Definition of Negative Half Derivative I know the definition of positive fractional derivative, which is given by $$D^{\alpha} f(x)=\frac{1}{\Gamma(1-\alpha)} \frac{d}{d x} \int_{0}^{x} \frac{f(t)}{(x-t)^{\alpha}} dt,\quad\quad \alpha\in(0,1)$$
But I encounter the negative half derivative $D^{-1/2}$ , which I am quite confused with.
My professor assigned a problem which let us solve the integral equation $$ D^{-1 / 2} h(t)=\int_{0}^{t} \frac{1}{\sqrt{\pi(t-s)}} h(s) \mathrm{d} s=g(t),$$ where $g$ is a nice function with $g(0)=0.$ And he gave a hint that lets us square $D^{-1/2}.$
I feel like the first equality is the definition of $D^{-1/2}$ but I am not sure. If it indeed is the definition of $D^{-1/2}$, can someone tell me the definition of $D^{-\beta}$ for any $\beta \gt0\,?$
Furthermore, can someone give me a further hint about the problem he assigned?
My attempt: Squaring $D^{-1/2}$, we get $$h^{-1}(t)=\int_0^t h(s)\mathrm{d}s=D^{-1/2}g(t)=\int_{0}^{t} \frac{1}{\sqrt{\pi(t-s)}} g(s) \mathrm{d} s\,.$$
Differentiating both sides w.r.t  $t$ and applying the Leibniz Integral Rule( but ${1\over \sqrt{t-t}}g(t) =\infty$, and it is invalided to use the Leibniz Integral Rule), we get $$h(t)=\frac{-1}{2\sqrt{\pi}}\int_0^t\frac{1}{(t-s)^{\frac{3}{2}}}g(s)\mathrm{d}s$$
I think it may not be simplified further, but I am not sure. Could the integral in the right-hand-side be calculated further?
Any help will be appreciated.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Lets use the notation $\ds{\hat{\mrm{f}}\pars{u} \equiv \int_{0}^{\infty}\mrm{f}\pars{t}\expo{-ut}\dd t}$:

\begin{align}
\int_{0}^{\infty}\expo{-ut}\int_{0}^{t}{\mrm{h}\pars{s} \over \root{\pi\pars{t - s}}}\dd s\,\dd t & = \hat{\mrm{g}}\pars{u}
\\[5mm]
{1 \over \root{\pi}}\int_{0}^{\infty}\mrm{h}\pars{s}
\int_{s}^{\infty}{\expo{-ut} \over \root{t - s}}\dd t\,\dd s & = \hat{\mrm{g}}\pars{u}
\\[5mm]
{1 \over \root{\pi}}\int_{0}^{\infty}\mrm{h}\pars{s}\expo{-us}\
\underbrace{\int_{0}^{\infty}t^{-1/2}\expo{-ut}\dd t}
_{\ds{=\ \root{\pi}u^{-1/2}}}\
\,\dd s & = \hat{\mrm{g}}\pars{u}
\\[5mm]
\int_{0}^{\infty}u^{-1/2}\,\mrm{h}\pars{s}\expo{-us}\
\,\dd s & = \hat{\mrm{g}}\pars{u}
\end{align}
Then,
\begin{align}
u^{-1/2}\,\mrm{h}\pars{s} & = \int_{c - \infty\ic}^{c + \infty\ic}
\hat{\mrm{g}}\pars{u}\expo{su}
{\dd u \over 2\pi\ic}
\end{align}
$$
\implies \bbx{\mrm{h}\pars{s} =
\root{u}\int_{c - \infty\ic}^{c + \infty\ic}
\hat{\mrm{g}}\pars{u}\expo{su}
{\dd u \over 2\pi\ic}}
$$
