What happens to poles lying on branch cuts in contour integration?

Question

Inverse the Laplace Transform $$\frac{1}{\sqrt{s}}\cdot\frac{1}{1 + s}$$ back to time domain requires evaluation of Bromwich integration: $$\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty} \frac{1}{\sqrt{s}}\cdot\frac{1}{1 + s} e^{st} ds$$ The $\sqrt{s}$ term causes a branch cut along the negative half $x$-axis, including the origin. My question is what happens to the point $(-1, 0)$: it is both a pole, and lies on the branch cut. More specifically, when constructing the contour integration path, could I just ignore the pole because the branch cut removes it (figure A), or I should deform the contour as half circles around the pole, once on each side of the branch cut (figure B), as if the pole is duplicated? Figure A and B: contour integration path

Answer 1

It would be the situation in B: you would deform around the pole. It works as follows.

The inverse Laplace transform is given by Cauchy's theorem. I present the parametrization of each piece of the contour, assuming that the radius of the semicircular detour about the pole $z=-1$ and the branch point $z=0$ is $\epsilon$:

$$\int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s} (1+s)} + e^{i \pi} \int_{\infty}^{1+\epsilon} dx \frac{e^{-t x}}{e^{i \pi/2} \sqrt{x} (1-x)}+i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{e^{t (-1+ \epsilon e^{i \phi})}}{\sqrt{e^{i \pi}+\epsilon e^{i \phi}} (\epsilon e^{i \phi})}\\+ e^{i \pi} \int_{1-\epsilon}^{\epsilon} dx \frac{e^{-t x}}{e^{i \pi/2} \sqrt{x} (1-x)}+i \epsilon \int_{\pi}^{-\pi} d\phi \, e^{i \phi} \frac{e^{t \epsilon e^{i \phi}}}{\sqrt{\epsilon e^{i \phi}} (1+\epsilon e^{i \phi})} +e^{-i \pi} \int_{\epsilon}^{1-\epsilon} dx \frac{e^{-t x}}{e^{-i \pi/2} \sqrt{x} (1-x)}\\+ i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{e^{t (-1+ \epsilon e^{i \phi})}}{\sqrt{e^{-i \pi}+\epsilon e^{i \phi}} (\epsilon e^{i \phi})}+ e^{-i \pi} \int_{1+\epsilon}^{\infty} dx \frac{e^{-t x}}{e^{-i \pi/2} \sqrt{x} (1-x)} = 0$$

Note that the integrals about the semicircular detours above and below the axis (the 3rd and the 7th integrals) cancel. In the limit as $\epsilon \to 0$, the integral about the branch point (the 5th integral) also vanishes. We are then left with, as $\epsilon \to 0$,

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s} (1+s)} + \frac1{2 \pi} PV \int_{\infty}^0 dx \frac{e^{-t x}}{\sqrt{x} (1-x)} - \frac1{2 \pi} PV \int_0^{\infty} dx \frac{e^{-t x}}{\sqrt{x} (1-x)} = 0$$

where $PV$ denotes the Cauchy principal value of the integral. Thus, the ILT is given by (subbing $x=u^2$)

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s} (1+s)} = \frac1{\pi} PV \int_{-\infty}^{\infty} du \, \frac{e^{-t u^2}}{1-u^2} $$

To evaluate the integral, we rewrite as

$$e^{-t} PV \int_{-\infty}^{\infty} du \, \frac{e^{t (1- u^2)}}{1-u^2} = e^{-t} I(t)$$

where

$$I'(t) = e^{t} PV \int_{-\infty}^{\infty} du \, e^{-t u^2} = \sqrt{\pi} t^{-1/2} e^{t} $$

and $I(0) = 0$. Thus,

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s} (1+s)} = e^{-t}\frac1{\pi} \sqrt{\pi} \int_0^t dt' \, t'^{-1/2} e^{t'} = e^{-t} \frac{2}{\sqrt{\pi}} \int_0^{\sqrt{t}} dv \, e^{v^2} $$

or, finally,

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s} (1+s)} = e^{-t} \operatorname{erfi}{\left (\sqrt{t} \right )} $$