Laplace transform with scaled time variable

Question

We know that by definition, $L{f(t)}=\int_{0}^{\infty}e^{-st}f(t)dt$. But how is $L{f(at)}=\int_{0}^{\infty}e^{-st}f(at)dt$? Shouldn't $e^{-st}$ be replaced by $e^{-s(at)}$ in the 2nd case since $at$ is the variable now? I don't understand how the definition fits here. This definition is used everywhere in proving $L{f(at)}=\frac{1}{a}F(\frac{s}{a})$

Answer 1

The Laplace transform is an operator acting on functions. Your confusion arises from the ambiguity (encouraged by poor notation) about which are the independent variables of that function and which are variables of integration in the transform expression.

It may help to introduce a dummy variable $u$ to clearly differentiate the parameter of $f$ from the variable of integration: $$[Lf(t)](s) = \int_0^{\infty} e^{-su} f(u)\,du.$$ Here we apply $L$ to the function $f(t)$, yielding a new function (now depending on $s$).

Your second example would be $$[Lf(at)](s) = \int_0^{\infty} e^{-su} f(au)\,du.$$ Here we are applying the Laplace transform to the new function $f(at)$. It might be even clearer to write it as $$[Lg(t)](s) = \int_0^{\infty} e^{-su} g(u)\,du, \qquad g(t) = f(at).$$

In your proposed expression involving $e^{-ast}$, you are trying to first write down the Laplace transform of $f(t)$, and then somehow "substitute in" $at$ for $t$. But nothing about the definition of the transform allows you to do that.