I am reading the book from Weil "Elliptic Functions according to Eisenstein and Kronecker", and in p 56 he makes the following reasoning
We write: $$ S_a(x,y,s) = \sum_\mu^*(x + \mu)^a|x+ \mu|^{-2s}e^{-2\pi i \mu y}$$ where $x$ and $y$ are real, and $a=0$ or $1$. Applying (The Mellin transform) formally, we get: $$ \Gamma(s)S_a(x,y,s)= \sum_\mu^* \exp[-t(x+\mu)^2 -2\pi i \mu y](x+\mu)^a t^{s-1} \, dt $$ On both sides, replace $y$ by $0$, $(x+\mu)^a$ by $|x+ \mu|^a$ and $s$ by $\operatorname{Re}(s)$; then both series become series with positive terms, and the left-hand side converges provided $\operatorname{Re}(s) > \frac{a+1}{2}$; therefore, if that is so, termwise integration is permissible in the right-hand side, and both sides are equal. Now take any $T>0$ and cut up the integral in the right-hand side into an integral $I_0$ on the interval $0 < t \leq T$ and an integral $I_\infty$ on $t> T$. The latter can be estimated by observing that, if $M$ is an integer $\geq |x|$, we have $$ |x+\mu|^a e^{-t(x + \mu)^2} \leq (n + 2M) e^{-tn^2}$$ for $\mu >M$, $n=\mu-M$, and also for $ \mu < -M$, $n= -\mu -M$. Using this, it is easy to verify that $I_\infty$ is absolutely convergent for all values of $s$, uniformly over every bounded domain in the $s$-plane; it is therefore an entire function of $s$.
For some context here, $s$ is a complex number and $\sum \limits_\mu ^*$ is the sum over all integers, unless $x$ is an integer. Now if my understanding of his claim about absolute convergence is correct, then since the series is absolutely convergent (provided $\operatorname{Re}(s) > \frac{a+1}{2}$) and making a change of variable $u=n^2t$ $$|I_\infty| \leq \int\limits_T^\infty \sum \limits_\mu ^* e^{-n^2t}(2M+N)^a t^{\operatorname{Re}(s) -1} \, dt \leq \sum \limits_\mu ^* \frac{(2M+n)^a}{n^{2 \operatorname{Re}(s)}} \int\limits_T^\infty e^{-u} u^{\operatorname{Re}(s)-1} \, du $$ I recognize the upper incomplete Gamma function $\Gamma(s,T) = \int\limits_T^\infty e^{-u} u^{s-1} \, du $ which converges for all real values of $s$ (which is the case here) but $I_\infty$ converges absolutely provided $\operatorname{Re}(s) > \frac{a+1}{2}$ as we have something like a sum on $\mu^{-(2 \operatorname{Re}(s) - a)}$.
My question is: Why does he say that it absolutely converges for all values of $s$? and that it defines an entire function of $s$? Is there something i am misunderstanding?
Thanks for any help!