I am trying to classify the singularities in the function $$\mathcal{L}^{-1}_t\left(\frac{e^{|x|\sqrt{\frac{s}{k}}}}{\sqrt{ks}}\right), \ k>0$$ as the final part of solving a PDE. To use the General Inversion Formula, $$f(t)=\frac{1}{2\pi I}\int_{\gamma-i\infty}^{\gamma+i\infty} e^{st}F(s) \ ds,$$ for $t\geq 0$, I need to calculate the residues of $$F(s)=\mathcal{L}^{-1}_t\left(\frac{e^{|x|\sqrt{\frac{s}{k}}}}{\sqrt{ks}}\right).$$ I think the only singular point is when $s=0$. Is this a simple pole?
I considered the equation $g(s)=\sqrt{ks}$. I used the definition below:
$z_0$ is a zero of order $k$ for $f(z)$ if and only if $f(z_0)=f'(z_0)=...=f^{k-1}(z_0)=0$ and $f^k(z_0)\neq 0$.
However, $g'(0)\rightarrow\infty$.
There is no deleted neighborhood of $s=0$ in which the function is defined and analytic. so we cannot talk about the type of singularity at $0$.