Inverse Laplace Transform of $e^{-\sqrt{s^2+s}}$. I've come across this problem trying to find an integral representation to a PDE (damped wave equation with initial conditions).
What I would like to do is compute $$f(t) = \mathcal{L}^{-1}\left[e^{-\sqrt{s^2+s}}\right](t)$$
via the bromwich integral,
$$f(t) = \frac1{2 \pi i} \int_{c- i \infty}^{c + i \infty} e^{-\sqrt{s^2+s}}e^{st}ds $$
My progress so far has been stunted by the fact that we have branch points at $s = 0$, and $s = -1$. My idea so far has been to make the branch cut the interval $[-1, 0]$ on the real axis, however this has been quite annoying as the typical shape of the contour takes the shape of a reversed $D$. My idea might be to do something of a dog bone contour to avoid the branch, but I've been having trouble keeping track of the evaluation.
Doing it this way, the sum of the integrals would have to be zero, there are no poles involved. Thankfully the outer arcing contours would go to zero, it's really the contour going around the sneaky branch cut that would be cumbersome. Perhaps a dog bone would work? 
My goal would be to even just come up with some sort of series/integral representation of the laplace transform. Any help is appreciated. 
 A: Well, we want to find:
$$\mathscr{P}\left(t\right):=\mathscr{L}_\text{s}^{-1}\left[\exp\left(-\sqrt{\text{s}^2+\text{s}}\right)\right]_{\left(t\right)}\tag1$$
Now, we first look at the series expansion of $\exp$:
$$\exp\left(-\sqrt{\text{s}^2+\text{s}}\right)=\sum_{\text{k}=0}^\infty\frac{\left(-\sqrt{\text{s}^2+\text{s}}\right)^\text{k}}{\text{k}!}=\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}\cdot\left(\sqrt{\text{s}^2+\text{s}}\right)^\text{k}}{\text{k}!}=$$
$$\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}\cdot\left(\left(\text{s}^2+\text{s}\right)^\frac{1}{2}\right)^\text{k}}{\text{k}!}=\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\text{k}!}\cdot\left(\text{s}^2+\text{s}\right)^\frac{\text{k}}{2}=$$
$$\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\text{k}!}\cdot\left(\text{s}\cdot\left(\text{s}+1\right)\right)^\frac{\text{k}}{2}=\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\text{k}!}\cdot\text{s}^\frac{\text{k}}{2}\cdot\left(\text{s}+1\right)^\frac{\text{k}}{2}\tag2$$
Using the 'convolution theorem' of the Laplace transform:
$$\mathscr{P}\left(t\right)=\mathscr{L}_\text{s}^{-1}\left[\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\text{k}!}\cdot\text{s}^\frac{\text{k}}{2}\cdot\left(\text{s}+1\right)^\frac{\text{k}}{2}\right]_{\left(t\right)}=$$
$$\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\text{k}!}\cdot\mathscr{L}_\text{s}^{-1}\left[\text{s}^\frac{\text{k}}{2}\cdot\left(\text{s}+1\right)^\frac{\text{k}}{2}\right]_{\left(t\right)}=$$
$$\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\text{k}!}\cdot\int_0^t\mathscr{L}_\text{s}^{-1}\left[\text{s}^\frac{\text{k}}{2}\right]_{\left(t-\tau\right)}\cdot\mathscr{L}_\text{s}^{-1}\left[\left(\text{s}+1\right)^\frac{\text{k}}{2}\right]_{\left(\tau\right)}\space\text{d}\tau=$$
$$\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\text{k}!}\cdot\int_0^t\frac{\left(t-\tau\right)^{-\frac{\text{k}}{2}-1}}{\Gamma\left(-\frac{\text{k}}{2}\right)}\cdot\frac{\exp\left(-\tau\right)\cdot\tau^{-\frac{\text{k}}{2}-1}}{\Gamma\left(-\frac{\text{k}}{2}\right)}\space\text{d}\tau=$$
$$\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\Gamma^2\left(-\frac{\text{k}}{2}\right)\cdot\left(\text{k}!\right)}\cdot\int_0^t\left(t-\tau\right)^{-\frac{\text{k}}{2}-1}\cdot\exp\left(-\tau\right)\cdot\tau^{-\frac{\text{k}}{2}-1}\space\text{d}\tau\tag3$$
