Inverse Laplace transform of $\frac{1}{s\sinh{c\sqrt{s}}}$ I have some questions.
I have to take the inverse Laplace transform of the given function below by residue theorem: 
$\frac{1}{s\sinh{c\sqrt{s}}}$ 
where s is Laplacian variable and c is a real constant. Actually it has a very crowded numerator which I do not need to define. So far I know that it has a simple pole at s = zero and infinitely many poles at c√s = nπi
Questions:
1- What is the order of the poles at nπi? Is it 1/2?
2- At zero, what is the order? (Because to residue at zero, s also contributes)
3- What is the way of solving this question at nπi by residue theorem (without defining sinh function as exponential function)?
Thanks for any help.
 A: By the residue theorem we have (assuming $c,s>0$)
$$ \frac{\pi}{\sin(\pi z)}=\frac{1}{z}+\sum_{n\geq 1}\frac{2z(-1)^n}{z^2-n^2} \tag{1}$$
$$ \frac{1}{\sinh(z)}=\frac{1}{z}+\sum_{n\geq 1}\frac{2z(-1)^n}{z^2+\pi^2 n^2} \tag{2}$$
$$ \frac{1}{\sinh(cz)}=\frac{1}{cz}+\sum_{n\geq 1}\frac{2cz(-1)^n}{c^2z^2+\pi^2 n^2} \tag{3}$$
$$ \frac{1}{\sinh(c\sqrt{s})}=\frac{1}{c\sqrt{s}}+\sum_{n\geq 1}\frac{2c\sqrt{s}(-1)^n}{c^2 s+\pi^2 n^2} \tag{4}$$
$$ \frac{1}{\sqrt{s}\sinh(c\sqrt{s})}=\frac{1}{c s}+\sum_{n\geq 1}\frac{2c(-1)^n}{c^2 s+\pi^2 n^2} \tag{5}$$
$$\mathcal{L}^{-1}\left( \frac{1}{\sqrt{s}\sinh(c\sqrt{s})}\right)=\frac{1}{c }+\sum_{n\geq 1}\frac{2}{c}(-1)^n \exp\left(-\frac{n^2 \pi^2 x}{c^2}\right) =\color{blue}{\tfrac{1}{c}\,\vartheta_4\left(0;e^{-\pi^2 x/c^2}\right)}\tag{6}$$
so the answer in this case is given by a Jacobi Theta function. Similarly,
$$\mathcal{L}^{-1}\left( \frac{1}{s\sinh(c\sqrt{s})}\right)=\frac{2\sqrt{x}}{c\sqrt{\pi}}+\frac{4}{\pi^{3/2}}\sum_{n\geq 1}\frac{(-1)^n}{n}\,F\left(\frac{\pi n \sqrt{x}}{c}\right)\tag{7} $$
where $F$ is Dawson function. The last series is related with an integral involving the $\vartheta_4$ function.
A: Regarding Jack D'Aurizio's comment and a typo had seen in the first post, I think you want to find inverse Laplace transform of $f(s)=\dfrac{1}{\sqrt{s}\sinh c\sqrt{s}}$. Then from
$$\dfrac{1}{\sqrt{s}\sinh c\sqrt{s}}=\dfrac{1}{\sqrt{s}\left((c\sqrt{s})+\dfrac{(c\sqrt{s})^3}{3!}+\dfrac{(c\sqrt{s})^5}{5!}+\cdots\right)}=\dfrac{1}{cs\left(1+\dfrac{c^2s}{3!}+\dfrac{c^4s^2}{5!}+\cdots\right)}$$
shows this function has a simple pole in $s=0$, also for other poles we have $\sinh c\sqrt{s}=0$ which determines poles $s_k=-\dfrac{k^2\pi^2}{c^2}$. So the residues of $e^{st}f(s)$ at these poles are
\begin{align}
\lim_{s\to0}(s-0)\dfrac{e^{st}}{\sqrt{s}\sinh c\sqrt{s}} 
&= \dfrac{e^{0}}{c.1} \\
&= \dfrac{1}{c} \\
\lim_{s\to s_k}(s-s_k)\dfrac{e^{st}}{\sqrt{s}\sinh c\sqrt{s}} 
&= \lim_{s\to s_k}e^{st}\dfrac{1+ts+t\dfrac{k^2\pi^2}{c^2}}{\dfrac{1}{2\sqrt{s}}\left(\sinh c\sqrt{s}+c\sqrt{s}\cosh c\sqrt{s}\right) } \hspace{2cm}\text{With L'Hopital}\\
&= e^{s_kt}\dfrac{2}{c}(-1)^k\\
&= \dfrac{2(-1)^k}{c}e^{-\dfrac{k^2\pi^2}{c^2}t}
\end{align}
hence
$${\cal L}^{-1}\left(\dfrac{1}{\sqrt{s}\sinh c\sqrt{s}}\right)=\color{blue}{\dfrac{1}{c}+\dfrac{2}{c}\sum_{k=1}^\infty (-1)^ke^{-\dfrac{k^2\pi^2}{c^2}t}}$$
