# Inverse Laplace transform of fractional power

I want to find the inverse Laplace transform of \begin{align} F(s)=\frac{1}{\sqrt{(s-a)(s-b)}} \end{align} Searching the net did not yield anything. I read post-1, post-2, site-1, but to no avail. I thought it would be easy if I could square the whole thing, i.e. finding inverse L.T. of $F^2(s)$ would be easy. But then how do I get back $\mathbb{L^{-1}}[F]$ knowing $\mathbb{L^{-1}}[F^2]$? Or is there another simpler way?

• Do you know complex analysis and the inverse transform in terms of the Bromwich integral? Commented Mar 24, 2017 at 7:48
• @Fabian I am an engineer.
– Deep
Commented Mar 24, 2017 at 8:09
• I just fixed my answer. Now, it's $\texttt{0 k}$. Commented Mar 26, 2017 at 0:53

The result is $$e^{(a+b) t/2} I_0\Bigl((a-b) t/2\Bigr)$$ with $I_0$ the modified Bessel function of the first kind. If you know about Bessel function and you know about complex analysis, I can give you a proof using the inverse Laplace transform in terms of the Bromwich integral. Otherwise, you have to find a large table of Laplace transforms (involving also Bessel functions) to do the job. See, e.g., the second to last formula here.

• +1 Thanks a lot. Is it possible to generalize it to $1/\sqrt{(s-a_1)(s-a_2)...(s-a_n)}$. Or should I post this as a separate question?
– Deep
Commented Mar 24, 2017 at 8:12
• No. For $n\geq 3$ the resulting functions do not have a name. Commented Mar 24, 2017 at 12:25

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

$\ds{\int_{c\ -\ \infty\ic}^{c\ +\ \infty\ic}{\exp\pars{st} \over \root{\pars{s - a}\pars{s - b}}}\,{\dd s \over 2\pi\ic}:\ {\large ?} .\qquad}$ I'll assume $\ds{a,b \in \mathbb{R}.\qquad \left\{\begin{array}{l} \ds{t > 0} \\[2mm] \ds{c > \max\braces{a,b}} \end{array}\right.}$

Note that \begin{align} &\left.\int_{c\ -\ \infty\ic}^{c\ +\ \infty\ic}{\expo{st} \over \root{\pars{s - a}\pars{s - b}}}\,{\dd s \over 2\pi\ic}\right\vert_{\ t\ >\ 0} \\[5mm] = &\ \left\{\begin{array}{lcl} \ds{\exp\pars{{a + b \over 2}\,t} \int_{1^{+}\ -\ \infty\ic}^{1^{+}\ +\ \infty\ic}{\exp\pars{s\tau} \over \root{s^{2} - 1}}\,{\dd s \over 2\pi\ic}} & \ds{\quad}\mbox{if} & \ds{a \not= b} \\[5mm] \ds{\int_{a^{+}\ -\ \infty\ic}^{a^{+}\ -\ \infty\ic} {\exp\pars{st} \over s - a}\,{\dd s \over 2\pi\ic} = \expo{at}} & \ds{\quad}\mbox{if} & \ds{a = b} \end{array}\right.\label{1}\tag{1} \\[5mm] &\ \mbox{where}\ds{\quad\tau \equiv {\verts{a - b} \over 2}\,t} > 0. \ds{\quad\Theta}\ \mbox{is the}\ Heaviside\ Step\ Function. \end{align}

With $\ds{\quad -\pi < \,\mrm{arg}\pars{\bracks{s \pm 1}^{-1/2}} < \pi}$: \begin{align} &\int_{1^{+}\ -\ \infty\ic}^{1\ +\ \infty\ic}{\exp\pars{s\tau} \over \root{s^{2} - 1}}\,{\dd s \over 2\pi\ic} = \int_{1^{+}\ -\ \infty\ic}^{1\ +\ \infty\ic} \pars{s - 1}^{-1/2}\pars{s + 1}^{-1/2}\exp\pars{s\tau}\,{\dd s \over 2\pi\ic} \\[1cm] = &\ \require{cancel}\cancel{-\int_{-\infty}^{-1} \bracks{\pars{1 - s}^{-1/2}\expo{-\pi\ic/2}} \bracks{\pars{-s - 1}^{-1/2}\expo{-\ic\pi/2}}\exp\pars{s\tau}\,{\dd s \over 2\pi\ic}} \\[5mm] &\ - \int_{-1}^{1} \bracks{\pars{1 - s}^{-1/2}\expo{-\ic\pi/2}} \,\pars{s + 1}^{-1/2}\exp\pars{s\tau}\,{\dd s \over 2\pi\ic} \\[5mm] &\ - \int_{1}^{-1} \bracks{\pars{1 - s}^{-1/2}\expo{\ic\pi/2}}\pars{s + 1}^{-1/2}\exp\pars{s\tau} \,{\dd s \over 2\pi\ic} \\[5mm] &\ - \cancel{\int_{-1}^{-\infty}\bracks{\pars{1 - s}^{-1/2}\expo{\pi\ic/2}} \bracks{\pars{-s - 1}^{-1/2}\expo{\ic\pi/2}}\exp\pars{s\tau}\,{\dd s \over 2\pi\ic}} \\[1cm] = &\ {1 \over \pi}\int_{-1}^{1}{\exp\pars{-s\tau} \over \root{1 - s^{2}}}\,\dd s \,\,\,\stackrel{s\ =\ \sin\pars{\theta}}{=}\,\,\, {1 \over \pi}\int_{-\pi/2}^{\pi/2}\exp\pars{-\sin\pars{\theta}\tau}\,\dd\theta \\[5mm] = &\ {2 \over \pi}\int_{0}^{\pi/2}\cosh\pars{\sin\pars{\theta}\tau}\,\dd\theta = {1 \over \pi}\int_{0}^{\pi}\cos\pars{\sin\pars{\theta}\,\ic\tau}\,\dd\theta = \,\mrm{J}_{0}\pars{\ic\tau} = \bbx{\ds{\,\mrm{I}_{0}\pars{\tau}}} \label{2}\tag{2} \end{align} $\ds{\mrm{J}_{\nu}}$ ( first kind ) and $\ds{\mrm{I}_{\nu}}$ ( modified first kind ) are Bessel Functions. See $\ds{\mathbf{9.1.18}}$ in A & S Table.
With \eqref{1} and \eqref{2}, the final result is given by \begin{align} &\left.\int_{c\ -\ \infty\ic}^{c\ +\ \infty\ic}{\exp\pars{st} \over \root{\pars{s - a}\pars{s - b}}}\,{\dd s \over 2\pi\ic} \right\vert_{\ c\ >\ \max\braces{a,b},\,\,\, t\ >\ 0} \\[5mm] = &\ \bbox[15px,#ffe,border:1px dotted navy]{\ds{\left\{\begin{array}{lcl} \ds{\exp\pars{{a + b \over 2}\,t}\,\mrm{I}_{0}\pars{{\verts{a - b} \over 2}\,t}} & \ds{\quad\mbox{if}\quad} & \ds{a \not= b} \\[3mm] \ds{\exp\pars{at}} & \ds{\quad\mbox{if}\quad} & \ds{a = b} \end{array}\right.}} \end{align}

• @Fabian Fixed. Thanks for your remarks. I guess I was almost sleeping last night. One of the problems was a division by $\left\vert a - b\right\vert/2$ which was causing a propagating trouble when $a = b$. I guess it's fine right now. Commented Mar 26, 2017 at 0:52
• @Fabian Thanks. So many details to check !!!... Commented Mar 26, 2017 at 20:45