finding the inverse Laplace transform of $\frac{1}{z\sqrt{z+1}}$ i know that the inverse Laplace transform is given by $$2\pi i \left\{\sum\space\text{ of the residues at the poles of}\space e^{zt}f(z)\right\}- \frac{1}{2 \pi i}\int \text{ along the branch cut}$$
$$f(z) = \frac{1}{z\sqrt{z+1}}$$
there is a branch point at $z=-1$ and there is also a singularity at $z=-1$ and a pole at $z = 0$
i want to know if i include the residue at the pole $z=-1$ even though its the branch point or if i simply include the residue at the pole $z=0$ only and subtract it from the integral across the branch cut.
i.e. 
$$\mathcal{L}^{-1}f(z) = 2\pi i(\operatorname{Res}(e^{zt}f(z);0)-\frac{1}{2\pi i}\int \text{ branch cut}$$
 A: There is no pole at $z=-1$; it is merely a branch point.  The Bromwich contour from which the ILT may be found must be deformed so as to avoid this branch point, like this:

You may show that the integrals over $C_2$, $C_4$, and $C_6$ all vanish.  The result is, letting $z=-1+e^{i \pi} u$ on $C_3$ and $z=-1+e^{-i \pi} u$ on $C_5$,
$$\int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{s \sqrt{1+s}} + e^{-i \pi/2} \int_{\infty}^0 du \frac{e^{-(1+u) t}}{(1+u) \sqrt{u}} \\ + e^{i \pi/2} \int_0^{\infty} du \frac{e^{-(1+u) t}}{(1+u) \sqrt{u}} = i 2 \pi$$
as the residue at the pole $z=0$ is $1$.
From this, you may rearrange to get that the ILT is
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{s \sqrt{1+s}} = 1-\frac1{\pi} \int_0^{\infty} du \frac{e^{-(1+u) t}}{(1+u) \sqrt{u}} $$
The integral may be evaluated by differentiating with respect to $t$ and subbing $u=v^2$.  The result, which I leave to the reader, is
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{s \sqrt{1+s}} = \operatorname{erf}{\sqrt{t}} $$
ADDENDUM
A little more detail on the evaluation of the integral on the RHS above.  Let
$$I(t) = \int_0^{\infty} du \frac{e^{-(1+u) t}}{(1+u) \sqrt{u}}  = \int_{-\infty}^{\infty} dv \frac{e^{-t (1+v^2)}}{1+v^2}$$
Then
$$I'(t) = -\int_{-\infty}^{\infty} dv\, e^{-t (1+v^2)} = \sqrt{\pi} t^{-1/2} e^{-t} $$
$$\implies I(t) = I(0) - \sqrt{\pi} \int_0^t dt' \, t'^{-1/2} e^{-t'} = \pi - 2 \sqrt{\pi} \int_0^{\sqrt{t}} du \, e^{-u^2} = \pi - \pi \, \operatorname{erf}{\sqrt{t}}$$
The result follows.
