$\newcommand{\+}{^{\dagger}}
\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
\newcommand{\dd}{{\rm d}}
\newcommand{\down}{\downarrow}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
\newcommand{\fermi}{\,{\rm f}}
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{{\rm i}}
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\isdiv}{\,\left.\right\vert\,}
\newcommand{\ket}[1]{\left\vert #1\right\rangle}
\newcommand{\ol}[1]{\overline{#1}}
\newcommand{\pars}[1]{\left(\, #1 \,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
\newcommand{\sech}{\,{\rm sech}}
\newcommand{\sgn}{\,{\rm sgn}}
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
\newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{{\rm g}\pars{t}\equiv t^{a}\quad\imp\quad
\hat{\rm g}\pars{s} = \int_{0}^{\infty}t^{a}\expo{-st}\,\dd t
={\Gamma\pars{a + 1} \over s^{a + 1}}}$
where $\ds{\Gamma\pars{z}}$ is the
Gamma Function ${\bf\mbox{6.1.1}}$. Also,
$\ds{{\rm g}\pars{t}
=\int_{\gamma - \infty\ic}^{\gamma + \infty\ic}
{\Gamma\pars{a + 1} \over s^{a + 1}}\,\expo{st}\,{\dd s \over 2\pi\ic}\,,
\qquad\gamma > 0}$.
\begin{align}
{\rm g}\pars{t}&
=\Gamma\pars{a + 1}\int_{\gamma - \infty\ic}^{\gamma + \infty\ic}
{\expo{st} \over s^{a + 1}}\,{\dd s \over 2\pi\ic}=\Gamma\pars{a + 1}\times
\\[3mm]&\bracks{%
-\int_{-\infty}^{0}\pars{-s}^{-a - 1}\expo{-\pars{a + 1}\pi\ic}\expo{st}
{\dd s \over 2\pi\ic}
-\int_{0}^{-\infty}\pars{-s}^{-a - 1}\expo{\pars{a + 1}\pi\ic}\expo{st}
{\dd s \over 2\pi\ic}}
\\[3mm]&=\Gamma\pars{a + 1}\bracks{%
\expo{-\pi a\ic}\int_{0}^{\infty}s^{-a - 1}\expo{-st}
{\dd s \over 2\pi\ic}
-\expo{\pi a\ic}\int_{0}^{\infty}s^{-a - 1}\expo{-st}{\dd s \over 2\pi\ic}}
\\[3mm]&=-\,{1 \over \pi}\,\Gamma\pars{a + 1}\,
{\expo{\pi a\ic} - \expo{-\pi a\ic} \over 2\ic}
\int_{0}^{\infty}s^{-a - 1}\expo{-st}\dd s
\\[3mm]&=-\,{\Gamma\pars{a + 1} \over \pi}\,\sin\pars{\pi a}t^{a}\
\underbrace{\int_{0}^{\infty}s^{-a - 1}\expo{-s}\dd s}_{\ds{=\ \Gamma\pars{-a}}}
\end{align}
$$
{\rm g}\pars{t}
={\Gamma\pars{1 + a}\Gamma\pars{-a}\sin\pars{-\pi a} \over \pi}\,t^{a}
$$
With Euler Reflection Formula
${\bf\mbox{6.1.17}}$,
$\ds{{\Gamma\pars{1 + a}\Gamma\pars{-a}\sin\pars{-\pi a} \over \pi} = 1}$
such that
$$
\color{#44f}{\large{\rm g}\pars{t} = t^{a}}
$$