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

  • $\begingroup$ Yah im sorry ,its just a mistake​ when I typing $\endgroup$ – Tasneem May 17 '17 at 3:11

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

  • $\begingroup$ What's the meaning of $0^+ + \infty i$ ? $\endgroup$ – Tasneem May 17 '17 at 3:19
  • $\begingroup$ @Tasneem It's a short-cut for $\displaystyle\lim_{\epsilon \to 0^{+}}\int_{\epsilon - \infty\,\mathrm{i}}^{\epsilon + \infty\,\mathrm{i}}$ because the integration path has to be 'at the right' of the integrand singularities. In this case, 'to the right' of zero. $\endgroup$ – Felix Marin May 17 '17 at 3:24


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.