How is the following derived?

Question

To give some context, I am viewing the following lecture which derives the Laplace Transform.

Lec 19 | MIT 18.03 Differential Equations, Spring 2006

At the 6:40 mark, the lecturer seems to skip over how the continuous analogue of the following discrete power series is derived.

$$\sum_{0}^{\infty} a(n)x^n$$

Afterwards, the lecturer presents the following integral.

$$\int_{0}^{\infty} a(t)x^tdt$$

I want to understand how the continuous analogue was derived as I am having a hard time of it myself.

(Disclaimer: I am an engineering student, not a mathematics student.)

Answer 1

Essentially, the idea is that $\int$ is the continuous analog of $\sum$ (recall that the integral is the limit of a Riemann sum as the difference $\Delta x$ tends to zero). So, the continuous analog of $\sum_0^\infty a(n)x^n$ will naturally be $\int_0^\infty a(t)x^tdt$, as long as we appropriately extend the domain of $a$ to $[0,\infty)$ (not just the nonnegative integers). The idea that the lecturer is trying to get across here is that the Laplace transform is in a sense just the continuous analog of a power series.

Note that you still have issues of convergence, etc. to deal with, so this isn't strictly speaking rigorous.

Answer 2

The answer is that you should not think in term of discrete analog. If you really want to then show if the series converges for $|x|< r\le1$ $$\sum_{n=0}^\infty a(n)x^n = \sum_{n=0}^\infty a(n)(-\log(x))\int_n^\infty x^t dt=-\log(x) \int_0^\infty (\sum_{n \le t} a(n)) x^tdt$$