Questions regarding the Riemann-Siegel $\theta$ Function My questions are a request, please, for help in understanding some comments in the wikipedia article discussing the Riemann-Siegel $\theta$ function http://en.wikipedia.org/wiki/Riemann%E2%80%93Siegel_theta_function:
The series for $\theta (t)$:
$$\theta(t) = -\frac{\gamma + \log \pi}t - \arctan 2t + \sum_{n = 1}^\infty (\frac{t}{2n} - \arctan (\frac{2t}{4n + 1}))$$
"For values with imaginary part between -1 and 1, the arctangent function is holomorphic, and it is easily seen that the series converges uniformly on compact sets in the region with imaginary part between -1/2 and 1/2, leading to a holomorphic function on this domain. It follows that the Z function is also holomorphic in this region, which is the critical strip."
-- First, is it correct that the principal branch of the $\arctan$ function is for values of t, Im $s$, imaginary part between $- 1$ and $1$?
-- Then, in consideration of the second term on the RHS, $t$ is thus constrained to be between $- 1/2$ and $1/2$?
-- a) Lastly, I am confused as to why the domain $- 1/2 < t < 1/2$ is referred to as "the critical strip." Especially since "t" is in the imaginary direction.
Then, b) presumably, it is assumed that $Re (s) = 1/2$, so in an open disk centered at that point with a radius $< 1/2$, $\arctan 2t$ would be holomorphic  - is this thinking correct?
c) And further, how is this region then the critical strip. I might guess by analytic continuation. If this is correct, I would also appreciate help as to how that is actually done.
Thanks very much.
 A: The $\arctan$ function may be rewritten as :
$$\arctan\;z=\frac i2\log\frac{i+z}{i-z}$$
which admits two branch points at $\,z=\pm i\,$ allowing to define $\,\arctan\;z\,$ in the whole complex plane except two cuts on the half-lines $(-\infty\,i,-i]$ and $[i,+\infty\,i)$. This implies that :


*

*The principal branch of the $\arctan$ function is holomorphic for values of $\,t\,$ such that $-1<\Im\,t<1$.

*From the second term $\,\arctan(2\,t)\,$ with $-1<2\;\Im\,t<1$ (since the other terms of $\arctan$ have smaller coefficients) and uniform convergence of the series for $\theta(t)$ on compact sets in the region $- \frac 12 < \;\Im\,t < \frac 12\;$ we get that $\theta(t)$ is holomorphic there.

*a) More exactly we want $- \frac 12 < \;\Im\,t < \frac 12\;$ but remember that $t$ appears in $\,s:=\frac 12+it\,$.
The usually real $\,t\,$ was extended here to complex values $\,t:=u+iv\,$ with $- \frac 12 < v < \frac 12\;$.
After substitution of $\,t\,$ our $\,s\,$ becomes $\,s=\frac 12-v+iu\,$.
Of course $\;0<\Re(s)=\frac 12-v<1\,$ while $\;\Im(s)=u\,$ can be any real value and we obtained simply the classical zeta critical strip (from the point of view of the variable $\,s$).
In the '$t$ space' this will correspond to the horizontal band $- \frac 12 < \;\Im\,t < \frac 12\;$.
b) and c) At this point we may use the definition $$Z(t):=\zeta\left(\frac 12+it\right)\;e^{i\,\theta(t)}$$ 
and the fact that $\zeta\left(\frac 12+it\right)$ and $\theta(t)\,$ are both holomorphic for any complex $t$ such that $- \frac 12 < \;\Im\,t < \frac 12\;$
(not just in a circle) to conclude that :
$Z(t)$ is holomorphic in the band $- \frac 12 < \;\Im\,t < \frac 12\;$ (i.e. in the critical strip using $\,s=\frac 12+it$).


Hoping this clarified things,  
