Let $$F(x) = \sum_{k\ge 1}\frac{1}{1+k^2x^2}$$ and its Mellin transform $$F(s) = \frac{\pi}{2}\frac{\zeta(s)}{\sin{\frac{\pi}{2}s}}, \space\space where \space\space (1<Re(s)<2)$$ And then, its asymptotics are $$F(x)_{x\to0+} \approx \frac{\pi}{2x}-\frac{1}{2}+O(x^M)\space (M\gt0),\space and \space\space F(x)_{x\to\infty} \approx \frac{\pi^2}{6x^2}-\frac{\pi^4}{90x^4}+\space...$$
Now I understand the poles of $F(s)$ are $0,1,2,4,6,8...$,
and $0,1 $ are related to $F(x)_{x\to0+}$ and $2,4,6,8...$ to $F(x)_{x\to\infty}$ , because of $(1<Re(s)<2)$, the right/left plane.
Q1) How can I find these 4 coefficients?
(I tried a typical partial fraction method, and managed to get only $\frac{\pi^2}{6}$)
(Also I tried to use residue theorem, but the contour integral was so frustrating at poles 0,1)
Q2) I thought the poles should be in the region $(1<Re(s)<2)$ for asymptotics. But it doesn't look like they should necessarily from the above-mentioned example. What's the difference between in and out of the region? What does it mean for asymptotics?