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How to find out the laplace transform of $$f(t) \cdot \sum_{i=1}^\infty a_i g_i(t),$$ w.r.t the variable $t$ on the domain $[0,\infty)$, where $a_i$'s are constants with value $0\le a_i\le1$.

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  • $\begingroup$ Do you mean $f$ multiplied by the sum, or is there an equality sign missing? Also, is there any information on $f$ and $g_i$? $\endgroup$ – 75064 May 17 '13 at 20:04
  • $\begingroup$ yes f is multiplied by the sum and $f,g_i$ are cts func on their domain. $\endgroup$ – Litun John May 18 '13 at 12:21
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Then, if they are just continuous functions, I only see 2 ways to get the Laplace transform of $h(t) = f(t) \sum_{i=1}^{\infty} a_i g_i(t) $:

  1. Considering $F(s) = \mathcal{L}\left\{ f(t) \right\}$, $G_i(s) =\mathcal{L}\left\{ g_i(t) \right\}$ for each $i \in \mathbb{N}$. Since the Laplace transform is a lineal operator, for every $n \in \mathbb{N}$ we have that $\mathcal{L}\left\{ \sum_{i=1}^n a_i g_i(t)\right\} = \sum_{i=1}^n a_i G_i(t)$. I'm not sure if there is any problem with linear combinations of infinitely many functions, so you may prove that $\mathcal{L}\left\{ \sum_{i=1}^{\infty} a_i g_i(t)\right\} = \sum_{i=1}^{\infty} a_i G_i(t)$ (as a limit, at least). And then, use the rule I gave you in the other answer.

  2. If you don't need the Laplace transform of $h(t)$ to depend on the Laplace transforms of $f$ and $g_i$ and you have more information about them, try to do it as a unique function. But for this you need to know explicitely (or at least, "better"), $f(t)$ and the $g_i(t)$ (or directly $h(t)$).

I'm working in a similar problem and I used the power series developement of one of the functions. Since $\mathcal{L}\left\{ t^n f(t) \right\} = (-1)^n \frac{d^n}{ds^n}\mathcal{L}\left\{ f(t) \right\}$, I obtained the Laplace transform of the product as a series on the derivatives of the Laplace transform of one of the functions in the product. But I don't think this is a good way.

I hope it has been helpfull!

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In general, if $F(s) = \mathcal{L} \left\{ f(t) \right\}$ and $G(s) = \mathcal{L} \left\{ g(t) \right\}$, the Laplace transform of the product can be found in that way:

$$\mathcal{L} \left\{ f(t)g(t) \right\} = \frac{1}{2\pi i}\lim_{T \to \infty} \int_{c -iT}^{c+iT} F(\sigma)G(s-\sigma)d\sigma = \frac{1}{2\pi i} \lim_{T \to \infty} \int_{c -iT}^{c+iT} F(s - \sigma)G(\sigma)d\sigma ,$$

where $c$ is any constant such that $ c + ia $ is in the region of convergence of F for every $a \in \mathbb{R}$.

But I think I'm not trully understanding your question. Have you said that $f,g_i$ are constant functions on their domain? Then you can express them as linear combinations of heavyside step functions and their Laplace transform will be easier to find.

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  • $\begingroup$ $f(t)$ and $g_i(t)$ are continuous functions of $t$ in the domain $[0,\infty)$. $\endgroup$ – Litun John Nov 25 '13 at 11:47
  • $\begingroup$ Can you point to a worked out example when $f$ and $g$ are not constant function? $\endgroup$ – Dilawar Apr 22 '15 at 9:38

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