Source: J. E. Santos, Cambridge Part IB Complex Methods, Example Sheet 3, Question 12.
Using the equality $\int_0^\infty e^{-x^2} d x
= \frac{1}{2}\sqrt{\pi}$, find the Laplace transform of $f(t) = t^{-1/2}$.
By integrating around a Bromwich keyhole contour, verify the inversion formula
for $f(t)$. What is the Laplace transform of $t^{1/2}$?
The integral can be evaluated easily by a change of variable $\tau^2 = t$:
\[\hat{F}(s) = \mathcal{L}[f(t)] = \int^\infty _ 0 dt \,e^{-st}t^{-1/2} = \int^\infty_0 d\tau \, 2e^{-s\tau^2} = \sqrt{\frac{\pi}{s}}.\tag{1.1}\]
Now, we shall verify the inverse Laplace transform of $\hat{F}(s)$. The function $\hat{F}(s)$ has a principal branch cut from $-\infty$ to $0$. To see that, write $s=re^{i\theta}$, and we find
\[s^{-1/2} = r^{-1/2}e^{-i\theta/2} = r^{-1/2}e^{-i\arg s/2} \quad \text{with} \quad \arg s\in (-\pi, \pi).\tag{1.2}\]
Therefore, just above the branch cut, we have $s^{-1/2} = -ir^{-1/2}$; just below the branch cut, we have $s^{-1/2} = ir^{-1/2}$.
The inverse Laplace transform is given by
\[f(t) = \frac{\sqrt{\pi}}{2\pi i}\int_{\mathcal{C}_\gamma}ds\,e^{st}s^{-1/2}= \frac{\sqrt{\pi}}{2\pi i}\int^{\gamma+i\infty}_{\gamma-\infty}ds\,e^{st}s^{-1/2}\tag{1.3}\]
with $\gamma\to 0^+$. We can choose to integrate the Bromwich keyhole contour, $\mathcal{C}_B = \Gamma_+ \cup \Gamma_- \cup \mathcal{C}_\epsilon \cup \mathcal{C}_\gamma \cup \mathcal{C}_+\cup \mathcal{C}_-$, where
\[\Gamma_\pm = \lim_{\epsilon\to 0}\{z\in(\pm i \epsilon -\infty, \pm i \epsilon]\}, \quad\mathcal{C}_\epsilon = \lim_{\epsilon\to0^+}\{z=\epsilon e^{i\theta}|\theta\in[-\pi/2, \pi/2]\},\tag{1.4}\]
and the large arcs are given by
\[\mathcal{C}_\pm = \lim_{R\to\infty} \{ z = R e^{i\theta} | \theta \in [\pm \tfrac{\pi}{2}, \pm \pi] \}.\tag{1.5}\]
Since there are no poles enclosed by this contour, we have
\[\oint_{\mathcal{C}_B}ds\,e^{st}s^{-1/2} = 0.\tag{1.6}\]
It is easy to see that the integral along $\mathcal{C}_\pm$ and $\mathcal{C}_\epsilon$ vanishes. The integral along $\Gamma_\pm$ is given by
\[I_\pm = \int_{\Gamma_\pm}ds\,e^{st}s^{-1/2} = \pm\int^0_{-\infty}dr\, r^{-1/2}e^{rt} = \mp i\int^\infty_0dr\,r^{-1/2}e^{-rt} = \mp i\sqrt{\frac{\pi}{t}}.\tag{1.7}\]
Thus, we find (notice the direction of the contour integration)
\[2\sqrt{\pi}if(t) +I_+ - I_- = 0 \implies f(t) = t^{-1/2}.\tag{1.8}\]