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

$\ds{\int_{c\ -\ \infty\ic}^{c\ +\ \infty\ic}{\exp\pars{st} \over
\root{\pars{s - a}\pars{s - b}}}\,{\dd s \over 2\pi\ic}:\ {\large ?}
.\qquad}$ 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}
