu x=200 to x=600 with step size 0.1]
* [https://pastebin.com/jvSvDP69 x=600 to x=1000 with step size 0.1]
* [https://pastebin.com/NkFKs3pB x=1000 to x=1300 with step size 0.1]
* [https://pastebin.com/k7vC6e7n x=1300 to x=1600 with step size 0.1]
* [https://gist.githubusercontent.com/p15-git-acc/3ada0ff0b9ec77e23cb7cace0dcb8691/raw/807e1b0a16356a9bbd2a5af872f71bc064830c38/gistfile1.txt x=20 to x=1000, adaptive mesh]
 
 
[https://ibb.co/fOroa7 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]].


<math>A+B-C</math> is a good approximation to <math>H_t</math> [https://terrytao.wordpress.com/2018/02/12/polymath15-third-thread-computing-and-approximating-h_t/#comment-492695 source] [https://terrytao.wordpress.com/2018/02/24/polymath15-fourth-thread-closing-in-on-the-test-problem/#comment-493282 source] [https://terrytao.wordpress.com/2018/02/24/polymath15-fourth-thread-closing-in-on-the-test-problem/#comment-493319 source]
Tables of <math>H'_t(x+iy)</math>:


{| border=1
* [https://drive.google.com/open?id=1mtzrJ_-hBMyt90gVzJdNCq4RU6Pz_dQQ x=0 to x=100 with step size 0.1]
|-
* [https://drive.google.com/open?id=1wSZJgoPp9-6C8CJqLr5g3E1uhEBHKmTJ x=100 to x=200 with step size 0.1]
! style="text-align:left;"| <math>x</math>
* [https://drive.google.com/open?id=1q3h0x8jSF0Z1KQ9iLx23JI1oCXfrx0RL x=200 to x=300 with step size 0.1]
! <math>\frac{|H_t-(A+B-C)|}{|B_0|}</math>
! <math>\frac{|H_t-(A^{eff}+B^{eff}-C^{eff})|}{|B_0^{eff}|}</math>
|-
|160
|0.06993270565802375041
|0.009155667752
|-
|320
|0.006716674125965016299
|0.0005529962481
|-
|480
|0.005332893070605698501
|0.0004966282128
|-
|640
|0.003363431256036816251
|0.0004482768972
|-
|800
|0.1548144749150572349
|0.002644344570
|-
|960
|0.03009229958121352990
|0.001270168744
|-
|1120
|0.004507664238680722472
|0.0009957229500
|-
|1280
|0.002283591962997851167
|0.0007024411378
|-
|1440
|0.01553727684468691873
|0.0007000473085
|-
|1600
|0.001778051951547709718
|0.0004882487218
|-
|1760
|0.02763769444052338578
|0.0002518910919
|-
|1920
|0.002108779890256530964
|0.0008378989413
|-
|2080
|0.02746770886040058927
|0.0004924765754
|-
|2240
|0.001567020041379128455
|0.0001171320991
|-
|2400
|0.01801417530687959747
|0.0002443802551
|-
|2560
|0.001359561117436848149
|0.0004569058755
|-
|2720
|0.008503327577240081269
|0.0006355966221
|-
|2880
|0.001089253262122934826
|0.0008864917365
|-
|3040
|0.003004181560093288747
|0.00004326840265
|-
|3200
|0.02931455383125538672
|0.0003598521453
|}


A closer look at the "spike" in error near <math>x=800 \approx 256 \pi \approx 804 </math>:
[https://drive.google.com/open?id=1ge0TD5hvs1O6BKLAz34JmTqxSrguvjsJ 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.


