Laplace transforms and error function

Question

Show that $\displaystyle\frac{e ^{-a \sqrt s}}{ (s \sqrt s) }$

Is the Laplace transform of :

$2 \sqrt \frac{t}{\pi} e^{\frac {-a^2}{4t}} - a \operatorname{erfc}\left( \frac {a}{2 \sqrt t} \right)$

Where, erfc is the complete error function

I really tried hardly to prove that , without any result ,I searched on the internet , some use series to find the laplace transform of erfc ,which I don't want to use, can anyone​ could help

Thanks in advanced

Answer 1

Simplification

You may want to use integration by parts on the difficult bit of the integral:

\begin{align*} \mathcal{L} \left[ \operatorname{erfc}\left( \frac{a}{2 \sqrt{t}} \right) \right] & = \int^\infty_0 \mathrm{e}^{-st} \operatorname{erfc}\left( \frac{a}{2 \sqrt{t}} \right) \, \mathrm{d}t \\ & = \left. -\frac{e^{-st}}{s} \operatorname{erfc}\left( \frac{a}{2 \sqrt{t}} \right) \right|^\infty_0+\frac{1}{s} \int^\infty_0 t^{-3/2} \mathrm{e}^{-s t + a^2/(4t)} \, \mathrm{d}t \\ & = \frac{1}{s} \int^\infty_0 t^{-3/2} \mathrm{e}^{-s t + a^2/(4t)} \, \mathrm{d}t, \end{align*} where I have taken into account the definition of the error function in order to compute its derivative and I have leveraged the values of $\operatorname{erfc}$ at $0$ and $\infty$. Now you are left with computing the last definite integral.

Let me know if this provides any help.

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{align} \mbox{Note that}\quad &\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic} {\expo{-a\root{s}}\over s\root{s}}\,\expo{ts}\,{\dd s \over 2\pi\ic} \\[5mm] & = -a\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic} {\expo{ts} \over s}\,{\dd s \over 2\pi\ic} + \int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic} {\expo{-a\root{s}} + a\root{s}\over s\root{s}}\,\expo{ts} \,{\dd s \over 2\pi\ic} \\[5mm] & = -a + \int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic} {\expo{-a\root{s}} + a\root{s}\over s\root{s}}\,\expo{ts} \,{\dd s \over 2\pi\ic}\label{1}\tag{1} \end{align}

The remaining integral, in \eqref{1}, is evaluated as follows:

\begin{align} &\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic} {\expo{-a\root{s}} + a\root{s} \over s\root{s}}\,\expo{ts}\,{\dd s \over 2\pi\ic} \\[1cm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}\,\,\,&\ -\int_{-\infty}^{-\epsilon} {\exp\pars{-a\root{-s}\expo{\ic\pi/2}} + a\root{-s}\expo{\ic\pi/2} \over s\root{-s}\expo{\ic\pi/2}}\,\expo{ts}\,{\dd s \over 2\pi\ic} \\[2mm] & - \int_{\pi}^{-\pi}{1 \over \epsilon^{3/2}\expo{3\ic\theta/2}}\, {\epsilon\expo{\ic\theta}\ic\,\dd\theta \over 2\pi\ic} \\[2mm] &\ -\int^{-\infty}_{-\epsilon} {\exp\pars{-a\root{-s}\expo{-\ic\pi/2}} + a\root{-s}\expo{-\ic\pi/2} \over s\root{-s}\expo{-\ic\pi/2}}\,\expo{ts}\,{\dd s \over 2\pi\ic} \\[1cm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}\,\,\,&\ -\int_{\epsilon}^{\infty} {\exp\pars{-a\root{s}\ic} + \ic a\root{s} \over s\root{s}}\,\expo{-ts}\,{\dd s \over 2\pi} + {2 \over \pi}\,\epsilon^{-1/2} \\[2mm] &\ - \int_{\epsilon}^{\infty} {\exp\pars{a\root{s}\ic} - \ic a\root{s} \over s\root{s}}\,\expo{-ts} \,{\dd s \over 2\pi} \\[1cm] = &\ -\int_{\epsilon}^{\infty} {\cos\pars{a\root{s}} \over s^{3/2}}\,\expo{-ts}\,{\dd s \over \pi} + {2 \over \pi}\,\epsilon^{-1/2} \\[1cm] \stackrel{\mrm{IBP}}{=}\,\,\,&\ -\,{2 \over \epsilon^{1/2}}\, \cos\pars{a\root{\epsilon}}\expo{-t\epsilon}\,{1 \over \pi} \\[2mm] &\ -\int_{\epsilon}^{\infty}{2 \over s^{1/2}} \bracks{-\sin\pars{a\root{s}}a\,{1 \over 2s^{1/2}}\,\expo{-ts} + \cos\pars{a\root{s}}\expo{-ts}\pars{-t}}\,{\dd s \over \pi} \\[2mm] & + {2 \over \pi}\,\epsilon^{-1/2} \\[1cm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\to}\,\,\,&\ {a \over \pi}\int_{0}^{\infty}{\sin\pars{a\root{s}} \over s}\,\expo{-ts}\,\dd s + {2t \over \pi}\int_{0}^{\infty}{\cos\pars{a\root{s}} \over s^{1/2}} \,\expo{-ts}\,\dd s \\[5mm] \stackrel{s\ \to\ s^{2}}{=}\,\,\,&\ {2a \over \pi}\ \underbrace{\int_{0}^{\infty}{\sin\pars{as} \over s}\,\expo{-ts^{2}}\,\dd s} _{\ds{{1 \over 2}\,\pi\,\mrm{erf}\pars{a \over 2\root{t}}}}\ +\ {4t \over \pi}\ \underbrace{\int_{0}^{\infty}\cos\pars{as}\expo{-ts^{2}}\,\dd s} _{\ds{{1 \over 2}\,\root{\pi \over t}\exp\pars{-\,{a^{2} \over 4t}}}} \\[5mm] = &\ a\,\mrm{erf}\pars{a \over 2\root{t}} + 2\root{t \over \pi}\exp\pars{-\,{a^{2} \over 4t}}\label{2}\tag{2} \end{align}

With \eqref{1} and \eqref{2}:

\begin{align} &\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic} {\expo{-a\root{s}}\over s\root{s}}\,\expo{ts}\,{\dd s \over 2\pi\ic} = -a + a\,\mrm{erf}\pars{a \over 2\root{t}} + 2\root{t \over \pi} \exp\pars{-\,{a^{2} \over 4t}} \\[5mm] = & \bbx{2\root{t \over \pi}\exp\pars{-\,{a^{2} \over 4t}} - a\,\mrm{erfc}\pars{a \over 2\root{t}}} \end{align}

Note that $\ds{\,\mrm{erfc}\pars{z} = 1 - \,\mrm{erf}\pars{z}}$.