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.

  • $\begingroup$ 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. $\endgroup$ – Ian Apr 16 at 3:20
  • $\begingroup$ In the end this analogy is not perfect, though. $\endgroup$ – Ian Apr 16 at 3:22
  • $\begingroup$ @Ian Woah, that explanation is close I suppose, can you explain it a bit more in rookie terms :) $\endgroup$ – Aravindh Vasu Apr 16 at 3:23
  • $\begingroup$ @Ian Can you please tell me, in what way? $\endgroup$ – Aravindh Vasu Apr 16 at 3:24
  • $\begingroup$ 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. $\endgroup$ – Ian Apr 16 at 3:25

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