{| border=1
|-
! style="text-align:left;"| <math>x</math>
! <math>\frac{|H_t-(A+B-C)|}{|B_0|}</math>
|-
|622.035345
|0.003667321
|-
|631.460123
|0.004268055
|-
|640.884901
|0.003284407
|-
|650.309679
|0.004453589
|-
|659.734457
|0.003872174
|-
|669.159235
|0.005048162
|-
|678.584013
|0.005009254
|-
|688.008791
|0.007418686
|-
|697.433569
|0.007464541
|-
|706.858347
|0.010692337
|-
|716.283125
|0.012938629
|-
|725.707903
|0.017830524
|-
|735.132681
|0.022428596
|-
|744.557459
|0.030907876
|-
|753.982237
|0.040060298
|-
|763.407015
|0.053652069
|-
|772.831793
|0.071092824
|-
|782.256571
|0.094081856
|-
|791.681349
|0.123108726
|-
|801.106127
|0.159299234
|-
|810.530905
|0.002870724
|}


In practice <math>E_1/B^{eff}_0</math> is smaller than <math>E_2/B^{eff}_0</math>, which is mostly dominated by the first term in the sum which is close to <math>\frac{t^2}{16 x} \log^2 \frac{x}{4\pi}</math>:


{| border=1
[https://drive.google.com/open?id=1855iryE-7uDDEyW7hXJ5-3Njomsz-mMH 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 [https://drive.google.com/open?id=1oGQ4HfXlEiC5WUnWHOzAt5EJQlEg9SRb an adaptive mesh plot]; here is a [https://drive.google.com/open?id=12zkFFXBF7H6Mjd1KBmLhZw10io3J_wer closeup near the origin].
|-
! style="text-align:left;"| <math>x</math>
! <math>E_1 / B^{eff}_0</math>
! <math>E_2 / B^{eff}_0</math>
! <math>\frac{t^2}{16x} \log^2 \frac{x}{4\pi}</math>
|-
|10^3
|<math>1.389 \times 10^{-3}</math>
|<math>2.341 \times 10^{-3}</math>
|<math>1.915 \times 10^{-4}</math>
|-
|10^4
|<math>1.438 \times 10^{-4}</math>
|<math>3.156 \times 10^{-4}</math>
|<math>4.461 \times 10^{-5}</math>
|-
|10^5
|<math>1.118 \times 10^{-5}</math>
|<math>3.574 \times 10^{-5}</math>
|<math>8.067 \times 10^{-6}</math>
|-
|10^6
|<math>7.328 \times 10^{-7}</math>
|<math>3.850 \times 10^{-6}</math>
|<math>1.273 \times 10^{-6}</math>
|-
|10^7
|<math>4.414 \times 10^{-8}</math>
|<math>4.197 \times 10^{-7}</math>
|<math>1.846 \times 10^{-7}</math>
|}


=== Estimation of <math>E_1,E_2</math> ===
Here is a [https://pastebin.com/TiFk6CfF 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.


...


=== Estimation of <math>E_3</math> ===
== Large negative values of <math>t</math> ==


Here we assume that <math>T_0 \geq 100</math>, which implies also <math>T'_0 \geq 100</math>.
See also [[Second attempt at computing H_t(x) for negative t]].


We first bound <math>w</math> by a Gaussian type quantity.
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.


We have
From equation (35) of the writeup we have
:<math>1 + \frac{\sigma^2}{(T'_0)^2} \leq \exp( \frac{\sigma^2}{(T'_0)^2})</math>
and
:<math>1 + \frac{(1-\sigma)^2}{(T'_0)^2} \leq \exp( \frac{(1-\sigma)^2}{(T'_0)^2})</math>
and thus
:<math>( 1 + \frac{\sigma^2}{(T'_0)^2} )^{1/2} (1 + \frac{(1-\sigma)^2}{(T'_0)^2})^{1/2} \leq \exp( \frac{1}{2} \frac{\sigma^2}{(T'_0)^2} +
\frac{1}{2} \frac{(1-\sigma)^2}{(T'_0)^2} )</math>
:<math> = \exp( \frac{(\sigma-1/2)^2}{(T'_0)^2} + \frac{1}{4 (T'_0)^2} ).</math>
Next, from calculus one can verify the bounds
:<math> \log(1+x^2) \leq 1.479 \sqrt{x}</math>
and
:<math> x - \mathrm{arctan}(x) \leq 0.230 x^2</math>
for any <math>x \geq 0</math>, and hence
:<math> \frac{(\sigma-1)_+}{4} \log (1 + \frac{\sigma^2}{(T'_0)^2}) \leq \frac{1}{4} 1.479 \frac{\sigma(\sigma-1)}{T'_0} 1_{\sigma \geq 1} </math>
:<math> \leq 0.37 \frac{(\sigma-1/2)^2}{T'_0} 1_{\sigma \geq 1}</math>
and
:<math>(\frac{T'_0}{2} \arctan \frac{\sigma}{T'_0} - \frac{\sigma}{2}) 1_{\sigma < 0} \leq \frac{T'_0}{2} 1_{\sigma<0} 0.230 (\frac{|\sigma|}{T'_0})^2 </math>
:<math> \leq 0.115 \frac{(\sigma-1/2)^2}{T'_0} 1_{\sigma < 0}.</math>
We conclude that
:<math> w(\sigma) \leq \exp(  \frac{(\sigma-1/2)^2}{(T'_0)^2} + \frac{1}{4 (T'_0)^2} + 0.37 \frac{(\sigma-1/2)^2}{T'_0} + \frac{1}{12(T'_0 - 0.33)}) </math>
:<math> \leq \exp(  0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} + \frac{1}{12(T'_0 - 3.33)}).</math>


