Can someone explain this Laplace Transform? Why is the inverse Laplace Transform of $$\frac{\sinh(x\sqrt{s})}{ s\cdot \sinh{\sqrt{s}}}$$ equal to $$ x + \frac{2}{\pi}\sum_{n=1}^{\infty} \frac{(-1)^n}{n}e^{-(n\pi)^2t}\sin{n\pi x} ?$$ 
Any help would be great, totally confused.
 A: Consider the poles of 
$$F(s) = \frac{\sinh(x\sqrt{s})}{ s\cdot \sinh{\sqrt{s}}} e^{st}$$
First we need to see if $s =0$ is a branch point or not 
$$F(s) = \frac{1}{s}\left( \frac{(x\sqrt{s})+(x\sqrt{s})^3/3!+\cdots}{(\sqrt{s})+(\sqrt{s})^3/3!+\cdots} \right) \left(1+(st)+\frac{(st)^2}{2!} + O(s^3)\right)$$
$$F(s) = \frac{x}{s} \left( \frac{1+x^2s/3!+\cdots}{1+s/3!+\cdots} \right) \left(1+(st)+\frac{(st)^2}{2!} + O(s^3)\right)$$
Hence we see that $s=0$ is a pole of order 1 with residue $x$. Now consider the other poles where the deliminator $\sinh(\sqrt{s})$ has zeros when $\sqrt{s} = n\pi i$ hence $s = -(n\pi)^2$ are the poles.
$$\sum_{s_0}\mathrm{Res}(F,s_0)= x+ \sum_{n=1}^\infty \mathrm{Res}(F,-(n\pi)^2)$$
$$\mathrm{Res}(F,-(n\pi)^2) = \lim_{s \to -(n\pi)^2} (s+(n\pi)^2) \frac{\sinh(x\sqrt{s})}{ s\cdot \sinh{\sqrt{s}}} e^{st} = 2 \lim_{s \to -(n\pi)^2} (s+(n\pi)^2) \frac{\sinh(x\sqrt{s})}{ \sqrt{s}\cdot \cosh{\sqrt{s}}} e^{st}$$
$$\mathrm{Res}(F,-(n\pi)^2)  = \frac{i \sin(n \pi x)}{ i (n\pi) \cdot \cos{(n\pi)}} e^{-(n\pi)^2 t} = \frac{(-1)^n}{n \pi }\sin(n \pi x) e^{-(n\pi)^2 t} $$
Collecting the results together we have 
$$\mathcal{L}^{-1}(f(s)) = x + \sum_{n \geq 1} \frac{(-1)^n}{n \pi }\sin(n \pi x) e^{-(n\pi)^2 t} $$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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\begin{equation}
\mbox{Note that}\quad
{1 \over \sinh\pars{\root{s}}} = {1 \over \root{s}} +
2\sum_{n = 1}^{\infty}\pars{-1}^{n}\color{#f00}{\root{s} \over s + n^{2}\pi^{2}}
\label{1}\tag{1}
\end{equation}
and
\begin{align}
&\bbox[#ffe,15px,border:1px dotted navy]{\ds{\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic}
{\sinh\pars{x\root{s}} \over s}\,\color{#f00}{\root{s} \over s + n^{2}\pi^{2}}\,
\exp\pars{ts}\,{\dd s \over 2\pi\ic}}}
\\[5mm] = &\
-\int_{-\infty}^{0}{\sinh\pars{x\root{-s}\ic} \over s}\,{\root{-s}\ic \over
s + n^{2}\pi^{2} + \ic 0^{+}}\,\exp\pars{ts}\,{\dd s \over 2\pi\ic}
\\[5mm] & -
\int_{0}^{-\infty}{\sinh\pars{-x\root{-s}\ic} \over s}\,{-\root{-s}\ic \over
s + n^{2}\pi^{2} - \ic 0^{+}}\,\exp\pars{ts}\,{\dd s \over 2\pi\ic}
\\[1cm] = &\
\int_{0}^{\infty}{\sin\pars{x\root{s}} \over s}\,{\root{s} \over
s - n^{2}\pi^{2} - \ic 0^{+}}\,\exp\pars{-ts}\,{\dd s \over 2\pi\ic}
\\[5mm] & -
\int_{0}^{\infty}{\sin\pars{x\root{s}} \over s}\,{\root{s} \over
s - n^{2}\pi^{2} + \ic 0^{+}}\,\exp\pars{-ts}\,{\dd s \over 2\pi\ic}
\\[1cm] & =
\int_{0}^{\infty}{\sin\pars{x\root{s}} \over \root{s}}\overbrace{\bracks{%
{1 \over s - n^{2}\pi^{2} - \ic 0^{+}} -
{1 \over s - n^{2}\pi^{2} + \ic 0^{+}}}}
^{\ds{2\pi\ic\,\delta\pars{s - n^{2}\pi^{2}}}}\,\exp\pars{-ts}
\,{\dd s \over 2\pi\ic}
\\[5mm] = &\
\bbox[#ffe,15px,border:1px dotted navy]{\ds{{\sin\pars{n\pi x} \over n\pi}\expo{-n^{2}\pi^{2}t}}}\qquad\qquad
\pars{~\mbox{note that}\
\lim_{n \to 0}\bracks{{\sin\pars{n\pi x} \over n\pi}\expo{-n^{2}\pi^{2}t}} = \color{#f00}{x}~}\label{2}\tag{2}
\end{align}

With \eqref{1} and \eqref{2}:
$$
\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic}
{\sinh\pars{x\root{s}} \over s\sinh\pars{\root{s}}}\,
\exp\pars{ts}\,{\dd s \over 2\pi\ic} =
\bbox[#ffe,15px,border:1px dotted navy]{\ds{x + {2 \over \pi}\sum_{n = 1}^{\infty}{\sin\pars{n\pi x} \over n}
\,\exp\pars{-\bracks{n\pi}^{2}t}}}
$$
