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On this post I got a comment to motivate the present question:

To prove a zero $\zeta(s_0)=0, s_0 \approx 1/2+iy$ is exactly on the critical line, it is enough to prove the real function $Z(t) = \zeta(1/2+it) e^{i \,\text{arg}(\Gamma(1/4+it/2)}\, \pi^{-it/2}$ changes of sign around $t=y$. We know $Z(t)$ is real for $t$ real thanks to the functional equation.

So I would like to ask for some help understanding where the expression

$$\large Z(t) = \zeta\left(\frac{1}{2}+it\right)\quad e^{i \;\text{arg}\left(\Gamma(\frac{1}{4}+\frac{it}{2})\right)}\quad \pi^{\frac{-it}{2}}\tag 1$$

comes from.

The terser version $Z(t)=\zeta\left(\frac{1}{2}+it\right)\; e^{i\theta(t)}$ becomes Eq. $(1)$ when $\theta(t)$ is unfolded:

$$\theta(t)= \Im\left(\ln\Gamma\left(\frac{1}{4}+\frac{it}{2} \right) \right)-\frac{t}{2}\ln(\pi)$$

The closest I got digging out online material for the amateur math curious is the following, probably related, equation of the Riemann functional:

$$\large \pi^{\frac{-s}{2}}\Gamma\left(\frac{s}{2}\right)\,\zeta(s)=\pi^{-\frac{1-s}{2}}\Gamma\left(\frac{1-s}{2}\right)\,\zeta(1-s)$$

J. M. ain't a mathematician
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Antoni Parellada
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  • $\xi(s) = \pi^{-s/2} \Gamma(s/2) \zeta(s)$. The functional equation is $\xi(1/2+s) = \xi(1/2-s)$. Together with $\xi(s) = \overline{\xi(\overline{s})}$ it means for $t$ real : $\xi(1/2+it) = \overline{\xi(1/2+it)}$ ie. $\xi(1/2+it)$ is real. Finally $\pi^{-s/2} \Gamma(s/2) $ is analytic and non-zero on $\Re(s) > 0$ so $\zeta(s)$ and $\xi(s)$ have the same (non-trivial) zeros. – reuns Jun 13 '17 at 19:28
  • And the functional equation comes from the Poisson summation formula for $\sum_{n=-\infty}^\infty e^{-\pi n^2 x}$ https://mathoverflow.net/questions/242907/xi-function-on-critical-strip-mellin-transform/242959#242959 – reuns Jun 13 '17 at 19:31
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    What do you want more ? Try computing the Fourier series of the $2$-periodic function $\theta(x) = \sum_{n=-\infty}^\infty e^{-\pi n^2 x}$ – reuns Jun 14 '17 at 09:38

1 Answers1

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The Riemann functional

$$\large \pi^{\frac{-s}{2}}\Gamma\left(\frac{s}{2}\right)\,\zeta(s)=\pi^{-\frac{1-s}{2}}\Gamma\left(\frac{1-s}{2}\right)\,\zeta(1-s)$$

relates $\zeta(s)$ to $\zeta(s-1)$, as a fraction expressed as

$$\chi(s) = \frac{\zeta(s)}{\zeta(s-1)} =\frac{\Gamma\left(\frac{1}{2}(1-s)\right)}{\Gamma\left(\frac{1}{2} s \right)}\; \pi^{s-1/2}$$

But we are interested in $s = \frac{1}{2} + it$. Therefore,

$$\begin{align} \chi\left(\frac{1}{2} + it \right) &=\frac{\Gamma\left(\frac{1}{2}\,\left(1 - \left(\frac{1}{2}+it\right)\right) \right)}{\Gamma \left( \frac{1}{2}\,\left(\frac{1}{2}+it \right)\right)} \,\pi^{\left(\frac{1}{2}+it\right)-\frac{1}{2}}\\[2ex] &= \frac{\Gamma\left(\frac{1}{4} - \frac{it}{2} \right)}{\Gamma \left( \frac{1}{4}+\frac{it}{2}\right)} \,\pi^{it} \end{align}$$

Now comes the definition of the Riemann-Siegel or Hardy Z-function:

$$Z(t) := \zeta\left(\frac{1}{2}+it\right)\;\left[\chi\left(\frac{1}{2} + it \right)\right]^{-1/2}$$