Now we work on <math>\nu</math>.  Observe that if <math>k \leq \frac{T'_0}{2.42 \pi} = \frac{a_0^2}{1.21}</math> then
:<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> (1.1)^{k+2} \frac{\Gamma(\frac{k+2}{2})}{a_0^{k+2}} = \frac{1.21 k}{2 a_0^2} \frac{\Gamma(\frac{k}{2})}{a_0^k} \leq \frac{1}{2} (1.1)^{k} \frac{\Gamma(\frac{k}{2})}{a_0^{k}},</math>
:<math> \sim \int_{\bf R} \xi(\frac{1+ix}{2} + i |t|^{1/2} v) e^{-v^2}\ dv. \quad (3.2)</math>
and hence
:<math> \sum_{2 \leq k \leq \frac{T'_0}{2.24 \pi}; k\ \mathrm{even}} (1.1)^{k} \frac{\Gamma(\frac{k}{2})}{a_0^{k}} \leq 2 (1.1)^2 \frac{\Gamma(\frac{2}{2})}{a_0^2} = \frac{2.42 \sqrt{\pi}}{a_0^2}</math>
and similarly
:<math> \sum_{3 \leq k \leq \frac{T'_0}{2.42 \pi}; k\ \mathrm{odd}} (1.1)^{k} \frac{\Gamma(\frac{k}{2})}{a_0^{k}} \leq 2 (1.1)^3 \frac{\Gamma(\frac{3}{2})}{a_0^2} = \frac{1.331}{a_0^3}</math>
and hence
:<math> \sum_{1 \leq k \leq \frac{T'_0}{2.42 \pi}} (1.1)^{k} \frac{\Gamma(\frac{k}{2})}{a_0^{k}} \leq \frac{1.1 \sqrt{\pi}}{a_0} + \frac{2.42}{a_0^2} + \frac{1.331 \sqrt{\pi}}{a_0^3} </math>
:<math> \leq \frac{1.1 \sqrt{\pi}}{a_0 - 1.25};</math>
also
:<math>(0.400 \frac{9^\sigma}{a_0} + 0.346 \frac{2^{3\sigma/2}}{a_0^2})1_{\sigma \geq 0} \leq 0.400 \times 9^\sigma (\frac{1}{a_0} + 0.865 \frac{1}{a_0^2})</math>
:<math> \leq 0.4 \frac{9^\sigma}{a_0 - 0.865}</math>
and hence (bounding <math>(0.9)^{\lceil -\sigma \rceil} \leq \frac{1}{1.1}</math>)
:<math> v(\sigma) \leq 1 + 0.400 \frac{9^\sigma}{a_0-0.865} + \frac{\sqrt{\pi}}{a_0-1.25} + \sum_{\frac{T'_0}{2.42 \pi} < k \leq 4-\sigma} \frac{(1.1)^{k-1}}{a_0^k} \Gamma(k/2).</math>
We conclude (using Fubini's theorem) that
:<math> \int_{-\infty}^\infty vwf(\sigma)\ d\sigma \leq \exp(\frac{1}{12(T'_0 - 3.33)}) (
(1 + \frac{\sqrt{\pi}}{a_0-1.25}) \int_{-\infty}^\infty f(\sigma) \exp(  0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} )\ d\sigma </math>
:<math> + \frac{0.4}{a_0-0.865} \int_{-\infty}^\infty 9^\sigma f(\sigma) \exp(  0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} )\ d\sigma </math>
:<math> + \sum_{k > \frac{T'_0}{2.42\pi}} \frac{(1.1)^{k-1}}{a_0^k} \Gamma(k/2) \int_{-\infty}^{4-k} f(\sigma) \exp(  0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} )\ d\sigma.)</math>
Now we estimate the integrals appearing in the right-hand side.  By symmetry we have
:<math>\int_{-\infty}^\infty f(\sigma) \exp(  0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} )\ d\sigma
= \frac{1}{\sqrt{\pi t}} \int_{-\infty}^\infty \exp( - \frac{(\sigma - (1+y)/2)^2}{t} + 0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} )\ d\sigma.</math>
:<math>= \frac{1}{\sqrt{\pi t}} \int_{-\infty}^\infty \exp( - \frac{\sigma^2}{t} + 0.37 \frac{(\sigma-y/2)^2}{T'_0-2.71} )\ d\sigma.</math>


