Question

If $\mathcal{L}\mathbf{\left\{}\mathbf{f}\mathbf{\right(}\mathbf{t}\mathbf{\left)}\mathbf{\right\}}\mathbf{=}\mathbf{F}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$ and$\mathbf{a}\mathbf{>}\mathbf{0}$ is a constant, show that

$\mathcal{L}\mathbf{\left\{}\mathbf{f}\mathbf{\right(}\mathbf{at}\mathbf{\left)}\mathbf{\right\}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{a}}\mathbf{F}\left(\frac{s}{a}\right)$

This result is known as the change of scale theorem.

Short answer

$\mathcal{L}\left\{\mathrm{f}\right(\mathrm{at}\left)\right\}=\frac{1}{\mathrm{a}}\mathrm{F}\left(\frac{\mathrm{s}}{\mathrm{a}}\right)$

Step-by-step solution

Definition of Laplace Transform

Let f be a function defined for $\mathbf{t}\mathbf{⩾}\mathbf{0}$. Then the integral

$\mathcal{L}\mathbf{\left\{}\mathbf{f}\mathbf{\right(}\mathbf{t}\mathbf{\left)}\mathbf{\right\}}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{x}}{\mathbf{e}}^{\mathbf{-}\mathbf{xt}}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{dt}$

is said to be the Laplace transform of f, provided that the integral converges.

Use Laplace Transform

The formula for Laplace transform

$\mathrm{F}\left(\mathrm{s}\right)=\mathcal{L}\left\{\mathrm{f}\right(\mathrm{t}\left)\right\}={\int }_0^{\infty }{\mathrm{e}}^{-\mathrm{st}}\mathrm{f}\left(\mathrm{t}\right)\text{d}\mathrm{t}$

Let$\mathrm{u}=\mathrm{at}\iff \mathrm{t}=\frac{\mathrm{u}}{\mathrm{a}}$

$\text{d}\mathrm{u}=\mathrm{a}\text{d}\mathrm{t}\iff \text{d}\mathrm{t}=\frac{\text{d}\mathrm{u}}{\mathrm{a}}$

Then,

$$\begin{gathered} \mathcal{L}\left\{f\right(at\left)\right\}={\int }_0^{\infty }{e}^{-st}f\left(at\right)\text{d}t \\ =\frac{1}{a}{\int }_0^{\infty }{e}^{-\frac{s}{a}u}f\left(u\right)\text{d}u \\ \stackrel{\left(1\right)}{=}\frac{1}{a}F\left(\frac{s}{a}\right) \end{gathered}$$

Therefore,$\mathcal{L}\left\{\mathrm{f}\right(\mathrm{at}\left)\right\}=\frac{1}{\mathrm{a}}\mathrm{F}\left(\frac{\mathrm{s}}{\mathrm{a}}\right)$.