The Riemann-Siegel theta function is defined by:
$ \theta(t) = arg\left(\Gamma(\frac{0.5+it}{2}) \right) - \frac{t}{2}\log{\pi}$
As in the following:
However the same article claims that $\theta(t)$ is increasing for $t>6.29$, $t\in\mathbb{R}$. Indeed graphs of the function from Wolfram confirm this.
I don't understand how the argument is being defined here. Clearly if one uses the "usual" method of defining a single valued argument by selecting an interval say $ (-\pi,\pi]$ or $(0,2\pi]$, then $\theta(t)$ is bounded above and decreasing for all $t$ above some sufficienctly large $n\in\mathbb{R}$.
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I had the same problem as you with that specific part of the Wikipedia page.
Fortunately, just below, they give the asymptotic expansion$$\theta(t)\sim\frac{t}{2} \log \left(\frac{t}{2 \pi }\right)-\frac{t}{2}-\frac{\pi}{8}+\frac{1}{48t}+\cdots$$$$\theta'(t)\sim \frac{1}{2} \log \left(\frac{t}{2 \pi }\right)-\frac{1}{48 t^2}\qquad \text{and} \qquad \theta''(t)\sim \frac{1}{2 t}+\frac{1}{24 t^3}\quad > 0 \quad \forall t>0$$
If we ignore the second term, the derivative cancels for $t=2\pi$; if we take this term into account, the minimum value is at $$t=\frac{1}{2 \sqrt{3 W\left(\frac{1}{48 \pi ^2}\right)}}\approx 6.28981$$ while the exact derivative$$\theta'(t)=-\frac{\log (\pi )}{2}+\frac{1}{4} \left(\psi \left(\frac{1}{4}-\frac{i t}{2}\right)+\psi \left(\frac{i t}{2}+\frac{1}{4}\right)\right)$$ cancels at $t=6.28984$
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