Using the Gaussian identity
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> \int_{-\infty}^\infty \exp( - (a\sigma^2 + b \sigma + c) )\ d\sigma = \sqrt{\pi} a^{-1/2} \exp( - c + \frac{b^2}{4a} ),</math>
:<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>
valid for any <math>a,b,c</math> with <math>a</math> positive, we can write the above expression as
:<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> (1 - \frac{0.37 t}{T'_0 - 2.71})^{-1/2} \exp( \frac{0.37 y^2}{4 (T'_0 - 2.71 - 0.37 t)} ).</math>
Similarly, since <math>9^\sigma</math> is larger for <math>\sigma \geq 1/2</math> than for <math>\sigma <1/2</math>, we have
:<math>\int_{-\infty}^\infty 9^\sigma f(\sigma) \exp(  0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} )\ d\sigma
\leq \frac{1}{\sqrt{\pi t}} \int_{-\infty}^\infty 9^\sigma \exp( - \frac{(\sigma - (1+y)/2)^2}{t} + 0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} )\ d\sigma.</math>
:<math>= \frac{3^{1+y}}{\sqrt{\pi t}} \int_{-\infty}^\infty \exp( - \frac{\sigma^2}{t} + 0.37 \frac{(\sigma-y/2)^2}{T'_0-2.71} + \sigma \log 9)\ d\sigma.</math>
:<math>= 3^{1+y} (1 - \frac{0.37 t}{T'_0 - 2.71})^{-1/2} \exp( 0.37 \frac{y^2}{4 (T'_0-2.71)} + \frac{b^2}{4 (\frac{1}{t} - \frac{0.37}{T'_0-2.71})} )</math>
where
where
:<math> b := - \log 9 + 0.37 \frac{y}{T'_0 - 2.71}.</math>
:<math> \tilde x := x - \pi |t|/4 = x + \frac{\pi t}{4}. \quad (3.5)</math>
If <math>T'_0 \geq 100</math> and <math>y \leq 1/2</math> then <math>|b| \leq \log 9</math>, thus the above integral is at most
Now from the definition of <math>\xi</math> and the Stirling approximation we have
:<math>= 3^{1+y} (1 - \frac{0.37 t}{T'_0 - 2.71})^{-1/2} \exp( 0.37 \frac{y^2}{4 (T'_0-2.71)} + \frac{t \log^2 9}{4 (1 - \frac{0.37 t}{T'_0-2.71})} ).</math>
:<math> \xi(s) \sim M_0(s) \zeta(s)\quad (3.6)</math>
Now we consider the integral
where <math>M_0</math> is defined in (6) of the writeup.  Thus
:<math> \int_{-\infty}^{4-k} f(\sigma) \exp( 0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} )\ d\sigma.</math>
:<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>
If we assume that <math>T_0 \geq 100</math>, then <math>4-k \leq 4 - \frac{100}{2.42 \pi} \leq -9</math> is negative, so this expression is at most
By Taylor expansion we have
:<math>\leq \frac{1}{\sqrt{\pi t}} \int_{-\infty}^{4-k} \exp( - \frac{(\sigma - (1-y)/2)^2}{t} + 0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} )\ d\sigma</math>
:<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> \leq \frac{1}{\sqrt{\pi t}} \int_{-\infty}^{4-k} \exp( - \sigma^2 (\frac{1}{t} - \frac{0.37}{T'_0-2.71}) )\ d\sigma.</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>
With <math>t \leq 0.4</math> and <math>T'_0 \geq 100</math>, one can verify numerically that
where <math>\alpha</math> is defined in equation (8) of the writeup.  We have the approximations
:<math>\frac{1}{t} - \frac{0.37}{T'_0-2.71} \geq 2 + \frac{1}{2} \log t</math>
:<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>
and so (since <math> \sigma^2 \geq 1 </math>) one can bound the above by
and
:<math> \leq \frac{1}{\sqrt{\pi}} \int_{-\infty}^{4-k} \exp( - 2 \sigma^2 )\ d\sigma</math>
:<math> \alpha'(\frac{1+i\tilde x}{2} ) \approx \frac{-i}{\tilde x} \quad (3.11)</math>
:<math> \leq \frac{1}{\sqrt{\pi}} \exp( - 2 (k - 4)^2 ) \frac{1}{4 (k - 4)}</math>
and hence
and so the contribution to <math>\int_{-\infty}^\infty vwf(\sigma)\ d\sigma)</math> is at most
:<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>
:<math> \frac{1}{4 (\frac{T'_0}{2.42\pi} - 4) \sqrt{\pi}} \sum_{k > \frac{T'_0}{2.42\pi}} c_k</math>
The two factors of <math>\exp( \pi |t|^{1/2} v/4 ) </math> cancel.  If we now write
where
:<math>N := \sqrt{\frac{\tilde x}{4\pi}}\quad (3.13)</math>  
:<math>c_k := \frac{(1.1)^{k-1}}{a_0^k} \Gamma(k/2) \exp( - 2(k-4)^2 ).</math>
and  
 
