Show $ g(s):=\int_{0}^{\infty}f(t)t^{s-1} dt$ is well defined and analytic. Let f be a real-valued function as following : for fixed $z \in \mathbb{C}$
$$f(t) = \sum_{a,b \in \mathbb{N} }e^{-\lvert az +b \rvert^2 t}$$
Let's define $$ g(s):=\int_{0}^{\infty}f(t)t^{s-1} dt$$
Then, show that $g(s)$ is a well-defined and analytic for all $s \in \mathbb{C}$ except $s=0, 1$

[ My attempt ]
First, I showed that $\int_{t_0}^{\infty}f(t)t^{s-1} dt$ is well-defined for $t_0 >0$.
We know that for each $a>0$ , $b>0$ $$ a^2+b^2 \geq 2ab $$
Therefore,  $$ \lvert az+b \rvert^2 = a^2\lvert z \rvert^2+b^2+2\text{Re}(z)ab \geq 2\lvert z \rvert ab+2\text{Re}(z)ab = (2\lvert z \rvert + 2\text{Re}(z)) ab =:k_1ab$$
so
$$f(t) = \sum_{a,b \in \mathbb{N}}e^{-\lvert az +b \rvert^2 t} \leq \sum_{a,b \in \mathbb{N} }e^{-k_1ab t}$$
Therefore, we can check that $f(t)t^{s-1}$ has rapidly decaying at $\infty$
However i'm stuck here. I can't prove the integral of $g$ is well defined at $0$ and analytic part.
How to prove this?
Thank you for your attention.
 A: You meant for $t >0,\Im(z)\ne 0$ $$f(t) = \sum_{a,b \in \mathbb{Z}^2 }e^{-\lvert az +b \rvert^2 t}$$ for $\Re(s) >1$ $$g(s):=\int_{0}^{\infty}(f(t)-1)t^{s-1} dt$$ is analytic, and it
extends to a meromorphic function on $\Bbb{C}$ with two simple poles at $0,1$.
For $v\in R^2$ look at
$$F_{M,t}(v)= \sum_{n\in Z^2} e^{-\pi ((v+n)^\top M (v+n))t}=\sum_{n\in Z^2} e^{-\pi \| M^{1/2} (v+n)\|^2 t}, \qquad M=\pmatrix{|z|^2&\Im(z)\\ \Im(z)& 1}$$
So that $$f(t)=F_{M,t/\pi}(0)$$
$F_{M,t}$ is $Z^2$ periodic, it has a Fourier series $$F_{M,t}(v)= \sum_{k\in Z^2} c_{M,t}(k) e^{2i\pi k^\top v}$$ $$ c_{M,t}(k)=\iint_{R^2} e^{-\pi \| M^{1/2} v\|^2 t}e^{-2i\pi k^\top v}dv= |\det(M)|^{-1/2} t^{-1} e^{-\pi \| M^{-\top/2} k\|^2/ t}$$
ie. $$F_{M,t}(0)= |\det(M)|^{-1/2} t^{-1} F_{M^{-\top},1/t}(0)$$
whence as $t\to 0$ $$F_{M,t}(0)  = |\det(M)|^{-1/2} t^{-1}+O(e^{-r/t})$$ so that
$$\int_0^\infty (F_{M,t}(0)-1+1_{t<1}-|\det(M)|^{-1/2} t^{-1} 1_{t<1}) t^{s-1}dt$$ $$=\pi^{-s} g(s)+\frac1s-|\det(M)|^{-1/2} \frac1{s-1}$$ is entire.
