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It doesn't have an Euler product.

Expanding $(1+n^{-2})^{-s} =\sum_{k=0}^\infty {-s \choose k} n^{-2k}$ you get $$F(s) =\sum_{n=0}^\infty (n^2+1)^{-s} = 1+\sum_{k=0}^\infty {-s \choose k} (\zeta(2s+2k)-1)$$ so it is meromorphic on the whole complex plane, with poles at $s = \frac{1}{2}-k, k \in \mathbb{N}$

Using $\Gamma(s) a^{-s} =\int_0^\infty x^{s-1}e^{-ax}dx$ and $\theta(x) = \sum_{n=1}^\infty e^{-x n^2}$ you have $$G(s) = (F(s)-1) \Gamma(s) = \int_0^\infty x^{s-1}e^{-x} \theta(x)dx, \qquad \Gamma(s) \zeta(2s) = \int_0^\infty x^{s-1}\theta(x) dx $$ where $\Gamma(s) \zeta(2s)-\frac{\sqrt{\pi}}{2(s-1/2)}+\frac{1}{2s}$ is entire.

From these poles location, we can deduce that for arbitrary large $N$ :

$\theta(x) = \frac{\sqrt{\pi}}{2}x^{-1/2}-\frac{1}{2}+o(x^N)$ as $x \to 0$ and hence $e^{-x}\theta(x) = \sum_{k\ge 0} \frac{x^k}{k!}(\frac{\sqrt{\pi}}{2}x^{-1/2}-\frac{1}{2})+o(x^N)$ as $x \to 0$

so that $$\lim_{s \to 1/2-k}(s+1/2-k)G(s)= \frac{\sqrt{\pi}}{2k!}, \qquad \lim_{s \to -k}(s-k)G(s)= \frac{-1/2}{(k+1)!}$$

It doesn't have an Euler product.

Expanding $(1+n^{-2})^{-s} =\sum_{k=0}^\infty {-s \choose k} n^{-2k}$ you get $$F(s) =\sum_{n=0}^\infty (n^2+1)^{-s} = 1+2^{-s}+\sum_{k=0}^\infty {-s \choose k} (\zeta(2s+2k)-1)$$ so it is meromorphic on the whole complex plane, with poles at $s = \frac{1}{2}-k, k \in \mathbb{N}$

Using $\Gamma(s) a^{-s} =\int_0^\infty x^{s-1}e^{-ax}dx$ and $\theta(x) = \sum_{n=1}^\infty e^{-x n^2}$ you have $$G(s) = (F(s)-1-2^{-s}) \Gamma(s) = \int_0^\infty x^{s-1}e^{-x} \theta(x)dx, \qquad \Gamma(s) \zeta(2s) = \int_0^\infty x^{s-1}\theta(x) dx $$ where $\Gamma(s) \zeta(2s)-\frac{\sqrt{\pi}}{2(s-1/2)}+\frac{1}{2s}$ is entire.

From these poles location, we can deduce that for arbitrary large $N$ :

$\theta(x) = \frac{\sqrt{\pi}}{2}x^{-1/2}-\frac{1}{2}+o(x^N)$ as $x \to 0$ and hence $e^{-x}\theta(x) = \sum_{k\ge 0} \frac{x^k}{k!}(\frac{\sqrt{\pi}}{2}x^{-1/2}-\frac{1}{2})+o(x^N)$ as $x \to 0$

so that $$\lim_{s \to 1/2-k}(s+1/2-k)G(s)= \frac{\sqrt{\pi}}{2k!}, \qquad \lim_{s \to -k}(s-k)G(s)= \frac{-1/2}{(k+1)!}$$

It doesn't have an Euler product.

Expanding $(1+n^{-2})^{-s} =\sum_{k=0}^\infty {-s \choose k} n^{-2k}$ you get $$F(s) =\sum_{n=0}^\infty (n^2+1)^{-s} = 1+\sum_{k=0}^\infty {-s \choose k} (\zeta(2s+2k)-1)$$ so it is meromorphic on the whole complex plane, with poles at $s = \frac{1}{2}-k, k \in \mathbb{N}$

Using $\Gamma(s) a^{-s} =\int_0^\infty x^{s-1}e^{-ax}dx$ and $\theta(x) = \sum_{n=0}^\infty e^{-x n^2}$ you have $$G(s) = (F(s)-1) \Gamma(s) = \int_0^\infty x^{s-1}e^{-x} \theta(x)dx, \qquad \Gamma(s) \zeta(2s) = \int_0^\infty x^{s-1}\theta(x) dx $$ where $\Gamma(s) \zeta(2s)-\frac{\sqrt{\pi}}{2(s-1/2)}+\frac{1}{2s}$ is entire.

From these poles location, we can deduce that for arbitrary large $N$ :

$\theta(x) = \frac{\sqrt{\pi}}{2}x^{-1/2}-\frac{1}{2}+o(x^N)$ as $x \to 0$ and hence $e^{-x}\theta(x) = \sum_{k\ge 0} \frac{x^k}{k!}(\frac{\sqrt{\pi}}{2}x^{-1/2}-\frac{1}{2})+o(x^N)$ as $x \to 0$

so that $$\lim_{s \to 1/2-k}(s+1/2-k)G(s)= \frac{\sqrt{\pi}}{2k!}, \qquad \lim_{s \to -k}(s-k)G(s)= \frac{-1/2}{(k+1)!}$$

It doesn't have an Euler product.

