@5716, to be clear, are you saying that it doesn't follow from (18) that alpha_{0}(s) has a holomorphic CONTINUATION to Re(sigma) > Theta ?

There are two separate assertions one can consider here, so let's be careful about it.

1) The function {Re(s)>1} -> R \subset C defined by formula (18) is not holomorphic.

2) Let U \subset R^2 be the subset of pairs (x,y) with x>1 and let V \subset C \times R be the open of pairs (s,y) where Re(s)>1. The function U -> C defined by (x,y) -> \alpha_0(x+iy) does admit an extension to a function V -> C that is holomorphic in the first variable; let me call this extension \alpha_1(s,y). In both of these, the paper convincingly argues that you can perfectly well analytically continue to x>\Theta or Re(s)>\Theta.

However, \alpha_1(s,y) from 2) has essentially nothing to do with formula (18), though they coincide with \sigma is real.

Now, the argument suggested in the paper to show f(\sigma) has an analytic extension to Re(\sigma)>\Theta needs to show that the integral \int_{t \in R} |\alpha_1(\sigma,t)|dt converges uniformly on compact sets K \subset {Re(\sigma)>\Theta}.

To do this, you need to estimate the size of the function \alpha_1(\sigma,t). The analysis in the paper applies when \sigma is real, but this alone is not sufficient to deduce the assertion on analytic continuation of f(\sigma). Things would be easy if you had to estimate the function from 1) (i.e., defined by equation (18)), and you have written posts correctly saying this is trivial from what's written. But the problem is you need to estimate the function \alpha_1 from 2).