# Laplace transform of $x^2$

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.

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Try an Integration by parts –  Siméon Jan 5 '13 at 12:02

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...

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Thank you SOOO much! –  KillaKem Jan 5 '13 at 12:16

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

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Can you please expand a little more? –  levap Jan 5 '13 at 13:08
@levap if we get $F(z) = x , G(z) = x^{n-1}$ then both formula are true together and no thing to contradiction. –  Hoseyn Heydari Jan 5 '13 at 13:41