Inverse Laplace Transform of ${1/s^{a+1}}\int_0^\infty e^{-t}x^a$ by contour integration In question 1) we get Laplace transform of $$ g(t) = t^a $$ is:
$$\hat g(t)= {1/s^{a+1}}\int_0^\infty e^{-t}x^a$$
then I was stuck at question 2) which asks me to evaluate the inverse laplace transform of $ \hat g(p) $ which is 
$$
{1/2\pi i}\int_0^\infty e^{pt}\hat g(p)dp
$$
I know the answer should be $ t^a $ as the inverse transform comes back to itself, but I cannot figure out how to make the contour integration. I tried to apply Cauchy's residue theorem to eliminate the $ 1/2 \pi i $ but was stuck then. Thanks a lot for help!
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$\ds{{\rm g}\pars{t}\equiv t^{a}\quad\imp\quad
     \hat{\rm g}\pars{s} = \int_{0}^{\infty}t^{a}\expo{-st}\,\dd t
     ={\Gamma\pars{a + 1} \over s^{a + 1}}}$
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

$$
\color{#44f}{\large{\rm g}\pars{t} = t^{a}}
$$
