themobius6s

2022-09-08

If Fi(t) and fi(s) are Laplace pairs, show that:
$\mathcal{L}\left\{\sum _{i=1}^{n}{F}_{i}\left(t\right)\right\}=\sum _{i=1}^{n}\mathcal{L}\left\{{F}_{i}\right\}=\sum _{i=1}^{n}{f}_{i}\left(s\right)$

Ufumanaxi

It comes from the linearity of the integral
$\begin{array}{rl}\mathcal{L}\left\{\sum _{i=1}^{n}{F}_{i}\left(t\right)\right\}& ={\int }_{0}^{\mathrm{\infty }}\left(\sum _{i=1}^{n}{F}_{i}\left(t\right)\right){\mathrm{e}}^{st}\mathrm{d}t\\ & ={\int }_{0}^{\mathrm{\infty }}\left({F}_{1}\left(t\right)+\cdots +{F}_{n}\left(t\right)\right){\mathrm{e}}^{st}\mathrm{d}t\\ & ={\int }_{0}^{\mathrm{\infty }}{F}_{1}\left(t\right){\mathrm{e}}^{st}\mathrm{d}t+\cdots +{\int }_{0}^{\mathrm{\infty }}{F}_{n}\left(t\right){\mathrm{e}}^{st}\mathrm{d}t\\ & =\mathcal{L}\left\{{F}_{1}\left(t\right)\right\}+\cdots +\mathcal{L}\left\{{F}_{n}\left(t\right)\right\}\\ & =\sum _{i=1}^{n}\mathcal{L}\left\{{F}_{i}\left(t\right)\right\}\\ & =\sum _{i=1}^{n}{f}_{i}\left(s\right)\end{array}$

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