Controlling H t-A-B/B 0

As computed in Effective bounds on H_t - second approach, there is an effective bound

[math]\displaystyle{ |H^{eff} - A^{eff} - B^{eff}| \leq E_1 + E_2 + E_3 }[/math]

where

[math]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ 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]\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{ \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]\displaystyle{ 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]\displaystyle{ 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 \lt 0} + \frac{1}{12(T'_0 - 0.33)}) }[/math]

[math]\displaystyle{ 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 \lt 0} }[/math]

[math]\displaystyle{ a_0 := \sqrt{\frac{T'_0}{2\pi}} }[/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]

[math]\displaystyle{ T'_0 := T_0 + \frac{\pi t}{8} }[/math]

Comparison between [math]\displaystyle{ H^{eff} = A^{eff}+B^{eff} }[/math], [math]\displaystyle{ A'+B' }[/math], and the effective error bound [math]\displaystyle{ E_1+E_2+E_3 }[/math] on [math]\displaystyle{ H - H^{eff} }[/math] at some points of [math]\displaystyle{ x }[/math] source:

[math]\displaystyle{ x }[/math]	[math]\displaystyle{ |H^{eff}/B'_0| }[/math]	[math]\displaystyle{ |(A'+B')/B'_0| }[/math]	[math]\displaystyle{ |(H^{eff}-(A'+B'))/B'_0| }[/math]	[math]\displaystyle{ |(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]\displaystyle{ E_3 }[/math] error dominates the other two source:

[math]\displaystyle{ x }[/math]	[math]\displaystyle{ \frac{E_3}{E_1+E_2} }[/math]
10000	9.11
15000	14.97
20000	19.26
50000	32.39
100000	42.99
[math]\displaystyle{ 10^7 }[/math]	87.23

[math]\displaystyle{ A^{eff}+B^{eff} }[/math] is a good approximation to [math]\displaystyle{ H_t }[/math], [math]\displaystyle{ A+B-C }[/math] is better, and [math]\displaystyle{ A^{eff}+B^{eff}-C^{eff} }[/math] is excellent source source source source

[math]\displaystyle{ \frac{|H_t-(A^{eff}+B^{eff})|}{|B_0^{eff}|} }[/math]	[math]\displaystyle{ x }[/math]	[math]\displaystyle{ \frac{|H_t-(A+B-C)|}{|B_0|} }[/math]	[math]\displaystyle{ \frac{|H_t-(A^{eff}+B^{eff}-C^{eff})|}{|B_0^{eff}|} }[/math]
160	0.1675083979955609185	0.06993270565802375041	0.009155667752
320	0.2776948344513698276	0.006716674125965016299	0.0005529962481
480	0.1675667240356922231	0.005332893070605698501	0.0004966282128
640	0.1635077306008453928	0.003363431256036816251	0.0004482768972
800	0.2045038601879677257	0.1548144749150572349	0.002644344570
960	0.1031837988358064657	0.03009229958121352990	0.001270168744
1120	0.07541968034203085865	0.004507664238680722472	0.0009957229500
1280	0.1339118061014743863	0.002283591962997851167	0.0007024411378
1440	0.07958929988050262854	0.01553727684468691873	0.0007000473085
1600	0.07700542235140914608	0.001778051951547709718	0.0004882487218
1760	0.09568042045936396570	0.02763769444052338578	0.0002518910919
1920	0.06385275621986742745	0.002108779890256530964	0.0008378989413
2080	0.09421231232885752514	0.02746770886040058927	0.0004924765754
2240	0.05888587520703223358	0.001567020041379128455	0.0001171320991
2400	0.07899341548208345822	0.01801417530687959747	0.0002443802551
2560	0.06283843631123482445	0.001359561117436848149	0.0004569058755
2720	0.05966972584730198272	0.008503327577240081269	0.0006355966221
2880	0.07980560515423855917	0.001089253262122934826	0.0008864917365
3040	0.04636072344121703969	0.003004181560093288747	0.00004326840265
3200	0.09664223832561922043	0.02931455383125538672	0.0003598521453

