Discussion
If \(L^{-1}\{F(s)\} = f(t)\), then
\(L^{-1}\{F^{(n)}(s)\} =\) is :
where \(L^{-1}\) be the inverse Laplace transform
Answer: B
\(L^{-1}\{F^{(n)}(s)\} =\) \((-1)^{n}t^{n}f(t)\).
Differentiate \(F(s) = \int_{0}^{\infty} e^{-st}f(t)\,dt\) under the integral sign: each derivative in \(s\) brings down a factor \(-t\), so
\[ F^{(n)}(s) = \int_{0}^{\infty} e^{-st}(-t)^{n}f(t)\,dt \]
That is, \(L\{(-1)^{n}t^{n}f(t)\} = F^{(n)}(s)\), and taking inverses gives the answer. For example \(L\{1\} = 1/s\) and \(\tfrac{d}{ds}(1/s) = -1/s^{2} = L\{-t\}\).
Option A misses the sign, option C misses the factor \(t^{n}\), and option D confuses differentiation in \(s\) with differentiation in \(t\).
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