Difference between revisions of "Effective bounds on H t"

From Polymath1Wiki
Jump to: navigation, search
(Created page with "== Explicit upper bounds on integrals == We will need effective upper bounds on various integrals that occur as error terms, with explicit constants. Here is a basic tool to...")
(No difference)

Revision as of 11:32, 12 February 2018

Explicit upper bounds on integrals

We will need effective upper bounds on various integrals that occur as error terms, with explicit constants. Here is a basic tool to do this:

Lemma 1 Let [math]\phi: [a,b] \to {\bf C}[/math] be a smooth function on a compact interval [math][a,b][/math]. Let [math]\psi: [a,b] \to {\bf C}[/math] be a measurable function. Let [math]I[/math] denote the integral [math]I := \int_a^b e^{\phi(x)} \psi(x)\ dx[/math].

1. If [math]\mathrm{Re} \phi(x) \lt 0[/math] for all [math]a \leq x \leq b[/math], then

[math] |I| \leq e^{\mathrm{Re} \phi(a)} \sup_{a \leq x \leq b} \frac{|\psi(x)|}{|\mathrm{Re} \phi'(x)|}.[/math]

2. If [math]\mathrm{Re} \phi(x) \gt 0[/math] for all [math]a \leq x \leq b[/math], then

[math] |I| \leq e^{\mathrm{Re} \phi(b)} \sup_{a \leq x \leq b} \frac{|\psi(x)|}{|\mathrm{Re} \phi'(x)|}.[/math]

3. If there is an point [math]x_0 \in (a,b)[/math] such that [math]\mathrm{Re} \phi'(x)[/math] is negative for [math]x \gt x_0[/math] and positive for [math]x \lt x_0[/math] with [math]\mathrm{Re} \phi''(x_0) \neq 0[/math] (thus [math]\mathrm{Re} \phi[/math] has a non-degenerate maximum at [math]x_0[/math]), then

[math] |I| \leq 2\sqrt{\pi} e^{\mathrm{Re} \phi(x_0)} \sup_{a \leq x \leq b: x \neq x_0} \frac{|\psi(x)| \sqrt{\mathrm{Re} \phi(x_0) - \mathrm{Re} \phi(x)}}{|\mathrm{Re} \phi'(x)|}.[/math]

4. With the same hypotheses as part 3, we also have

[math] |I| \leq \sqrt{\pi} e^{\mathrm{Re} \phi(x_0)} \sup_{a \leq x \leq b: x \neq x_0} \frac{|\psi(x)|}{|\mathrm{Re} \phi'(x)| \sqrt{\mathrm{Re} \phi(x_0) - \mathrm{Re} \phi(x)}}.[/math]

Proof Write [math]\Phi := \mathrm{Re} \phi[/math]. To prove part 1, we may normalise [math]\Phi(a)=0[/math] and the supremum to be [math]1[/math], then [math]\Phi[/math] is decreasing with [math]\Phi(b) \lt \Phi(a)=0[/math]. By the triangle inequality and change of variables we then have

[math] |I| \leq -\int_a^b e^{\Phi(x)} \Phi'(x)\ dx = \int_{\Phi(b)}^0 e^{-y}\ dy \leq 1[/math]

as desired. Part 2 is proven similarly.

To prove Part 3, we may normalise [math]\Phi(x_0) = x_0 = 0[/math] and the supremum to be 1, then [math]\Phi[/math] is negative on the rest of [math][a,b][/math] and by Taylor expansion we may write [math]\Phi(x) = - f(x)^2[/math] for some smooth [math]f: [a,b] \to {\bf R}[/math] with [math]f(0)=0[/math] and [math]f'(x) \gt 0[/math] for all [math]x \in [a,b][/math]. For any [math]x \in [a,b] \backslash \{x_0\}[/math], we have

[math] |\psi(x)| \leq \frac{|\Phi'(x)|}{\sqrt{-\Phi(x)}} = \frac{2 |f(x)| f'(x)}{|f(x)|} = 2 f'(x)[/math]

and hence by the triangle inequality and change of variables

[math] |I| \leq 2 \int_a^b e^{-f(x)^2} f'(x)\ dx = 2 \int_{f(a)}^{f(b)} e^{-y^2}\ dy \leq 2 \sqrt{\pi}[/math]

as desired.

Part 4 is proven similarly to Part 3, except that the upper bound of [math]|\psi|[/math] is now [math] 2 f(x)^2 f'(x)[/math], and one uses the identity [math]\int_{-\infty}^{\infty} e^{-y^2} y^2\ dy = \frac{1}{2} \sqrt{\pi}[/math]. [math]\Box[/math]

Note that one can use monotone convergence to send [math]b[/math] to infinity in part 1, and similarly send [math]a[/math] to negative infinity in part 2. In parts 3 and 4 one can send either [math]a[/math] or [math]b[/math] or both to infinity. The bounds can be tight, as can be seen by setting [math]\psi(x)=1[/math] (for parts 1,2,3) or [math]\psi(x) = x^2[/math] (for part 4) and [math]\phi(x)[/math] equal to [math]-x[/math] (for part 1), [math]x[/math] (for part 2), or [math]-x^2[/math] (for parts 3,4), and sending as many endpoints of integration to infinity as possible.

