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
Complex Analysis – Poles on Branch Cuts in Contour Integration
cauchy-principal-valuecomplex-analysiscontour-integrationlaplace transform
Related Solutions
Consider $f(\omega) = \omega^{-3/4} (x-\omega)^{-3/4} (1-\omega)^{-3/4}$ and $ I(x) = \oint_{\vert \omega \vert = x} f(\omega) \mathrm{d} \omega$.
The function $f(\omega)$ is discontinuous at $\omega = -x$ along $\vert \omega \vert = x$ and has integrable singularity at $\omega = x$. Making a change of variables, $\omega = x z$: $$ I(x) = \frac{1}{\sqrt{x}} \oint_{\vert z \vert = 1} z^{-3/4} (1-z)^{-3/4} (1-x z)^{-3/4} \mathrm{d} z = \frac{1}{\sqrt{x}} \oint_{\vert z \vert = 1} h\left(z\right) \, \mathrm{d} z $$
Let $\mathcal{C}$ denote the circle $\vert z \vert = 1$. Let $C_{-1, \delta}$ denote segment of $\mathcal{C}$ which crosses the negative axis and of length $2 \pi \delta$, with $\pi \delta$ above and $\pi \delta$ below the negative axis.
It is clear that integral along $\mathcal{C}_{-1,\delta}$ is vanishing as $\delta \to 0$: $$ \begin{eqnarray} \left\vert \int_{\mathcal{C}_{-1,\delta}} z^{-3/4} (1-z)^{-3/4} (1-x z)^{-3/4} \mathrm{d} z \right\vert &\le& (2(1+x))^{-3/4} \left\vert \int_{\mathcal{C}_{-1,\delta}} z^{-3/4} \mathrm{d} z \right\vert \\ &=& \left(2 (1+x) \right)^{-3/4} \frac{8\sqrt{2}}{7} \sin\left(\frac{7 \delta}{8}\right) \left( \cos\left(\frac{7 \delta}{8}\right) - \sin\left(\frac{7 \delta}{8}\right) \right) \\ &\le& \sqrt{2} \delta \left(2 (1+x) \right)^{-3/4} \end{eqnarray} $$
Let's complete $\mathcal{C} \backslash \mathcal{C}_{-1,\delta}$ with integration along $(-1,0)$ above the axis and then along $(0,-1)$ below the axis so as to complete the contour, and call the completed contour $\mathcal{L}$. Then $$ \begin{eqnarray} \oint_{\mathcal{C}} h\left(z\right) \, \mathrm{d} z &=& \oint_{\mathcal{L}} h\left(z\right) \, \mathrm{d} z + \int_0^1 \left( h\left(-y - i \epsilon \right) - h\left(-y + i \epsilon \right) \right)\, \mathrm{d} y \\ &=& \oint_{\mathcal{L}} h\left(z\right) \, \mathrm{d} z + \left( \mathrm{e}^{i \frac{3 \pi}{4}} - \mathrm{e}^{-i \frac{3 \pi}{4}} \right) \int_0^1 y^{-3/4}(1+y)^{-3/4}(1+x y)^{-3/4} \mathrm{d} y \end{eqnarray} $$ The claim is that the principal value of $\oint_\mathcal{L} h(z) \mathrm{d} z = 0$, so we get $$ I(x) = \frac{2 i \sin\left( 3/4 \pi \right)}{\sqrt{x}} \int_0^1 y^{-3/4}(1+y)^{-3/4}(1+x y)^{-3/4} \mathrm{d} y $$
Now, let's check this with quadratures:
Notice that the purported answer you gave in your post can not be correct, as it is not purely imaginary.
The result below assumes $t>0$ (as usual for Laplace transforms).
In the Bromwich contour $\gamma$ has to be chosen large enough that it is to the right of all singularities (poles and branch points) so $\gamma = 0^+$ is perfectly valid. The singularities of $\log s/(1+s)$ are a pole at $s=-1$ and two branch points at $s=0$ and $s=\infty$ which we connect via a branch cut along the negative real line.
Then we can deform the contour further to a path which starts at $-\infty -i 0^+$. Runs along the negative real line just below the branch cut. Ends at $0-i 0^+$ in a little semi-circle and then runs back from $0+i 0^+$ to $-\infty +i0^+$ just above the branch cut.
The Bromwich integral thus is given by $$f(t)=\frac1{2\pi i} \int_{-\infty}^0\!dx\, \left( \frac{\log (x-i0^+)}{1+x-i0^+ } - \frac{\log (x+i0^+)}{1+x+i0^+ } \right) e^{x t} $$ as the small circle around the branch point at $0$ does not contribute ($|z|\log z \to 0$ for $|z|\to0$).
In the remaining integral, we use $\log(x \pm i 0^+) = \log |x| \pm i \pi$ valid for $x<0$: $$\begin{align} f(t) &= \frac1{2\pi i} \int_{-\infty}^0\!dx\, \left( \frac{\log |x|-i\pi}{1+x-i0^+ } - \frac{\log |x|+i\pi}{1+x+i0^+} \right) e^{x t}\\ &= \frac1{2\pi i} \overbrace{\int_{-\infty}^0\!dx\, \log |x| \underbrace{\left( \frac1{1+x-i0^+ } - \frac1{1+x+i0^+}\right)}_{2\pi i \delta(x+1)} e^{x t}}^{=0}\\ &\quad -\frac12 \int_{-\infty}^0\!dx \underbrace{\left( \frac1{1+x-i0^+ } + \frac1{1+x+i0^+}\right)}_{2\mathcal{P}\,(1+x)^{-1}} e^{x t} \\ &= -\int_{-\infty}^0\!dx \,\mathcal{P} \frac{e^{x t}}{1+x} =- e^{-t} \int_{-\infty}^t\!ds \,\mathcal{P} \frac{e^{s}}{s}\\ &=- e^{-t} \mathop{\rm Ei}(t) \end{align}$$ with $s=(1+x)t$ and Ei the exponential integral.
Best Answer
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,