# Laplace Transform: Continuous analogue of Power series

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 $$0 and hence we consider $$\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, you are looking to include values of $a$ over the continuum so you can look at $\sum_n a(hn) e^{-shn}$. This sum scales like $1/h$ for small $h$ so you multiply by $h$ to scale it down to obtain convergence. So the series you get is in powers of $e^{-sh}$ and the coefficients are $ha(hn)$. This $h$ corresponds to $dt$ in the limit. – Ian Apr 16 at 3:20
• In the end this analogy is not perfect, though. – Ian Apr 16 at 3:22
• @Ian Woah, that explanation is close I suppose, can you explain it a bit more in rookie terms :) – Aravindh Vasu Apr 16 at 3:23
• @Ian Can you please tell me, in what way? – Aravindh Vasu Apr 16 at 3:24
• Mainly that in power series the coefficients are proportional to derivatives; when you replace them by a continuously varying quantity, that interpretation is essentially lost. – Ian Apr 16 at 3:25