Laplace transform of $x^a$ 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!!!
 A: You should use the definition of Laplace transform and of the Gamma function:
$$\mathcal{L}\{f(x)\}(s)=\int_0^\infty e^{-sx}f(x)dx,\hspace{10pt} \Gamma(\alpha)=\int_0^\infty e^{-t}t^{\alpha-1}dt$$
Just substitute $f(x)=x^a$ and change the variable $sx\mapsto t$.
A: By definition, the Laplace transform $\mathcal{L}(x^a)$ of the function $x\mapsto x^a$ is given by
$$
\mathcal{L}(x^a)(s) = \int_0^\infty \exp(-sx)x^a\mathrm{d}x.
$$
The Gamma function is defind by a similar integral, namely
$$
\Gamma(s)=\int_0^\infty \exp(-x)x^{s-1}\mathrm{d}x.
$$
The Laplace transform of $x^a$ can thus be computed by the variable transformation $x\mapsto x/s$.
A: \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*}
A: Use the definition of laplace transform 
ie. in your case ,
$$F(s) = \int_0^\infty x^a e^{-sx} dx$$ 
