Inverse Laplace of $\frac{\sinh{x\sqrt{s}}}{s^2\sinh{\sqrt{s}}}$ What is the inverse Laplace of $\frac{\sinh{x\sqrt{s}}}{s^2\sinh{\sqrt{s}}}$?
Using the residues, I can calculate the residues at $s_n=2n\pi i$, but I have problem in calculating residue at $s=0$. 
The final answer should be:
$f(t)=\frac{1}{6}x(x^2-1)+xt+\frac{2}{\pi ^3}\sum\limits_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^3}e^{-n^2\pi ^2 t}\sin{n\pi x}$
which I could get the last term from the residues at $s_n=2n\pi i$, but I have problem for the first two terms.

First of all, the poles you are thinking of are at $s=-n^2 \pi^2$.  ($s=0$ is a removable pole of the sinh terms.)
Also note that, even though there are square roots in the sinh terms, we need not worry about taking branches because any odd phase behavior will cancel in the fraction.  Therefore, what we have is a straightforward Bromwich contour encircling the poles.
The residue from the pole at $s_n=-n^2 \pi^2$ is simply
$$\frac1{s_n^2} \frac{\sinh{x \sqrt{s_n}}}{\cosh{\sqrt{s_n}}} 2 \sqrt{s_n} e^{s_n t} = \frac{2 (-1)^{n+1}}{n^3 \pi^3} e^{-n^2 \pi^2 t} \sin{n \pi x}$$
The residue at $s=0$ is a little trickier but still straightforward; it is equal to
$$\begin{align}\left [\frac{d}{ds} \frac{\sinh{x \sqrt{s}}}{\sinh{\sqrt{s}}} e^{s t} \right ]_{s=0}  &= \left [\frac{d}{ds} \frac{\sinh{x \sqrt{s}}}{\sinh{\sqrt{s}}} \right ]_{s=0} + t \left [ \frac{\sinh{x \sqrt{s}}}{\sinh{\sqrt{s}}} \right ]_{s=0}\\ &= \lim_{s\to0} \frac{x \sinh{\sqrt{s}} \cosh{x\sqrt{s}}-\sinh{x \sqrt{s}} \cosh{\sqrt{s}}}{2 \sqrt{s} \sinh^2{\sqrt{s}}} + t x \\ &= \lim_{s\to0} \frac{\frac{x}{2} \left (1+\frac{s}{6} \right ) \left (1+\frac{x^2 s}{2} \right )-\frac{x}{2} \left (1+\frac{x^2 s}{6} \right ) \left (1+\frac{s}{2} \right )}{s}+ t x \\ &= \frac16 (x^3-x) + x t \end{align}$$
The sought-after result follows.
