Finding the inverse Laplace transform of $\arctan \left(\frac{1}{s} \right)$using contour integration Using contour integration, I want to show that $$\mathcal{L}^{-1} \left\{\text{arctan} \left(\frac{1}{s}\right) \right\}(x)= \frac{\sin x }{x}.$$
In other words, I want to show that $$ \frac{1}{2 \pi i} \int_{a - i \infty}^{a + i \infty} \text{arctan}\left(\frac{1}{s}\right) e^{xs} \ ds =\frac{\sin x}{x} ,$$ where $a$ is a constant greater than the real parts of all the singularities of $\text{arctan}\left(\frac{1}{s}\right)$.
We can define $\arctan \left(\frac{1}{s} \right)$ in terms of the complex logarithm.
Specifically, $$\arctan \left(\frac{1}{s} \right) = \frac{i}{2} \left[\log\left(1-\frac{i}{s}\right) - \log \left(1+\frac{i}{s}\right) \right]. $$
If we use the principal branch of the logarithm, then we need a branch cut on the imaginary axis from $-i$ to $i$.
I don't understand how to close contour.
 A: 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.
A: 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, $$\begin{align} \mathcal{L}^{-1} \left\{\arctan \left(\frac{1}{s} \right) \right\}(x)  &= \frac{1}{2 \pi i} \int^{a+ i \infty}_{a - i \infty} \arctan \left(\frac{1}{s} \right) e^{xs} \, ds  \\ &= - \frac{1}{2 \pi i} \left(\pi \int_{1}^{0} e^{ixt} \, i \, dt  - \pi \int_{-1}^{0} e^{ixt} \, i \,  dt \right) \\ &= \frac{1}{2} \int_{-1}^{1} e^{ixt} \, dt \\ &= \frac{1}{2ix} \left(e^{ix} - e^{-ix} \right) \\ &= \frac{\sin(x)}{x}. \end{align}$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\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}