A closer look at the "spike" in error near [math]\displaystyle{ x=800 \approx 256 \pi \approx 804 }[/math]:

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

[math]\displaystyle{ x }[/math]	[math]\displaystyle{ E_1 / B^{eff}_0 }[/math]	[math]\displaystyle{ E_2 / B^{eff}_0 }[/math]	[math]\displaystyle{ \frac{t^2}{16x} \log^2 \frac{x}{4\pi} }[/math]
10^3	[math]\displaystyle{ 1.389 \times 10^{-3} }[/math]	[math]\displaystyle{ 2.341 \times 10^{-3} }[/math]	[math]\displaystyle{ 1.915 \times 10^{-4} }[/math]
10^4	[math]\displaystyle{ 1.438 \times 10^{-4} }[/math]	[math]\displaystyle{ 3.156 \times 10^{-4} }[/math]	[math]\displaystyle{ 4.461 \times 10^{-5} }[/math]
10^5	[math]\displaystyle{ 1.118 \times 10^{-5} }[/math]	[math]\displaystyle{ 3.574 \times 10^{-5} }[/math]	[math]\displaystyle{ 8.067 \times 10^{-6} }[/math]
10^6	[math]\displaystyle{ 7.328 \times 10^{-7} }[/math]	[math]\displaystyle{ 3.850 \times 10^{-6} }[/math]	[math]\displaystyle{ 1.273 \times 10^{-6} }[/math]
10^7	[math]\displaystyle{ 4.414 \times 10^{-8} }[/math]	[math]\displaystyle{ 4.197 \times 10^{-7} }[/math]	[math]\displaystyle{ 1.846 \times 10^{-7} }[/math]

Estimation of [math]\displaystyle{ E_1,E_2 }[/math]

...

Estimation of [math]\displaystyle{ E_3 }[/math]

Here we assume that [math]\displaystyle{ T_0 \geq 100 }[/math], which implies also [math]\displaystyle{ T'_0 \geq 100 }[/math].

We first bound [math]\displaystyle{ w }[/math] by a Gaussian type quantity.

We have

[math]\displaystyle{ 1 + \frac{\sigma^2}{(T'_0)^2} \leq \exp( \frac{\sigma^2}{(T'_0)^2}) }[/math]

and

[math]\displaystyle{ 1 + \frac{(1-\sigma)^2}{(T'_0)^2} \leq \exp( \frac{(1-\sigma)^2}{(T'_0)^2}) }[/math]

and thus

[math]\displaystyle{ ( 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]\displaystyle{ = \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]\displaystyle{ \log(1+x^2) \leq 1.479 \sqrt{x} }[/math]

and

[math]\displaystyle{ x - \mathrm{arctan}(x) \leq 0.230 x^2 }[/math]

for any [math]\displaystyle{ x \geq 0 }[/math], and hence

[math]\displaystyle{ \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]\displaystyle{ \leq 0.37 \frac{(\sigma-1/2)^2}{T'_0} 1_{\sigma \geq 1} }[/math]

and

[math]\displaystyle{ (\frac{T'_0}{2} \arctan \frac{\sigma}{T'_0} - \frac{\sigma}{2}) 1_{\sigma \lt 0} \leq \frac{T'_0}{2} 1_{\sigma\lt 0} 0.230 (\frac{|\sigma|}{T'_0})^2 }[/math]

[math]\displaystyle{ \leq 0.115 \frac{(\sigma-1/2)^2}{T'_0} 1_{\sigma \lt 0}. }[/math]

We conclude that

[math]\displaystyle{ 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]\displaystyle{ \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]\displaystyle{ v }[/math]. Observe that if [math]\displaystyle{ k \leq \frac{T'_0}{2.42 \pi} = \frac{a_0^2}{1.21} }[/math] then

[math]\displaystyle{ (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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \leq \frac{1.1 \sqrt{\pi}}{a_0 - 1.25}; }[/math]

also

[math]\displaystyle{ (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]\displaystyle{ \leq 0.4 \frac{9^\sigma}{a_0 - 0.865} }[/math]

and hence (bounding [math]\displaystyle{ (0.9)^{\lceil -\sigma \rceil} \leq \frac{1}{1.1} }[/math])

