Laplace Transform of Derivative

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:

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

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:

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

Proof

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

Consider:

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

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

This means that integration by parts can be invoked:

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

Here:

 $\displaystyle h$ $=$ $\displaystyle e^{-s t}$ $\displaystyle \leadsto \ \$ $\displaystyle h'$ $=$ $\displaystyle -s e^{-s t}$ $\displaystyle j\,'$ $=$ $\displaystyle \map {f'} t$ $\displaystyle \leadsto \ \$ $\displaystyle j$ $=$ $\displaystyle \map f t$

So:

 $\displaystyle \int_0^A e^{-s t} \map {f'} t \rd t$ $=$ $\displaystyle \bigintlimits {e^{-s t} \map f t} {t \mathop = 0} {t \mathop = A} + s \int_0^A e^{-s t} \map f t \rd t$

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

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

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

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

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

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

which produces:

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

$\blacksquare$

Examples

Example $1$

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