:<math>u := |t|/N^2 = 4\pi |t|/\tilde x,\quad (3.14)</math>  
Observe that
we conclude that
:<math>c_{k+2}/c_k = \frac{(1.1)^2}{a_0^2} \frac{k}{2} \exp( - 4 (k+5)  )</math>
:<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>
and this can be shown to be less than <math>1/2</math> if <math>T_0 \geq 100</math>, and <math>k > \frac{T'_0}{2.42 \pi}</math>.  Thus
If we formally write <math>\zeta(s) = \sum_n \frac{1}{n^s}</math> (ignoring convergence issues) we obtain
:<math>\sum_{k > \frac{T'_0}{2.42\pi}} c_k \leq 4 \sup_{\frac{T'_0}{2.42\pi} < k \leq \frac{T'_0}{2.42\pi}+2} a_k</math>
:<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> \leq 4 (\frac{1.1}{a_0})^{\frac{T'_0}{2.42\pi}} \Gamma( \frac{T'_0}{4.84\pi}+1 ) \exp( - 4 (\frac{T'_0}{2.42\pi}-4)^2 ).</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>
Writing
:<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 [https://en.wikipedia.org/wiki/Theta_function#Auxiliary_functions 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).


Putting all this together, we obtain
[[Category:Polymath15]]
:<math>\int_{-\infty}^\infty vwf(\sigma)\ d\sigma \leq
\exp(\frac{1}{12(T'_0 - 3.33)}) (1 - \frac{0.37 t}{T'_0 - 2.71})^{-1/2} \exp( \frac{0.37 y^2}{4 (T'_0 - 2.71 - 0.37 t)} ) \times </math>
:<math>
(1 + \frac{\sqrt{\pi}}{a_0-1.25} + \frac{0.4}{a_0-0.85}
3^{1+y} \exp( \frac{t \log^2 9}{4 (1 - \frac{0.37}{T'_0-2.71})} ) )</math>
:<math> + \varepsilon</math>
where <math>\varepsilon</math> is the exponentially small quantity
:<math> \varepsilon := \exp(\frac{1}{12(T'_0 - 3.33)})
\frac{1}{(\frac{T'_0}{2.42\pi} - 4) \sqrt{\pi}}
(\frac{1.1}{a_0})^{\frac{T'_0}{2.42\pi}} \Gamma( \frac{T'_0}{4.84\pi}+1 ) \exp( - 4 (\frac{T'_0}{2.42\pi}-4)^2 )</math>
which looks fearsome but is extremely negligible in practice.  For instance, one can check that
:<math> \varepsilon \leq \frac{10^{-10}}{a_0^2} \leq 0.4 (\frac{1}{a_0-0.85} - \frac{1}{a_0-1.25})</math>
whenever <math>T_0 \geq 100</math>, and hence
:<math>\int_{-\infty}^\infty vwf(\sigma)\ d\sigma \leq
(1 - \frac{0.37 t}{T'_0 - 2.71})^{-1/2} \exp( \frac{1}{12(T'_0 - 3.33)} + \frac{0.37 y^2}{4 (T'_0 - 2.71 - 0.37 t)} ) \times </math>
:<math>
(1 + \frac{\sqrt{\pi}}{a_0-1.25} + \frac{0.4}{a_0-1.25}
3^{1+y} \exp( \frac{t \log^2 9}{4 (1 - \frac{0.37}{T'_0-2.71})} ) ).</math>
To clean this up, we write
:<math>1 - \frac{0.37 t}{T'_0 - 2.71} = \exp( O_{\leq}( \frac{0.37 t}{T'_0 - 2.71 - 0.37 t} )</math>
