Finding the inverse Laplace transform of $\arctan \left(\frac{1}{s} \right)$using contour integration

Solution 1:

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.

Solution 2:

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, $$\begin{align} \mathcal{L}^{-1} \left\{\arctan \left(\frac{1}{s} \right) \right\}(x) &= \frac{1}{2 \pi i} \int^{a+ i \infty}_{a - i \infty} \arctan \left(\frac{1}{s} \right) e^{xs} \, ds \\ &= - \frac{1}{2 \pi i} \left(\pi \int_{1}^{0} e^{ixt} \, i \, dt - \pi \int_{-1}^{0} e^{ixt} \, i \, dt \right) \\ &= \frac{1}{2} \int_{-1}^{1} e^{ixt} \, dt \\ &= \frac{1}{2ix} \left(e^{ix} - e^{-ix} \right) \\ &= \frac{\sin(x)}{x}. \end{align}$$