Estimating [math]I_t(s,0)[/math]

For any [math]t \geq 0[/math], [math]s, b \in {\bf C}[/math] with [math]\mathrm{Im} s \gt \mathrm{Re} s \gt 0[/math], we consider the integral

[math]I_t(s,b) := \int_C \exp( s(1 + u - e^u) + \frac{t}{16} ((u+b)^2 - b^2) )\ du[/math]

where [math]C[/math] is any contour from [math]+i\infty[/math] to [math]+\infty[/math] that stays a bounded distance from the upper imaginary and right real axes. This integral appears in several places in the Riemann-Siegel formula for [math]H_t[/math], so it will be of importance to estimate it efficiently. Note that the integral is absolutely convergent on the contour [math]C[/math]

From a change of variables [math]x = \exp(su)[/math] we see that

[math]I_0(s,b) = \exp( s - s \log s ) \Gamma(s)[/math]

and hence by Stirling's formula

[math]I_0(s,b) = (1 + O(\frac{1}{|s|}) \sqrt{\frac{2\pi}{s}} [/math]

assuming for instance that [math]\mathrm{Re}(s) \geq 1/2[/math]. As the phase [math]1+u-e^u[/math] is stationary at [math]u=0[/math], one has the heuristic approximation

[math]I_t(s,b) \approx I_0(s,b) = \exp( s - s \log s ) \Gamma(s)[/math]

as long as [math]b[/math] is not too large. In this section we give effective bounds on this quantity in the [math]b=0[/math] case, that is to say we estimate

[math]I_t(s,0) = \int_C \exp( s(1 + u - e^u) + \frac{t}{16} u^2 )\ du.[/math]

Proposition 2 Let [math]t,s[/math] be above. Write [math]s = \sigma+iT[/math]. Let [math]\epsilon_-, \epsilon_+ \gt 0[/math] be parameters, and assume the inequalities

[math]1 - e^{\epsilon_+} \cos \epsilon_+, \frac{t}{8} \leq T \sin \epsilon_+[/math]
[math]\epsilon_- \leq 0.771; \quad \epsilon_+ \leq 1.292[/math]

1. We have

[math]|I_t(s,b)| \leq \sqrt{2} \frac{ \exp( -T (e^{-\epsilon_-} \sin \epsilon_- - \epsilon_- ) + \sigma (1 - \epsilon_- - e^{\epsilon_-} \cos \epsilon_- ) }{T ( 1 + e^{\epsilon_-} \sin \epsilon_- - e^{-\epsilon_-} \cos \epsilon_- )}[/math]
[math]+ \sqrt{2 \pi} \frac{\sqrt{T+\sigma}}{T-\sigma} \frac{2\sqrt{\epsilon_- - e^{-\epsilon_-} \sin \epsilon_-}}{1 + e^{-\epsilon_-} \sin \epsilon_- - e^{-\epsilon_-} \cos \epsilon_-}[/math]
[math]+ \frac{\exp(-T ( e^{\epsilon_+} \sin \epsilon_+ - \epsilon_+) + \sigma(1+\epsilon_+ - e^\epsilon_+ \cos \epsilon_+) - \epsilon_+)}{T \sin \epsilon_+ - \max(1 - e^{\epsilon_+} \cos \epsilon_+, \frac{t}{8})}.[/math]

2. We have

[math]|I_t(s,b) - I_0(s,b)| \leq ???[/math].

Remark For [math]T[/math] large, [math]\sigma,t[/math] bounded, and [math]\epsilon_+, \epsilon_-[/math] small, the requirements of the proposition are obeyed, and part 1 essentially yields the bounds

[math]|I_t(s,b)| = O( \frac{\exp( - T \epsilon_-^2 / 2 )}{\epsilon_- T} ) + (1 + O(\epsilon_-)) \sqrt{\frac{2\pi}{T}} + \frac{\exp( -T \epsilon_+^2 / 2 )}{\epsilon_+ T}.[/math]

Setting [math]\epsilon_-[/math] to be a large multiple of [math]T^{-1/2}[/math] and [math]\epsilon_+[/math] to be 1.292 we obtain a near-optimal bound of the shape

[math]|I_t(s,b)| \leq (1+o(1)) \sqrt{\frac{2\pi}}{T}}.[/math]


...