and note that <math>T'_0 - 2.71 - 0.37t \geq T'_0 - 3.33</math> to obtain
:<math>\int_{-\infty}^\infty vwf(\sigma)\ d\sigma \leq \exp( \frac{6 \times 0.37t + 1 + 3 \times 0.37 y^2}{12(T'_0 - 3.33)}) \times </math>
:<math>
(1 + \frac{1}{a_0-1.25} (\sqrt{\pi} + 1.2 \times 3^y \exp( \frac{t \log^2 9}{4 (1 - \frac{0.37 t}{T'_0-2.71})} ) ).</math>
We bound <math>(6 \times 0.37t + 1 + 3 \times 0.37 y^2)/12 \leq 0.181</math> and <math>1.2 \times 3^y \exp( \frac{t \log^2 9}{4(1 - \frac{0.37 t}{T'_0-2.71}) \leq 5.15</math> for <math>y \leq 1/2</math>, thus
:<math>\int_{-\infty}^\infty vwf(\sigma)\ d\sigma \leq \exp( \frac{0.181}{T'_0 - 3.33}) (1 + \frac{5.15}{a_0-1.25}).</math>
We conclude that
:<math>E_3 \leq \frac{1}{8} \sqrt{\pi} \exp( - \frac{t\pi^2}{64}) (T'_0)^{3/2} e^{-\pi T_0/4} \exp( \frac{0.181}{T'_0 - 3.33}) (1 + \frac{5.15}{a_0-1.25}).</math>

Latest revision as of 22:26, 1 January 2019

We are initially focusing attention on the following

Test problem For [math]\displaystyle{ t=y=0.4 }[/math], can one prove that [math]\displaystyle{ H_t(x+iy) \neq 0 }[/math] for all [math]\displaystyle{ 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]\displaystyle{ H_t(x+iy) }[/math] at [math]\displaystyle{ y=0.4 }[/math]) that one can show that [math]\displaystyle{ H_t(x+iy) \neq 0 }[/math] for all [math]\displaystyle{ y \geq 0.4 }[/math] as well. This would give a new upper bound

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

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

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

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

Thus one will seek upper bounds on the error [math]\displaystyle{ \frac{|H_t - A - B|}{|B_0|} }[/math] and lower bounds on [math]\displaystyle{ \frac{|A+B|}{|B_0|} }[/math] for various ranges of [math]\displaystyle{ x }[/math]. Numerically it seems that the RHS stays above 0.4 as soon as [math]\displaystyle{ 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]\displaystyle{ A+B }[/math], where

[math]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ s := \frac{1-y+ix}{2} }[/math]
[math]\displaystyle{ N := \lfloor \sqrt{\frac{\mathrm{Im} s}{2\pi}} \rfloor = \lfloor \sqrt{\frac{x}{4\pi}} \rfloor. }[/math]

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

[math]\displaystyle{ 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]\displaystyle{ \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]\displaystyle{ A'+B' }[/math], where

