Difference between revisions of "Riemann-Siegel formula"

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:<math> (1 + e^{-i\pi s}) \pi^{-\frac{1-s}{2}} \Gamma(\frac{1-s}{2}) \int_{\nwarrow 0} z^{s-1} e^{2\pi i z u}\ du = \pi^{-s/2} \Gamma(s/2) u^{-s}.</math>
 
:<math> (1 + e^{-i\pi s}) \pi^{-\frac{1-s}{2}} \Gamma(\frac{1-s}{2}) \int_{\nwarrow 0} z^{s-1} e^{2\pi i z u}\ du = \pi^{-s/2} \Gamma(s/2) u^{-s}.</math>
  
'''Proof''' ... <math>\Box</math>
+
'''Proof''' Using the duplication formula
 +
:<math> \Gamma(\frac{s}{2}) \Gamma(\frac{1+s}{2}) = 2^{1-s} \sqrt{\pi} \cos(\pi s/2)</math>
 +
and the reflection formula
 +
:<math> \Gamma(\frac{1-s}{2}) \Gamma(\frac{1+s}{2}) = \frac{\pi}{\cos(\pi s/2)}</math>
 +
one can rewrite the claim after some algebra as
 +
:<math>\int_{\nwarrow 0} z^{s-1} e^{2\pi i z u}\ du  = e^{\pi i s/2} (2\pi )^{-s} \Gamma(s) u^{-s}.</math>
 +
But by making the change of variables <math>w = -2\pi i zu</math> and shifting contours we see that the left-hand side is
 +
:<math>(-2\pi i z)^{-s} \int_0^\infty w^{s-1} e^{-w}\ dw</math>
 +
and the claim follows from the definition of the Gamma function. <math>\Box</math>
  
 
From this Lemma and Fubini (carefully verifying the absolute integrability) we have
 
From this Lemma and Fubini (carefully verifying the absolute integrability) we have
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Finally by reflecting the ray <math>\nwarrow 0</math> around the origin and then shifting slightly to the right we have
 
Finally by reflecting the ray <math>\nwarrow 0</math> around the origin and then shifting slightly to the right we have
 
:<math>C = \pi^{-\frac{1-s}{2}} \Gamma(\frac{1-s}{2}) \int_{0 \nwarrow 1} \frac{z^{s-1} e^{-i\pi z^2}}{e^{i\pi z} -e^{-i\pi z}}\ dz,</math>
 
:<math>C = \pi^{-\frac{1-s}{2}} \Gamma(\frac{1-s}{2}) \int_{0 \nwarrow 1} \frac{z^{s-1} e^{-i\pi z^2}}{e^{i\pi z} -e^{-i\pi z}}\ dz,</math>
where <math>0 \nwarrow 1</math> is a line in the direction <math>e^{3\pi i 4}</math> passing through <math>[0,1]</math>.  By analytic continuation we conclude the Riemann-Siegel formula
+
where <math>0 \nwarrow 1</math> is a line in the direction <math>e^{3\pi i 4}</math> passing through <math>[0,1]</math>.  By analytic continuation we conclude the Riemann-Siegel formula (see also equation (2.10.6) of [T1986])
 
:<math> \pi^{-s/2} \Gamma(s/2) \zeta(s) = - \pi^{-s/2} \Gamma(s/2) \int_{0 \nearrow 1} \frac{u^{-s} e^{i\pi u^2}}{e^{i\pi u} - e^{-i\pi u}}\ du -  
 
:<math> \pi^{-s/2} \Gamma(s/2) \zeta(s) = - \pi^{-s/2} \Gamma(s/2) \int_{0 \nearrow 1} \frac{u^{-s} e^{i\pi u^2}}{e^{i\pi u} - e^{-i\pi u}}\ du -  
 
\pi^{-\frac{1-s}{2}} \Gamma(\frac{1-s}{2}) \int_{0 \nwarrow 1} \frac{z^{s-1} e^{-i\pi z^2}}{e^{-i\pi z} -e^{i\pi z}}\ dz.</math>
 
