Finding the inverse Laplace transform of $\arctan \left(\frac{1}{s} \right)$using contour integration
I haven't worked out the integral along the contour yet, but here's a contour I had in mind:
This is definitely a sticky one.
The following answer uses the contour provided above by Ron Gordon.
From the Maclaurin series of $\arctan(s)$, we can deduce that $\arctan \left(\frac{1}{s} \right) \sim \frac{1}{s}$ as $|s| \to \infty$.
We can then use a modified version of Jordan's lemma to conclude that $\int \arctan \left(\frac{1}{s} \right) e^{xs} \, ds$ vanishes along the two big arcs of the contour as the radii of the arcs go to infinity.
Also, since $\lim_{s \to \pm i} (s \mp i) \arctan \left( \frac{1}{s}\right)=0$, we get no contributions from letting the radii of the small circles about the branch points go to zero.
Moving clockwise around the branch point at $s=1$, the value of $\arctan \left( \frac{1}{s} \right)$ increases by $\pi$.
And moving clockwise around the branch point at $s=-i$, the value of $\arctan \left(\frac{1}{s} \right)$ decreases by $\pi$.
Therefore, $$\begin{align} \mathcal{L}^{-1} \left\{\arctan \left(\frac{1}{s} \right) \right\}(x) &= \frac{1}{2 \pi i} \int^{a+ i \infty}_{a - i \infty} \arctan \left(\frac{1}{s} \right) e^{xs} \, ds \\ &= - \frac{1}{2 \pi i} \left(\pi \int_{1}^{0} e^{ixt} \, i \, dt - \pi \int_{-1}^{0} e^{ixt} \, i \, dt \right) \\ &= \frac{1}{2} \int_{-1}^{1} e^{ixt} \, dt \\ &= \frac{1}{2ix} \left(e^{ix} - e^{-ix} \right) \\ &= \frac{\sin(x)}{x}. \end{align}$$
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\int_{0^{+}\ -\ \infty\ic} ^{0^{+}\ +\ \infty\ic}\arctan\pars{1 \over s}\expo{st}\,{\dd s \over 2\pi\ic}} = \int_{0}^{1}\int_{0^{+}\ -\ \infty\ic} ^{0^{+}\ +\ \infty\ic} {s\expo{st} \over s^{2} + x^{2}}\,{\dd s \over 2\pi\ic}\,\dd x \\[5mm] & = \int_{0}^{1}\pars{{-\ic x\expo{-\ic tx} \over -\ic x - \ic x} + {\ic x\expo{-\ic tx} \over \ic x + \ic x}}\dd x = \int_{0}^{1}\cos\pars{tx}\dd x \\[5mm] = &\ \bbx{\ds{\sin\pars{t} \over t}} \\ & \end{align}