Laplace transform of $x^2$

Question

I can't seem to be able to understand why $$\mathcal{L}(x^n)(s) = \frac{n}{s} \mathcal{L}(x^{n-1})(s)$$ This one line has got me stuck! I know that $$\mathcal{L}(f(x)g(x)) \neq F(s)G(s)$$ so how could this line be valid.Please help.

Answer 1

By parts:

$$u=x^n\,\,\Longrightarrow u'=nx^{n-1}\\v'=e^{-sx}\,\,\Longrightarrow v=-\frac{e^{-sx}}{s}$$

$$\int\limits_0^\infty e^{-sx}x^n\,dx=\left.-\frac{x^ne^{-sx}}{s}\right|_0^\infty+\frac{n}{s}\int\limits_0^\infty x^{n-1}e^{-sx}\,dx$$

Now just check the first term in the RHS above is zero...

Answer 2

$\mathcal{L}(x)(s) \neq \frac{n}{s}$ then no contradiction.