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I am reposting this from MO, since, I guess, my question might have been too elementary for MO and did not receive any reaction at all for the past 30 hours. Here is the link.

Here is my post:

I am reading "On the Structure of the Selberg class, I: $0\leq d\leq1$" by Kaczorowki and Perelli. In section 5, they state certain Lemma 5.1 that I am trying to understand. Let me give short introduction to the statement.

Let $\lambda_j$ be positive numbers with $\lambda_j>0$ and $\mu_j$ be complex numbers with $\Re(\mu_j)\geq 0$, $1\leq j\leq r$. Denote $\Lambda:=\sum_{j=1}^r\lambda_j$ and $\xi:=2 \sum_{j=1}^r(\mu_j- \frac{1}{2})$. Suppose that $\Lambda=\frac{1}{2}$. We observe the function $$\tilde{h}(w,s):=\prod_{j=1}^r\frac{\Gamma(\lambda_j (1-s-w)+ \bar{\mu_j})}{\Gamma(\lambda_j (s+w)+ \mu_j)}$$ It can be easily shown that for all $s$ with $|\Re(s)| < N $ for some natural number $N$ the $w$-poles of $\tilde{h}$ lie on the right side of the line $\Re(w)=-N- \frac{1}{2}$ with positive distance from it.

Fix again $\Re(w)=-N-\frac{1}{2}$. Now the Lemma 5.1 in question tells us that $$\tilde{h}(w,s)=E(w,s) \frac{\Gamma(\frac{1}{2}(1-s-w)+\frac{1}{2}\bar{\xi}+\frac{1}{2})}{\Gamma(\frac{1}{2}(s+w)+ \frac{1}{2}\xi+ \frac{1}{2})}(1+f(w,s)),$$ where $E$ is an entire and non-vanishing in both variables and $f$ is holomorphic for $0<\Re(s)<2$ and for each such $s$ $f(w,s)=O(\frac{1}{w})$ when $|w|\to\infty$.

The above representation of $\tilde{h}$ is easily derived via Stirling's approximation formula and the asymptotics of $f$ is just a byproduct of this. Also the factor $ \frac{1}{2} $ in front of the variables corresponds to the number $ \Lambda $ if we observe it in a more general context. What I don't understand is why $f(w,s)$ is holomorphic on the strip $0<\Re(s)<2$.

We already know that $\tilde{h}(w,s)$ is holomorphic on $\{ \Re(w)=-N- \frac{1}{2} \} \times \{ -N <\Re(s)< N \}$. There is a pole cancellation between the Gamma-Factors on the RHS only if $2\Re(\xi_F)$ is a negative odd number, which is not assumed. We take a look at the poles of $\Gamma(\frac{1}{2}(s+w)+ \frac{1}{2}\xi+ \frac{1}{2})$. Since $\Re(w)=-N-\frac{1}{2}$, we have: $$\Re(s_{pol})=-2m-\Re(\xi)-1+N+\frac{1}{2}=-2m-\Re(\xi)-\frac{1}{2}+N, m\in\mathbb{N}_0 $$ Thus we get only that $\tilde{h}(w,s)$ is holomorphic between vertical lines $-2m-\Re(\xi)-\frac{1}{2}+N$.

Because of the positive distance of the $w$-poles of $\tilde{h}(w,s)$ from the line $\Re(w)=-N-\frac{1}{2}$ one can actually relax the condition for $s$ to $-N-1 < \Re(s) < N+1 $.

Also, it does not seem to me to be possible to specify any reasonable pole cancellation between $\Gamma(\lambda_j (s+w)+ \mu_j)$, $1\leq j\leq r$ and $\Gamma(\frac{1}{2}(s+w)+ \frac{1}{2}\xi+ \frac{1}{2})$ due to the quite general character of the $\lambda_j$'s.

Thus my question is: What is the reason/argument for $f(w,s)$ being holomorphic on the strip $0< \Re(s)<2$?

EDIT: I have edited my post to reflect the corrections mentioned in my comments.

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Ok, here is something I did not realize due to my mind being jammed with all that stuff... For fixed real part of $w$, the $s$-poles of the gamma-factor in question have a "period" of "2m", but this still does not yield the strip between $0$ and $2$... – ex-falso-quodlibet Sep 24 '11 at 2:51
And the sign of $\Re(\xi)$ is actually not known, I forgot about the $1/2$ being substracted from the $\mu_j$'s... – ex-falso-quodlibet Sep 24 '11 at 3:00

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