\pi^{-\frac{1-s}{2}} \Gamma(\frac{1-s}{2}) \int_{0 \nwarrow 1} \frac{z^{s-1} e^{-i\pi z^2}}{e^{-i\pi z} -e^{i\pi z}}\ dz.</math>

Latest revision as of 13:49, 10 March 2018

Lemma 1 For any complex number [math]z[/math], one has

[math] \int_{0 \nearrow 1} \frac{e^{i\pi u^2 + 2\pi i z u}}{e^{i\pi u} - e^{-i\pi u}}\ du = \frac{e^{i\pi z} - e^{-i\pi z^2}}{e^{i \pi z} - e^{-i\pi z}}[/math]

where [math]0 \nearrow 1[/math] denotes a line passing through the line segment [math][0,1][/math] oriented in the direction [math]e^{i\pi/4}[/math].

Proof Denote the left-hand side by [math]F(z)[/math]. Observe that

[math]F(z) - F(z-1) = \int_{0 \nearrow 1} e^{i\pi u^2 + 2\pi i z u - i \pi u}\ du[/math]
[math]= e^{-i \pi (z^2 - z + 1/4)} \int_{0 \nearrow 1} e^{i \pi (u + z-\frac{1}{2})^2}\ du[/math]
[math]= e^{-i \pi (z^2 - z)}[/math]

while from the residue theorem one has

[math]F(z) = 1 + \int_{-1 \nearrow 0} \frac{e^{i\pi u^2 + 2\pi i z u}}{e^{i\pi u} - e^{-i\pi u}}\ du [/math]
[math] = 1 + \int_{0 \nearrow 1} \frac{e^{i\pi (u-1)^2 + 2\pi i z (u-1)}}{e^{i\pi (u-1)} - e^{-i\pi (u-1)}}\ du [/math]
[math] = 1 + e^{-2\pi i z} F(z-1).[/math]

The claim then follows from elementary algebra. [math]\Box[/math]

We can rearrange the above lemma as

[math] \frac{e^{i\pi z}}{e^{i\pi z} -e^{-i\pi z}} = \int_{0 \nearrow 1} \frac{e^{i\pi u^2 + 2\pi i z u}}{e^{i\pi u} - e^{-i\pi u}}\ du + \frac{e^{-i\pi z^2}}{e^{i\pi z} - e^{-i\pi z}}.[/math]

Now let [math]s[/math] be a complex number with [math]\mathrm{Re} s \gt 1[/math]. Multiplying both sides of the above equation by [math](1 + e^{-i\pi s}) \pi^{-\frac{1-s}{2}} \Gamma(\frac{1-s}{2}) z^{s-1}[/math] and integrating on the ray [math]\nwarrow 0[/math] from [math]0[/math] in the direction [math]e^{3\pi i/4}[/math], we have

[math] A = B + C[/math]

where

[math]A := (1 + e^{-i\pi s}) \pi^{-\frac{1-s}{2}} \Gamma(\frac{1-s}{2}) \int_{\nwarrow 0} \frac{z^{s-1} e^{i\pi z}}{e^{i\pi z} -e^{-i\pi z}}\ dz[/math]
[math]B := (1 + e^{-i\pi s}) \pi^{-\frac{1-s}{2}} \Gamma(\frac{1-s}{2}) \int_{\nwarrow 0} \int_{0 \nearrow 1} \frac{e^{i\pi u^2 + 2\pi i z u} z^{s-1}}{e^{i\pi u} - e^{-i\pi u}}\ du dz[/math]
[math]C := (1 + e^{-i\pi s}) \pi^{-\frac{1-s}{2}} \Gamma(\frac{1-s}{2}) \int_{\nwarrow 0} \frac{z^{s-1} e^{-i\pi z^2}}{e^{i\pi z} -e^{-i\pi z}}\ dz.[/math]

Lemma 2 For any [math]u[/math] to the right of the line [math]e^{i\pi/4} {\bf R}[/math], We have

