Laplace transform is considered as the continuous analogue of the power series, $$A(x)=\sum_{n=0}^\infty a(n)x^n \rightarrow A(x)= \int_0^\infty a(t)x^t\mathbf {dt} $$ sub $\,\, x^t=e^{(\ln\,x)^t} $ As the integral converges when $0<x<1,\,\, ln \, x <0$ and hence we consider $\mathbf -s=\ln \, x$ $$\Rightarrow A(x)=\sum_{n=0}^\infty a(n)x^n \rightarrow A(s)= \int_0^\infty a(t)e^{-st}\mathbf {dt}$$
My question is; How did $\,\,\mathbf {dt}\,\,$ kick in? When we consider a sum, $F(x)=\sum_{n=0}^t f(x)\Delta x $ , limit of this sum as $\Delta x \rightarrow 0 \,\,and \,\,t\rightarrow \infty$ is equal to $F(x)=\int_{n=0}^\infty f(x) \mathbf {dx}\, $. Here, the $\mathbf {\Delta x}$ changed to $\mathbf {dx}$, during the limiting process. But then while considering the continuous analogue of power series of did $\mathbf {dt}$ come in. This question would be naive, but then I'm new to calculus.
Well I know this reply is pretty late, but
$$A(x)=\sum_{n=0}^\infty a(n)x^n$$ could be written as $$A(x)=\sum_{n=0}^\infty a(n)x^n \Delta n.$$
Note that $\Delta n$ in this case equals $1$ (the discrete case).