# Proof for this Laplace transform

## Main Question or Discussion Point

Can someone help me out with the proof of the Laplace transform of the function tn?

I did have a go at this one.
$$L[t^n] = \int_0^{\infty} t^n e^{-st}dt$$

$$= t^n\frac{-e^{-st}}{s}\vert_0^{\infty} + {1\over s}\int_0^{\infty} nt^{n-1}e^{-st}dt$$

$$={n\over s}\int_0^{\infty} t^{n-1}e^{-st}dt$$

$$={n\over s}L[t^{n-1}]$$

I am supposed to arrive at the result:
$$L[t^n] = \frac{(n+1)!}{s^{n+1}}$$

## Answers and Replies

Just an addendum, should I prove this using induction?

$$L(t^{n}) = \int_{0}^{\infty} t^{n}e^{-st}dt$$

Let $$v=st$$ to give $$L(t^{n}) = \frac{1}{s^{n+1}}\int_{0}^{\infty}v^{n}e^{-v}dv = \frac{1}{s^{n+1}}\Gamma(n+1)$$

$$\Gamma(n+1) = n!$$, giving the result $$L(t^{n}) = \frac{n!}{s^{n+1}}$$. This isn't what you've got, but check by using n=1 to see whose correct.

$$L(t) = \int_{0}^{\infty}te^{-st}dt =\left[ \frac{te^{-st}}{-s}\right]_{0}^{\infty} - \int_{0}^{\infty}\frac{1}{-s}e^{-st}dt = \frac{1}{s^{2}}$$

Hence $$L(t^{1}) = \frac{1!}{s^{2}}$$, so the formula you were trying to derive isn't correct.

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