Can someone explain this Laplace Transform?

Question

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.

Answer 1

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

$$\mathcal{L}^{-1}(f(s)) = x + \sum_{n \geq 1} \frac{(-1)^n}{n \pi }\sin(n \pi x) e^{-(n\pi)^2 t} $$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

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