Show Laplace[f(at)] = (1/a) F(s/a)]

[Saladsamurai]

Homework Statement

Show that if

$$\text{L}\[f(t)] = F(s) \text{ then } \text{L}\[f(at)] = \frac{1}{a}F(\frac{s}{a})$$

Homework Equations

Definition of Laplace

The Attempt at a Solution

By definition,

$$L[f(at)] = \int_0^\infty f(at)e^{-st}dt$$

I was given a hint to let u = at --> dt = du/a so we have

$$L[f(at)] = \frac{1}{a}\int_0^\infty f(u)e^{-\frac{s}{a}u}\,du$$

Now it looks like I am about done, but I am not sure how to proceed? I believe I now need to show that the if by definition

$$L[f(t)] = F(s) = \int_0^\infty f(t)e^{-st}dt$$

then the integral

$$\int_0^\infty f(t)e^{-\frac{s}{a}t}\,dt = F(\frac{s}{a})$$

Seems simple enough, but I am not sure how to show it.

 

[hunt_mat]

Differentiate w.r.t a and obtain a differentiatal equation in a and solve it.

Mat

 

[fzero]

Define $$\tilde{s} = s/a$$, then by definition the integral is $$F(\tilde{s})$$.

 

[Saladsamurai]

Define $$\tilde{s} = s/a$$, then by definition the integral is $$F(\tilde{s})$$.

Man, I knew I was almost there. Thanks! That works perfectly.

Edit: For completeness of the thread: Letting $\tilde{s} = s/a,$

$$L[f(at)] = \frac{1}{a}\int_0^\infty f(u)e^{-\frac{s}{a}u}\,du = \frac{1}{a}\int_0^\infty f(u)e^{-\tilde{s}u}\,du$$

which is an integral transform that sends f from the u domain to the $tilde{s}$ domain:

$$\int_0^\infty f(u)e^{-\tilde{s}u}\,du = F(\tilde{s}) = F(\frac{s}{a})$$

so,

$$L[f(at)] = \frac{1}{a}\int_0^\infty f(u)e^{-\frac{s}{a}u}\,du = \frac{1}{a}\int_0^\infty f(u)e^{-\tilde{s}u}\,du = \frac{1}{a}F(\frac{s}{a})$$