Proof We begin with part 1. By shifting the contour we have

[math]I_t(s,0) = I_{t,1}(s,0) + I_{t,2}(s,0) + I_{t,3}(s,0)[/math]

where

[math]I_{t,j}(s,0) := \int_{C_j} \exp( s(1 + u - e^u) + \frac{t}{16} u^2 )\ du[/math]

for [math]j=1,2,3[/math], [math]C_1[/math] is the diagonal line parameterised by

[math]x \mapsto x - ix: -\infty \leq x \leq -\epsilon_-;[/math]

[math]C_2[/math] is the diagonal line parameterised by

[math]x \mapsto x - ix: -\epsilon_- \leq x \leq \epsilon_+;[/math]

and [math]C_3[/math] is the horizontal line parameterised by

[math]x \mapsto x - i\epsilon_+: \epsilon_+ \leq x \leq \infty;[/math]

We will use Lemma 1.3 to estimate the first integral, Lemma 1.2 to estimate the second, and Lemma 1.1 to estimate the third.

We begin with the [math]C_3[/math] integral, which we can write as

[math] \int_{\epsilon_+}^\infty \exp( \phi_3(x) )\ dx[/math]

where

[math]\phi_3(x) := (\sigma+iT) (1 + x - i\epsilon_+ - e^x e^{-i\epsilon_+}) + \frac{t}{16} (x - i\epsilon_+)^2.[/math]

The real part [math]\Phi_3[/math] of [math]\phi_3[/math] is

[math]\Phi_3(x) = \sigma( 1 + x - e^x \cos \epsilon_+ ) + T (\epsilon_+ - e^x \sin \epsilon_+ ) + \frac{t}{16} (x^2 - \epsilon_+^2)[/math]

and the first derivative is

[math]\Phi'_3(x) = \sigma( 1 - e^x \cos \epsilon_+ ) - T e^x \sin \epsilon_+ + \frac{t}{8} x.[/math]

Bounding [math]1 - e^x \cos \epsilon_+ \leq 1 - e^{\epsilon_+} \cos \epsilon_+[/math] and using [math]1 + x \leq e^x[/math] we have

[math]\Phi'_3(x) \leq - (T \sin \epsilon_+ - \max(1 - e^{\epsilon_+} \cos \epsilon_+, \frac{t}{8})) e^x[/math].

Bounding [math]e^x \geq e^{\epsilon_+}[/math] and applying Lemma 1.1, we conclude that

[math]|I_{t,3}(s,0)| \leq \frac{\exp( \Phi_3(\epsilon_+) - \epsilon_+ )}{T \sin \epsilon_+ - \max(1 - e^{\epsilon_+} \cos \epsilon_+, \frac{t}{8})}.[/math]

Note that

[math]\Phi_3(\epsilon_+) = -T ( e^{\epsilon_+} \sin \epsilon_+ - \epsilon_+) + \sigma(1+\epsilon_+ - e^\epsilon_+ \cos \epsilon_+)[/math].

Now we estimate [math]I_{t,1}(s,0)[/math]. We write this term as

[math](1-i) \int_{-\infty}^{-\epsilon_-} \exp( \phi_1(x) )\ dx [/math]

where

[math]\phi_1(x) := (\sigma+iT) (1 + x - ix - e^x e^{-ix}) + \frac{t}{4} (x - ix)^2.[/math]

The real part [math]\Phi_1[/math] of [math]\phi_1[/math] is

[math]\Phi_1(x) = \sigma( 1 + x - e^x \cos x ) + T (x - e^x \sin x ) [/math]

and the first derivative is

[math]\Phi'_1(x) = \sigma( 1 - e^x \cos x + e^x \sin x ) + T (1 - e^x \sin x - e^x \cos x ).[/math]

The expression [math]1 - e^x \cos x + e^x \sin x[/math] is non-negative for [math]x \leq 0[/math]. The expression [math]1 - e^x \sin x - e^x \cos x[/math] exceeds [math]1[/math] for [math]x \leq -0.771[/math] and is decreasing for [math]-0.771 \leq x \leq 0[/math], hence

[math]1 - e^x \sin x - e^x \cos x \geq 1 + e^{\epsilon_-} \sin \epsilon_- - e^{-\epsilon_-} \cos \epsilon_- [/math]

and hence by Lemma 1.3

[math]|I_{t,1}(s,0)| \leq \sqrt{2} \frac{\exp(\Phi_1(-\epsilon_-))}{T ( 1 + e^{\epsilon_-} \sin \epsilon_- - e^{-\epsilon_-} \cos \epsilon_- )}.[/math]

Note that

[math]\Phi_1(-\epsilon_-) = -T (e^{-\epsilon_-} \sin \epsilon_- - \epsilon_- ) + \sigma (1 - \epsilon_- - e^{\epsilon_-} \cos \epsilon_- ).[/math]

Finally we estimate [math]I_{t,2}(s,0)[/math]. We write this as

[math](1-i) \int_{-\epsilon_-}^{\epsilon_+} \exp( \phi_2(x) )\ dx [/math]

where [math]\phi_2 = \phi_1[/math]. The real part [math]\Phi_2 = \Phi_1[/math] and its derivative are as before.