[math]\displaystyle{ 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} \lt k \leq 4-\sigma} \frac{(1.1)^{k-1}}{a_0^k} \Gamma(k/2). }[/math]

We conclude (using Fubini's theorem) that

[math]\displaystyle{ \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]\displaystyle{ + \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]\displaystyle{ + \sum_{k \gt \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]\displaystyle{ \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]\displaystyle{ = \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]\displaystyle{ \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]\displaystyle{ a,b,c }[/math] with [math]\displaystyle{ a }[/math] positive, we can write the above expression as

[math]\displaystyle{ (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]\displaystyle{ 9^\sigma }[/math] is larger for [math]\displaystyle{ \sigma \geq 1/2 }[/math] than for [math]\displaystyle{ \sigma \lt 1/2 }[/math], we have

[math]\displaystyle{ \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]\displaystyle{ = \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]\displaystyle{ = 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]\displaystyle{ b := - \log 9 + 0.37 \frac{y}{T'_0 - 2.71}. }[/math]

If [math]\displaystyle{ T'_0 \geq 100 }[/math] and [math]\displaystyle{ y \leq 1/2 }[/math] then [math]\displaystyle{ |b| \leq \log 9 }[/math], thus the above integral is at most

[math]\displaystyle{ = 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]\displaystyle{ \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]\displaystyle{ T_0 \geq 100 }[/math], then [math]\displaystyle{ 4-k \leq 4 - \frac{100}{2.42 \pi} \leq -9 }[/math] is negative, so this expression is at most

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ t \leq 0.4 }[/math] and [math]\displaystyle{ T'_0 \geq 100 }[/math], one can verify numerically that

[math]\displaystyle{ \frac{1}{t} - \frac{0.37}{T'_0-2.71} \geq 2 + \frac{1}{2} \log t }[/math]

and so (since [math]\displaystyle{ \sigma^2 \geq 1 }[/math]) one can bound the above by

[math]\displaystyle{ \leq \frac{1}{\sqrt{\pi}} \int_{-\infty}^{4-k} \exp( - 2 \sigma^2 )\ d\sigma }[/math]

[math]\displaystyle{ \leq \frac{1}{\sqrt{\pi}} \exp( - 2 (k - 4)^2 ) \frac{1}{4 (k - 4)} }[/math]

and so the contribution to [math]\displaystyle{ \int_{-\infty}^\infty vwf(\sigma)\ d\sigma) }[/math] is at most

[math]\displaystyle{ \frac{1}{4 (\frac{T'_0}{2.42\pi} - 4) \sqrt{\pi}} \sum_{k \gt \frac{T'_0}{2.42\pi}} c_k }[/math]

where

[math]\displaystyle{ c_k := \frac{(1.1)^{k-1}}{a_0^k} \Gamma(k/2) \exp( - 2(k-4)^2 ). }[/math]

Observe that

[math]\displaystyle{ 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]\displaystyle{ 1/2 }[/math] if [math]\displaystyle{ T_0 \geq 100 }[/math], and [math]\displaystyle{ k \gt \frac{T'_0}{2.42 \pi} }[/math]. Thus

[math]\displaystyle{ \sum_{k \gt \frac{T'_0}{2.42\pi}} c_k \leq 4 \sup_{\frac{T'_0}{2.42\pi} \lt k \leq \frac{T'_0}{2.42\pi}+2} a_k }[/math]

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ (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]\displaystyle{ + \varepsilon }[/math]

where [math]\displaystyle{ \varepsilon }[/math] is the exponentially small quantity

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ T_0 \geq 100 }[/math], and hence

[math]\displaystyle{ \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]\displaystyle{ (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]\displaystyle{ 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]\displaystyle{ T'_0 - 2.71 - 0.37t \geq T'_0 - 3.33 }[/math] to obtain

[math]\displaystyle{ \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]\displaystyle{ (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]\displaystyle{ (6 \times 0.37t + 1 + 3 \times 0.37 y^2)/12 \leq 0.181 }[/math] and [math]\displaystyle{ 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]\displaystyle{ y \leq 1/2 }[/math], thus

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