Laplace Transform of Derivative with Finite Discontinuities

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Theorem

Let $f: \R \to \R$ or $\R \to \C$ be a continuous function, differentiable on any interval of the form $0 < t \le A$.

Let $f$ be of exponential order $a$.

Let $f'$ be piecewise continuous with one-sided limits on said intervals.

Let $\laptrans f$ denote the Laplace transform of $f$.

Let $f$ have a finite number of jump discontinuities at $t = a_i$ for $i = 1, 2, \ldots, n$.

Then:

$\laptrans {\map {f'} t} = s \laptrans {\map f t} - \map f 0 - \ds \sum_{i \mathop = 1}^n e^{-a_i s} \paren {\map f {a_i^+} - \map f {a_i^-} }$


Proof


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The proof is found similarly to the proof in Laplace Transform of Derivative but requires breaking the integral into $n$ integrals of finite range between the discontinuities and one improper integral from $a_n^+$ to $+\infty$. Integration by parts yields the summation when evaluating $e^{-s t}x(t)$ and the remaining integrals can all be recombined into one improper integral that is identically equal to $s \laptrans {\map f t}$.

Specifically, as in the proof in Laplace Transform of Derivative,

\(\ds \laptrans {\map {f'} t}\) \(=\) \(\ds \int_0^{\mathop \to +\infty} e^{-s t} \map {f'} t \rd t\) Definition of Laplace Transform
\(\ds \) \(=\) \(\ds \lim_{A \mathop \to +\infty} \int_0^A e^{-s t} \map {f'} t \rd t\) Definition of Improper Integral on Closed Interval Unbounded Above

Breaking the integral into $n+1$ integrals over continuous intervals gives

$\ds \lim_{A \mathop \to +\infty} \int_0^A e^{-s t} \map {f'} t \rd t = \int_0^{a_1^-} e^{-s t} \map {f'} t \rd t + \int_{a_1^+}^{a_2^-} e^{-s t} \map {f'} t \rd t + \cdots + \int_{a_{n-1}^+}^{a_n^-} e^{-s t} \map {f'} t \rd t + \lim_{A \mathop \to +\infty} \int_{a_n^+}^A e^{-s t} \map {f'} t \rd t$

Integration by parts on all integrals yields

$\ds \lim_{A \mathop \to +\infty} \int_0^A e^{-s t} \map {f'} t \rd t = \bigintlimits {e^{-s t} \map f t} {t \mathop = 0} {t \mathop = a_1^-} + s \int_0^{a_1^-} e^{-s t} \map {f} t \rd t + \bigintlimits {e^{-s t} \map f t} {t \mathop = a_1^+} {t \mathop = a_2^-} + s \int_{a_1^+}^{a_2^-} e^{-s t} \map {f} t \rd t + \cdots + \bigintlimits {e^{-s t} \map f t} {t \mathop = a_{n-1}^+} {t \mathop = a_n^-} + s \int_{a_{n-1}^+}^{a_n^-} e^{-s t} \map {f} t \rd t + \lim_{A \mathop \to +\infty} \bigintlimits {e^{-s t} \map f t} {t \mathop = a_n^+} {t \mathop = A} + \lim_{A \mathop \to +\infty} s \int_{a_n^+}^A e^{-s t} \map {f} t \rd t$

These terms combine to form

$\ds \lim_{A \mathop \to +\infty} \int_0^A e^{-s t} \map {f'} t \rd t = \sum_{i \mathop = 1}^n e^{-a_i s} \paren {\map f {a_i^+} - \map f {a_i^-} } + \lim_{A \mathop \to +\infty} \bigintlimits {e^{-s t} \map f t} {t \mathop = 0} {t \mathop = A} + \lim_{A \mathop \to +\infty} s \int_0^A e^{-s t} \map {f} t \rd t$

where the proof in Laplace Transform of Derivative simplifies the terms other than the summation and gives the final result

$\ds \lim_{A \mathop \to +\infty} \int_0^A e^{-s t} \map {f'} t \rd t = s \laptrans {\map f t} - \map f 0 - \ds \sum_{i \mathop = 1}^n e^{-a_i s} \paren {\map f {a_i^+} - \map f {a_i^-} }$


Sources


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