Expanding $(1+n^{-2})^{-s} =\sum_{k=0}^\infty {-s \choose k} n^{-2k}$ you get $$F(s) =\sum_{n=0}^\infty (n^2+1)^{-s} = 1+\sum_{k=0}^\infty {-s \choose k} (\zeta(2s+2k)-1)$$ so it is meromorphic on the whole complex plane, with poles at $s = \frac{1}{2}-k, k \in \mathbb{N}$

Using $\Gamma(s) a^{-s} =\int_0^\infty x^{s-1}e^{-ax}dx$ and $\theta(x) = \sum_{n=1}^\infty e^{-x n^2}$ you have $$G(s) = (F(s)-1) \Gamma(s) = \int_0^\infty x^{s-1}e^{-x} \theta(x)dx, \qquad \Gamma(s) \zeta(2s) = \int_0^\infty x^{s-1}\theta(x) dx $$ where $\Gamma(s) \zeta(2s)-\frac{\sqrt{\pi}}{2(s-1/2)}+\frac{1}{2s}$ is entire.

From these poles location, we can deduce that for arbitrary large $N$ :

$\theta(x) = \frac{\sqrt{\pi}}{2}x^{-1/2}-\frac{1}{2}+o(x^N)$ as $x \to 0$ and hence $e^{-x}\theta(x) = \sum_{k\ge 0} \frac{x^k}{k!}(\frac{\sqrt{\pi}}{2}x^{-1/2}-\frac{1}{2})+o(x^N)$ as $x \to 0$

so that $$\lim_{s \to 1/2-k}(s+1/2-k)G(s)= \frac{\sqrt{\pi}}{2k!}, \qquad \lim_{s \to -k}(s-k)G(s)= \frac{-1/2}{(k+1)!}$$

It doesn't have an Euler product.

Expanding $(1+n^{-2})^{-s} =\sum_{k=0}^\infty {-s \choose k} n^{-2k}$ you get $$F(s) =\sum_{n=0}^\infty (n^2+1)^{-s} = 1+\sum_{k=0}^\infty {-s \choose k} (\zeta(2s+2k)-1)$$ so it is meromorphic on the whole complex plane, with poles at $s = \frac{1}{2}-k, k \in \mathbb{N}$

Using $\Gamma(s) a^{-s} =\int_0^\infty x^{s-1}e^{-ax}dx$ and $\theta(x) = \sum_{n=0}^\infty e^{-x n^2}$ you have $$G(s) = (F(s)-1) \Gamma(s) = \int_0^\infty x^{s-1}e^{-x} \theta(x)dx, \qquad \Gamma(s) \zeta(2s) = \int_0^\infty x^{s-1}\theta(x) dx $$ where $\Gamma(s) \zeta(2s)-\frac{\sqrt{\pi}}{2(s-1/2)}+\frac{1}{2s}$ is entire.

From these poles location, we can deduce that for arbitrary large $N$ :

$\theta(x) = \frac{\sqrt{\pi}}{2}x^{-1/2}-\frac{1}{2}+o(x^N)$ as $x \to 0$ and hence $e^{-x}\theta(x) = \sum_{k\ge 0} \frac{x^k}{k!}(\frac{\sqrt{\pi}}{2}x^{-1/2}-\frac{1}{2})+o(x^N)$ as $x \to 0$

so that $$\lim_{s \to 1/2}(s-1/2)G(s)=\frac{\sqrt{\pi}}{2},\qquad \lim_{s \to 1/2-k}(s-1/2-k)G(s)= \frac{\sqrt{\pi}}{2k!}, \qquad \lim_{s \to 1/2-k}(s-k)G(s)= \frac{-1/2}{(k+1)!}$$

It doesn't have an Euler product.

Expanding $(1+n^{-2})^{-s} =\sum_{k=0}^\infty {-s \choose k} n^{-2k}$ you get $$F(s) =\sum_{n=0}^\infty (n^2+1)^{-s} = 1+\sum_{k=0}^\infty {-s \choose k} (\zeta(2s+2k)-1)$$ so it is meromorphic on the whole complex plane, with poles at $s = \frac{1}{2}-k, k \in \mathbb{N}$

Using $\Gamma(s) a^{-s} =\int_0^\infty x^{s-1}e^{-ax}dx$ and $\theta(x) = \sum_{n=0}^\infty e^{-x n^2}$ you have $$G(s) = (F(s)-1) \Gamma(s) = \int_0^\infty x^{s-1}e^{-x} \theta(x)dx, \qquad \Gamma(s) \zeta(2s) = \int_0^\infty x^{s-1}\theta(x) dx $$ where $\Gamma(s) \zeta(2s)-\frac{\sqrt{\pi}}{2(s-1/2)}+\frac{1}{2s}$ is entire.

From these poles location, we can deduce that for arbitrary large $N$ :

$\theta(x) = \frac{\sqrt{\pi}}{2}x^{-1/2}-\frac{1}{2}+o(x^N)$ as $x \to 0$ and hence $e^{-x}\theta(x) = \sum_{k\ge 0} \frac{x^k}{k!}(\frac{\sqrt{\pi}}{2}x^{-1/2}-\frac{1}{2})+o(x^N)$ as $x \to 0$

so that $$\lim_{s \to 1/2-k}(s+1/2-k)G(s)= \frac{\sqrt{\pi}}{2k!}, \qquad \lim_{s \to -k}(s-k)G(s)= \frac{-1/2}{(k+1)!}$$