Inverse Laplace Transform of F(ks)
Formulas1
Assuming that the Laplace transform $\mathcal{L} \left\{ f(t) \right\} = \displaystyle \int _{0} ^\infty e^{-st}f(t)dt = F(s)$ of the function $f(t)$ exists and is $s>a \ge 0$ for a positive number $k> 0$ then the inverse Laplace transform of $F(ks)$ is as follows.
$$ \mathcal{L^{-1}} \left\{ F(ks) \right\} =\dfrac{1}{k}f\left(\frac{t}{k}\right),\quad s>\frac{a}{k} $$
Derivation
1
The Laplace transform of $f(ct)$
$$ \mathcal{L} \left\{ f(ct) \right\} =\dfrac{1}{c}F\left(\dfrac{s}{c}\right), \quad s>ca $$
By substituting $\dfrac{1}{k}$ for $c$ in the above equation
$$ \begin{align*} \mathcal{L} \left\{ f\left(\frac{t}{k}\right) \right\} &= kF(ks) \\ \implies F(ks) &= \dfrac{1}{k} \mathcal{L} \left\{ f\left(\frac{t}{k}\right) \right\} =\mathcal{L} \left\{\dfrac{1}{k} f\left(\frac{t}{k}\right) \right\} \end{align*} $$
Therefore
$$ \mathcal{L^{-1}} \left\{ F(ks) \right\} =\dfrac{1}{k}f\left(\frac{t}{k}\right) $$
Since the condition when $c$ was $s>ca$, the condition naturally changes to $s>\dfrac{a}{k}$.
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2
By the definition of the Laplace transform,
$$ \mathcal{L} \left\{ \frac{1}{k} f\left( \frac{t}{k} \right) \right\} =\dfrac{1}{k}\int _{0} ^\infty e^{-st}f \left( \frac{t}{k} \right) dt $$
Let us replace $\dfrac{t}{k}=\tau$ with. Then $st=sk\tau$ and $dt=kd\tau$ are, so
$$ \begin{align*} \mathcal{L} \left\{ \frac{1}{k} f\left( \frac{t}{k} \right) \right\} &=\int _{0} ^\infty e^{-sk\tau}f \left(\tau \right) d\tau \\ &= F(ks) \end{align*} $$
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See Also
William E. Boyce, Boyce’s Elementary Differential Equations and Boundary Value Problems (11th Edition, 2017), p263 ↩︎