Substituting,

$$\begin{align} Z(t) &= \zeta \left(\frac{1}{2}+it\right)\left( \frac{\Gamma\left(\frac{1}{4} - \frac{it}{2} \right)}{\Gamma \left( \frac{1}{4}+\frac{it}{2}\right)} \right)^{-1/2}\; \,\pi^{\frac{-it}{2}}\\[2ex] &=\zeta \left(\frac{1}{2}+it\right)\color{blue}{ \frac{\Gamma^{1/2} \left( \frac{1}{4}+\frac{it}{2}\right)}{\Gamma^{1/2}\left(\frac{1}{4} - \frac{it}{2} \right)}} \;\pi^{\frac{-it}{2}}\\[2ex] &= \zeta \left(\frac{1}{2}+it\right)\color{blue}{\sqrt{\frac{\Gamma \left( \frac{1}{4}+\frac{it}{2}\right)}{\Gamma\left(\frac{1}{4} - \frac{it}{2} \right)}}} \;\pi^{\frac{-it}{2}}\\[2ex] &= \zeta \left(\frac{1}{2}+it\right)\color{blue}{e^{i\,\arg\left(\Gamma\left(\frac{1}{4}+\frac{it}{2}\right)\right)}} \;\pi^{\frac{-it}{2}} \end{align}$$

So, how to see that

$$\color{blue}{\sqrt{\frac{\Gamma \left( \frac{1}{4}+\frac{it}{2}\right)}{\Gamma\left(\frac{1}{4} - \frac{it}{2} \right)}}}=\color{blue}{e^{i\,\arg\left(\Gamma\left(\frac{1}{4}+\frac{it}{2}\right)\right)}}$$?

The argument of a complex number $z = x + i y=\rho\ e^{i\theta}$

$$\theta = \arg(z)= \frac{1}{i}\log\sqrt{\frac{z}{\bar z}}=\frac{\log z - \log \bar z}{2i}$$

so we can modify

$$\frac{\Gamma^{1/2} \left( \frac{1}{4}+\frac{it}{2}\right)}{\Gamma^{1/2}\left(\frac{1}{4} - \frac{it}{2} \right)}= \sqrt{\frac{\Gamma \left( \frac{1}{4}+\frac{it}{2}\right)}{\Gamma\left(\frac{1}{4} - \frac{it}{2} \right)}}$$

Now $\Gamma\left(\frac{1}{4} - \frac{it}{2}\right)$ is the complex conjugate of $\Gamma\left(\frac{1}{4} + \frac{it}{2}\right)$, acknowledging the comment by Daniel Fischer on this post:

From whichever definition of $\Gamma$ you use, it should be easy to see that $\Gamma(x) \in \mathbb R$ for positive real $x$. So the two entire meromorphic functions $f \colon z \mapsto \Gamma(z)$ and $g \colon z \mapsto \overline{\Gamma(\overline{z})}$ coincide on a set that has an accumulation point in the domain where they are holomorphic. By the identity theorem, $f \equiv g$.

Hence,

$$\arg\left(\Gamma\left(\frac{1}{4} + \frac{it}{2}\right)\right)=\frac{1}{i}\;\log\left(\sqrt{\frac{\Gamma \left( \frac{1}{4}+\frac{it}{2}\right)}{\Gamma\left(\frac{1}{4} - \frac{it}{2} \right)}}\right)$$

and

$$\exp\left[i\;{\arg\left(\Gamma\left(\frac{1}{4} + \frac{it}{2}\right)\right)}\right] =\exp\left[i\;{\frac{1}{i}\;\log\left(\sqrt{\frac{\Gamma \left( \frac{1}{4}+\frac{it}{2}\right)}{\Gamma\left(\frac{1}{4} - \frac{it}{2} \right)}}\right)}\right]$$

Finally,

$$\color{blue}{e^{i\,\arg\left(\Gamma\left(\frac{1}{4}+\frac{it}{2}\right)\right)}}=\color{blue}{\sqrt{\frac{\Gamma \left( \frac{1}{4}+\frac{it}{2}\right)}{\Gamma\left(\frac{1}{4} - \frac{it}{2} \right)}}}\quad\square$$

Now, of course, the most intriguing part, after all this algebra is the origin of the definition of the Hardy Z-function. So back to square one!

Antoni Parellada
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    Let $\Lambda(s) = \pi^{-s/2}\Gamma(s/2) \zeta(s)$. From the definition it is continuous and for $t$ real : $\Lambda(1/2+it)=\overline{\Lambda(1/2-it)}$. Now the functional equation says that $\Lambda(1/2+it) = \Lambda(1/2-it)$. Thus $\Lambda(1/2+it) = \overline{\Lambda(1/2+it)}$ and $\Lambda(1/2+it)$ as well as $Z(t) =\frac{\Lambda(1/2+it)}{|\pi^{-(1/2+it)/2} \Gamma((1/2+it)/2)|} = \zeta(1/2+it) e^{i \theta(t)}$ is real. Thus $\zeta(s)$ has a zero at $s = 1/2+ib$ exactly on the critical line whenever $Z(t)$ changes of sign around $t = b$. – reuns Jun 20 '17 at 02:33