For [math]x[/math] between -0.771 and 1.292, one can check that

[math]|1 - e^x \cos x + e^x \sin x| \leq |1 - e^x \sin x - e^x \cos x|[/math]

so

[math]|\Phi'_2(x)| \geq (T-\sigma) |1 - e^x \sin x - e^x \cos x|.[/math]

Similarly one can check that in this range that

[math]|1 + x - e^x \cos x| \leq |x - e^x \sin x|[/math]

and so

[math]\Phi_2(0) - \Phi_2(x) \leq (T+\sigma) (e^x \sin x - x).[/math]

The quantity

[math] \frac{\sqrt{(e^x \sin x-x)}}{|1 - e^x \sin x - e^x \cos x|} [/math]

is decreasing for [math]-0.771 \leq x \leq 0[/math], equals [math]1/2[/math] at [math]x=0[/math], and lies below 1/2 for [math]0 \lt x \lt 1.292[/math], hence is bounded by

[math] \frac{\sqrt{\epsilon_- - e^{-\epsilon_-} \sin \epsilon_-}}{1 + e^{-\epsilon_-} \sin \epsilon_- - e^{-\epsilon_-} \cos \epsilon_-}.[/math]

We thus have from Lemma 3.3 that the contribution of this integral is at most

[math] \sqrt{2 \pi} \frac{\sqrt{T+\sigma}}{T-\sigma} \frac{2\sqrt{\epsilon_- - e^{-\epsilon_-} \sin \epsilon_-}}{1 + e^{-\epsilon_-} \sin \epsilon_- - e^{-\epsilon_-} \cos \epsilon_-}.[/math]

Combining these estimates we obtain part 1.


\frac{\sqrt{T+\sigma}}{T-\sigma} \exp( \frac{t}{16} \max( \epsilon (\alpha-\beta), \epsilon' (\beta-\alpha) ) )\quad (7.8).</math>

Thus

Estimating [math]I_t(s,b)[/math]

Estimating [math]F_{t,N}[/math]

From (5.2) we have

[math]F_{t,N}(s) = \sum_{n=1}^N \frac{2}{(\pi n^2)^{s/2}} \exp( \frac{s+4}{2} \log \frac{s+4}{2} - \frac{s+4}{2} + \frac{t}{16} \log^2 \frac{s+4}{2\pi n^2} ) I_t( \frac{s+4}{2}, \log \frac{s+4}{2\pi n^2} ) - \frac{3}{(\pi n^2)^{s/2}} \exp( \frac{s+2}{2} \log \frac{s+2}{2} - \frac{s+2}{2} \frac{t}{16} \log^2 \frac{s+2}{2\pi n^2} ) I_t( \frac{s+2}{2}, \log \frac{s+2}{2\pi n^2} )[/math]


[math] \tilde \psi(u) = \frac{t}{8} (u+b) \frac{\exp( \frac{t}{16} ((u+b)^2 -b^2) )}{e^u - 1} - \frac{e^u (\exp( \frac{t}{16} ((u+b)^2 -b^2) )-1)}{(e^u-1)^2} \quad (7.9).[/math]

This expression is tractable as long as [math]e^u-1[/math] stays away from zero. Otherwise we can do the following. First observe from the fundamental theorem of calculus that

[math]\exp( \frac{t}{16} ((u+b)^2 -b^2) ) - 1 = u \int_0^1 \frac{t}{8} (\theta u+b) \exp(\frac{t}{16} ((\theta u+b)^2 -b^2) )\ d\theta [/math]

and

[math]\frac{u}{e^u-1} = \int_0^1 \frac{1}{1 + \sigma(e^u-1)}\ d\sigma[/math]

and hence

[math]\tilde \psi(u) = \int_0^1 \int_0^1 \frac{d}{du} [ (\theta u+b) \frac{\exp(\frac{t}{16} ((\theta u+b)^2 -b^2) )}{1+\sigma(e^u-1)} ]\ d\sigma d\theta.[/math]
[math]\tilde \psi(u) = \int_0^1 \int_0^1 \frac{\exp(\frac{t}{16} ((\theta u+b)^2 -b^2) )}{1+\sigma(e^u-1)} [ \theta + \frac{t}{8} (\theta u + b)^2 - (\theta u+b) \frac{\sigma e^u}{1+\sigma(e^u-1)} ]\ d\sigma d\theta.[/math]