TITLE: Laplace transform of $x^a$
QUESTION [5 upvotes]: How to prove that the Laplace transform of $x^a$ is: 
$$\mathcal{L}\{x^a\}(s)=\frac{\Gamma(a+1)}{ s^{a+1}}$$
Also how to prove that the inverse Laplace transform of $\frac{\Gamma(a+1)}{ s^{a+1}}$ is $x^a$? Thanks a lot!!!

REPLY [4 votes]: \begin{align*}
\mathcal L\left\{ x^{a}\right\}&=\int\limits_0^\infty x^{a}e^{-{s}x}\text dx
\\\text{let } x=\frac u{s}\qquad\quad\text dx=\frac{\text du}{s}&
\\x=0\rightarrow u=0\qquad x=\infty\rightarrow u=\infty&\\
\\&=\int\limits_0^\infty \left(\frac u{s}\right)^{a}e^{-{s}\frac u{s}}\frac{\text du}{s}
\\&=\frac{1}{{s}^{{a}+1}}\int\limits_0^\infty u^{a}e^{-u} {\text du}\\
\\&=\frac{\Gamma({a}+1)}{{s}^{{a}+1}}&\qquad a\in\mathbb C\big|\Re(a)\geq-1\\
\\&\color{grey}{=\frac{{a}!}{{s}^{{a}+1}},}&\qquad \color{grey}{{a}\in\mathbb Z\geq0}
\end{align*}

\begin{align*}
\mathcal L\left\{ x^{a}\right\}=\frac{\Gamma({a}+1)}{{s}^{{a}+1}}\\
% \mathcal L^{-1}\Big\{\mathcal L\left\{ x^{a}\right\}\Big\}=\mathcal L^{-1}\left\{\frac{\Gamma({a}+1)}{{s}^{{a}+1}}\right\}\\
x^{a}=\mathcal L^{-1}\left\{\frac{\Gamma({a}+1)}{{s}^{{a}+1}}\right\}\\
\end{align*}