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

Question

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?

score 5 · Accepted Answer · answered May 22 '19 at 23:38

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.

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

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