Change of argument in Laplace Transform

Question

The Laplace transform of a function $f=f(x)$ has the following definition: $$\mathcal{L}({f(x)})=\int^{\infty}_0e^{-sx}f(x)dx \tag{1}$$

However, when $f=f(ax-c)$, where $a$ and $c$ are arbitrary constants, we have

$$\mathcal{L}(f(ax-c))=\int^{\infty}_0e^{-sx}f(ax-c)dx \tag{2}$$

with which I got very confused, why should $(2)$ not be $$\mathcal{L}(f(ax-c))=\int^{\infty}_0e^{-s(ax-c)}f(ax-c)d(ax-c) \tag{3}$$ or equivalently $$\mathcal{L}(f(ax-c))=a\int^{\infty}_0e^{-s(ax-c)}f(ax-c)dx \tag{3}$$

instead?

Answer 1

The Laplace transform of a function $f(x)$ is itself not a function of $x$.

You can write $$\mathcal{L}(f(x))(s) = F(s) = \int_{0}^\infty e^{-s\tau} f(\tau) d\tau$$

where $\tau$ is a dummy index of integration.

So when you write $g(x) = f(ax - c)$, the Laplace transform $G(s)$ of $g(x)$ is \begin{align*} G(s) &= \int_{0}^\infty e^{-s \lambda} g(\lambda) d\lambda\\ &= \int_{0}^\infty e^{-s \lambda} f(a \lambda - c) d\lambda \end{align*}