Inverse Laplace transform of | is
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Now the inverse Laplace transform of 2 (s−1) is 2e1 t. Less straightforwardly, the inverse Laplace transform of 1 s2 is t and hence, by the first shift theorem, that of 1 (s−1)2 is te1 t.
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F(s)=L(f)=∫∞0e−stf(t)dt. f=L−1(F
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