Laplace Transform of Derivative

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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 $0 \le t \le A$.

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

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

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


Then $\laptrans f$ exists for $\map \Re s > a$, and:

$\laptrans {\map {f'} t} = s \laptrans {\map f t} - \map f 0$


Discontinuity at $t = 0$

Let $f$ fail to be continuous at $t = 0$, but let:

$\ds \lim_{t \mathop \to 0} \map f t = \map f {0^+}$

exist.

Then $\laptrans f$ exists for $\map \Re s > a$, and:

$\laptrans {\map {f'} t} = s \laptrans {\map f t} - \map f {0^+}$


Discontinuity at $t = a$

Let $f$ have a jump discontinuity at $t = a$.

Then:

$\laptrans {\map {f'} t} = s \laptrans {\map f t} - \map f 0 - e^{-a s} \paren {\map f {a^+} - \map f {a^-} }$


Finite Discontinuities at $t = a_i$ for $i = 1, 2, \ldots, n$

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

\(\ds \laptrans {\map {f'} t}\) \(=\) \(\ds \int_0^{\mathop \to +\infty} e^{-s t} \map {f'} t \rd t\) 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


Consider:

$\ds \int_0^A e^{-s t} \map {f'} t \rd t$

By hypothesis, $f'$ is piecewise continuous with one-sided limits.

So by Piecewise Continuous Function with One-Sided Limits is Darboux Integrable, this integral exists.

This means that integration by parts can be invoked:

$\ds \int h j\,' \rd t = h j - \int h' j \rd t$

Here:

\(\ds h\) \(=\) \(\ds e^{-s t}\)
\(\ds \leadsto \ \ \) \(\ds h'\) \(=\) \(\ds -s e^{-s t}\)
\(\ds j\,'\) \(=\) \(\ds \map {f'} t\)
\(\ds \leadsto \ \ \) \(\ds j\) \(=\) \(\ds \map f t\)


So:

\(\ds \int_0^A e^{-s t} \map {f'} t \rd t\) \(=\) \(\ds \bigintlimits {e^{-s t} \map f t} {t \mathop = 0} {t \mathop = A} + s \int_0^A e^{-s t} \map f t \rd t\)

Now, take the limit as $t = A \to +\infty$:

\(\ds \laptrans {\map {f'} t}\) \(=\) \(\ds \lim_{A \mathop \to +\infty} e^{-s A} \map f A - \map f 0 + s \laptrans {\map f t}\)

Recall that $f$ is of exponential order $a$:

\(\ds \size {\map f A}\) \(<\) \(\ds K e^{a t}\)
\(\ds \leadsto \ \ \) \(\ds \size {\map f A} \size {e^{-s A} }\) \(<\) \(\ds K e^{a t} \size {e^{-s A} }\)
\(\ds \leadsto \ \ \) \(\ds \size {e^{-s A} \map f A}\) \(<\) \(\ds \size {K e^{a t} e^{-s A} }\) Modulus of Product, Exponential Tends to Zero and Infinity
\(\ds \) \(=\) \(\ds \size {K e^{\paren {a - s} t} }\) Exponent Combination Laws, definition of $A$
\(\ds \) \(=\) \(\ds \size {K e^{\paren {a - \map \Re s - i \map \Im s} t} }\)
\(\ds \) \(=\) \(\ds \size {K e^{\paren {a - \map \Re s} t} }\) Modulus of Exponential is Modulus of Real Part
\(\ds \) \(=\) \(\ds K e^{\paren {a - \map \Re s} t}\) Exponential Tends to Zero and Infinity

This implies, from Complex Exponential Tends to Zero and the Squeeze Theorem:

$\ds \lim_{A \mathop \to +\infty} e^{-s A} \map f A = 0$

which produces:

$\laptrans {\map {f'} t} = s \laptrans {\map f t} - \map f 0$

$\blacksquare$


Examples

Example $1$

$\laptrans {-3 \sin 3 t} = \dfrac {-9} {s^2 + 9}$


Also see


Sources