View source for 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
|-
! style="text-align:left;"| <math>x</math>
! <math>|H^{eff}/B'_0|</math>
! <math>|(A'+B')/B'_0|</math>
! <math>|(H^{eff}-(A'+B'))/B'_0|</math>
! <math>|(H^{eff}-(A'+B'))/B'_0| + |(E_1+E_2+E_3)/B'_0|</math> 
|-
|10000
|0.52
|0.52
|0.0006
|0.039
|-
|12131
|1.28
|1.28
|0.0004
|0.033
|-
|15256
|0.97
|0.97
|0.0003
|0.027
|-
|18432
|0.68
|0.68
|0.0003
|0.023
|-
|20567
|0.98
|0.98
|0.0004
|0.022
|-
|30654
|1.93
|1.93
|0.0004
|0.016
|}

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]:

{| border=1
|-
! style="text-align:left;"| <math>x</math>
! <math>\frac{E_3}{E_1+E_2}</math>
|-
|10000
|9.11
|-
|15000
|14.97
|-
|20000
|19.26
|-
|50000
|32.39
|-
|100000
|42.99
|-
|<math>10^7</math>
|87.23
|}

<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]

{| border=1
|-
! style="text-align:left;"| <math>x</math>
! <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>:

{| 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
|-
! 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> ===

...

=== Estimation of <math>E_3</math> ===

Here we assume that <math>T_0 \geq 100</math>, which implies also <math>T'_0 \geq 100</math>.

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

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> (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>
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
:<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>
valid for any <math>a,b,c</math> with <math>a</math> positive, we can write the above expression as
:<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
:<math> b := - \log 9 + 0.37 \frac{y}{T'_0 - 2.71}.</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
:<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>
Now we consider the integral
:<math> \int_{-\infty}^{4-k} f(\sigma) \exp(  0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} )\ d\sigma.</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
:<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> \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>
With <math>t \leq 0.4</math> and <math>T'_0 \geq 100</math>, one can verify numerically that
:<math>\frac{1}{t} - \frac{0.37}{T'_0-2.71} \geq 2 + \frac{1}{2} \log t</math>
and so (since <math> \sigma^2 \geq 1 </math>) one can bound the above by
:<math> \leq \frac{1}{\sqrt{\pi}} \int_{-\infty}^{4-k} \exp( - 2 \sigma^2 )\ d\sigma</math>
:<math> \leq \frac{1}{\sqrt{\pi}} \exp( - 2 (k - 4)^2 ) \frac{1}{4 (k - 4)}</math>
and so the contribution to <math>\int_{-\infty}^\infty vwf(\sigma)\ d\sigma)</math> is at most
:<math> \frac{1}{4 (\frac{T'_0}{2.42\pi} - 4) \sqrt{\pi}}  \sum_{k > \frac{T'_0}{2.42\pi}} c_k</math>
where
:<math>c_k := \frac{(1.1)^{k-1}}{a_0^k} \Gamma(k/2) \exp( - 2(k-4)^2 ).</math>

Observe that
:<math>c_{k+2}/c_k = \frac{(1.1)^2}{a_0^2} \frac{k}{2} \exp( - 4 (k+5)  )</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
:<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> \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>

Putting all this together, we obtain
:<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>