Riemann-Siegel formula

Lemma 1 For any complex number [math]\displaystyle{ z }[/math], one has

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

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

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

[math]\displaystyle{ = e^{-i \pi (z^2 - z)} }[/math]

while from the residue theorem one has

[math]\displaystyle{ 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]\displaystyle{ = 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]\displaystyle{ = 1 + e^{-2\pi i z} F(z-1). }[/math]

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

We can rearrange the above lemma as

[math]\displaystyle{ \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]\displaystyle{ s }[/math] be a complex number with [math]\displaystyle{ \mathrm{Re} s \gt 1 }[/math]. Multiplying both sides of the above equation by [math]\displaystyle{ (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]\displaystyle{ \nwarrow 0 }[/math] from [math]\displaystyle{ 0 }[/math] in the direction [math]\displaystyle{ e^{3\pi i/4} }[/math], we have

[math]\displaystyle{ A = B + C }[/math]

where

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

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

and the reflection formula

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ w = -2\pi i zu }[/math] and shifting contours we see that the left-hand side is

[math]\displaystyle{ (-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]\displaystyle{ \Box }[/math]

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

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

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

[math]\displaystyle{ 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]\displaystyle{ 0 \nwarrow 1 }[/math] is a line in the direction [math]\displaystyle{ e^{3\pi i 4} }[/math] passing through [math]\displaystyle{ [0,1] }[/math]. By analytic continuation we conclude the Riemann-Siegel formula

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ N }[/math]; similarly

[math]\displaystyle{ \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]