# If $f(x) = \sum_{n=0}^{\infty} a_n x^n$ converges for all $x\geq0$, show that $\mathcal{L}\{f\}(s) = \sum_{n=0}^{\infty} \frac{a_n n!}{s^{n+1}}$

If $$f(x) = \sum_{n=0}^{\infty} a_n x^n$$ converges for all $$x\geq0$$,with $$|a_n| \leq \frac{K a^n}{n!}$$ for all $$n \in \mathbf{N}$$ and some constant $$K > 0$$.

I need to show that $$\mathcal{L}\{f\}(s) = \sum_{n=0}^{\infty} \frac{a_n n!}{s^{n+1}}$$

Could someone give me a hint?

## 1 Answer

Assuming that $$s>a$$ we have $$\int_0^{\infty} |\sum_n a_nx^{n}e^{-sx}|dx \leq \sum_n Ka^{n} x^{n}/(n!)e^{-sx}=K\int_0^{\infty} e^{-s(x-a)}dx <\infty$$. This allows us to interchange the infinite sum and the integral. Hence the Laplace transform is equal to $$\sum a_nx^{n}\int_0^{\infty} e^{-sx} x^{n}dx$$. I leave it to you to compute the last integral using integration by parts.