[math] (1 + e^{-i\pi s}) \pi^{-\frac{1-s}{2}} \Gamma(\frac{1-s}{2}) \int_{\nwarrow 0} z^{s-1} e^{2\pi i z u}\ du = \pi^{-s/2} \Gamma(s/2) u^{-s}.[/math]

Proof Using the duplication formula

[math] \Gamma(\frac{s}{2}) \Gamma(\frac{1+s}{2}) = 2^{1-s} \sqrt{\pi} \cos(\pi s/2)[/math]

and the reflection formula

[math] \Gamma(\frac{1-s}{2}) \Gamma(\frac{1+s}{2}) = \frac{\pi}{\cos(\pi s/2)}[/math]

one can rewrite the claim after some algebra as

[math]\int_{\nwarrow 0} z^{s-1} e^{2\pi i z u}\ du = e^{\pi i s/2} (2\pi )^{-s} \Gamma(s) u^{-s}.[/math]

But by making the change of variables [math]w = -2\pi i zu[/math] and shifting contours we see that the left-hand side is

[math](-2\pi i z)^{-s} \int_0^\infty w^{s-1} e^{-w}\ dw[/math]

and the claim follows from the definition of the Gamma function. [math]\Box[/math]

From this Lemma and Fubini (carefully verifying the absolute integrability) we have

[math]B = \pi^{-s/2} \Gamma(s/2) \int_{0 \nearrow 1} \frac{u^{-s} e^{i\pi u^2}}{e^{i\pi u} - e^{-i\pi u}}\ du.[/math]

Similarly, using the geometric series formula

[math] \frac{e^{i\pi z}}{e^{i\pi z}-e^{-i\pi z}} = -\sum_{n=1}^\infty e^{2\pi i n z}[/math]

and Fubini again one has

[math]A = -\pi^{-s/2} \Gamma(s/2) \zeta(s).[/math]

Finally by reflecting the ray [math]\nwarrow 0[/math] around the origin and then shifting slightly to the right we have

[math]C = \pi^{-\frac{1-s}{2}} \Gamma(\frac{1-s}{2}) \int_{0 \nwarrow 1} \frac{z^{s-1} e^{-i\pi z^2}}{e^{i\pi z} -e^{-i\pi z}}\ dz,[/math]

where [math]0 \nwarrow 1[/math] is a line in the direction [math]e^{3\pi i 4}[/math] passing through [math][0,1][/math]. By analytic continuation we conclude the Riemann-Siegel formula (see also equation (2.10.6) of [T1986])

[math] \pi^{-s/2} \Gamma(s/2) \zeta(s) = - \pi^{-s/2} \Gamma(s/2) \int_{0 \nearrow 1} \frac{u^{-s} e^{i\pi u^2}}{e^{i\pi u} - e^{-i\pi u}}\ du - \pi^{-\frac{1-s}{2}} \Gamma(\frac{1-s}{2}) \int_{0 \nwarrow 1} \frac{z^{s-1} e^{-i\pi z^2}}{e^{-i\pi z} -e^{i\pi z}}\ dz.[/math]

From the residue theorem we can also write

[math]\int_{0 \nearrow 1} \frac{u^{-s} e^{i\pi u^2}}{e^{i\pi u} - e^{-i\pi u}}\ du =\sum_{n=1}^N \frac{1}{n^s} + \int_{N \nearrow N+1} \frac{u^{-s} e^{i\pi u^2}}{e^{i\pi u} - e^{-i\pi u}}\ du[/math]

for any natural number [math]N[/math]; similarly

[math]\int_{0 \nwarrow 1} \frac{z^{s-1} e^{-i\pi z^2}}{e^{-i\pi z} -e^{i\pi z}}\ dz = \sum_{m=1}^M \frac{1}{m^{1-s}} + \int_{M \nwarrow M+1} \frac{z^{s-1} e^{-i\pi z^2}}{e^{-i\pi z} -e^{i\pi z}}\ dz.[/math]