[math]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ s := \frac{1-y+ix}{2} }[/math]
[math]\displaystyle{ 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]\displaystyle{ A^{eff} + B^{eff} }[/math] was introduced:

[math]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ \alpha_1(s) := \frac{1}{2s} + \frac{1}{s-1} + \frac{1}{2} \log \frac{s}{2\pi} }[/math]
[math]\displaystyle{ N := \lfloor \sqrt{ \frac{T'}{2\pi}} \rfloor }[/math]
[math]\displaystyle{ T' := \frac{x}{2} + \frac{\pi t}{8}. }[/math]

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

[math]\displaystyle{ 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]\displaystyle{ s' := \frac{1-y}{2} + iT' = \frac{1-y+ix}{2} + \frac{\pi i t}{8} }[/math]
[math]\displaystyle{ a := \sqrt{\frac{T'}{2\pi}} }[/math]
[math]\displaystyle{ p := 1 - 2(a-N) }[/math]
[math]\displaystyle{ \sigma := \mathrm{Re} s' = \frac{1-y}{2} }[/math]
[math]\displaystyle{ U := \exp( -i (\frac{T'}{2} \log \frac{T'}{2\pi} - \frac{T'}{2} - \frac{\pi}{8} )) }[/math]
[math]\displaystyle{ 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]\displaystyle{ C^{eff} }[/math] by the very slightly different quantity

[math]\displaystyle{ \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]\displaystyle{ A^{toy} + B^{toy} }[/math], where

[math]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ N := \lfloor \sqrt{\frac{x}{4\pi}} \rfloor. }[/math]

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

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

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

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

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

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

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


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

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

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

Controlling |A+B|/|B_0|

See Controlling A+B/B_0.

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

Here is a table of analytic lower bounds for [math]\displaystyle{ A^{eff}+B^{eff}/B^{eff}_0 }[/math] for [math]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ |H_t-A^{eff}-B^{eff}/B_0^{eff}| }[/math] and [math]\displaystyle{ E_1+E_2+E_3^*/|B_0^{eff}| }[/math] for [math]\displaystyle{ x \leq 1600 }[/math].

Small values of x

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


Here are some snapshots of [math]\displaystyle{ 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]\displaystyle{ H'_t(x+iy) }[/math]:

Here is a table of [math]\displaystyle{ x }[/math], pari/gp prec, [math]\displaystyle{ 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]\displaystyle{ H_t/B_0 }[/math] for a rectangle [math]\displaystyle{ \{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]\displaystyle{ H_t }[/math] in a rectangle. It can eliminate zeros in the rectangle [math]\displaystyle{ \{0 \leq x \leq 1000, 0.4 \leq y \leq 0.45\} }[/math] when t = 0.4.


Large negative values of [math]\displaystyle{ t }[/math]

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

We heuristically compute [math]\displaystyle{ H_t(x) }[/math] in the regime where [math]\displaystyle{ x }[/math] is large and [math]\displaystyle{ t }[/math] is large and negative with [math]\displaystyle{ |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]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ \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]\displaystyle{ \xi }[/math] function, it is convenient to shift the v variable by [math]\displaystyle{ \pi |t|^{1/2}/8 }[/math], thus

[math]\displaystyle{ 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]\displaystyle{ \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]

where

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

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

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

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

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

[math]\displaystyle{ \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]

and

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

and hence

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

[math]\displaystyle{ N := \sqrt{\frac{\tilde x}{4\pi}}\quad (3.13) }[/math]

and

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

we conclude that

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

[math]\displaystyle{ 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]\displaystyle{ \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]\displaystyle{ v }[/math] integral to obtain

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

[math]\displaystyle{ 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]

Writing

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

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

[math]\displaystyle{ H_t(x) \sim \theta_{10}(N, \frac{iu - 8\pi}{16 \pi} )\quad (3.26) }[/math]
[math]\displaystyle{ \sim \theta( N + \frac{1}{2} \frac{iu - 8\pi}{16 \pi}, \frac{iu - 8\pi}{16 \pi})\quad (3.27) }[/math]
[math]\displaystyle{ \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]\displaystyle{ H_t(x) }[/math] is (by the functional equation).

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