I have the following Laplace transform problem:
If $G(at + b) = F(t)$, determine the Laplace transform of $G$ in terms of $\mathcal{L}\{F\} = \overline{f}(s)$ and a finite integral.
The author's solution is as follows:
This problem illustrates the difficulty in deriving a linear translation plus scaling property for Laplace transforms. The zero in the bottom limit is the culprit. Direct integration yields:
$$\mathcal{L}\{ G(t) \} = \int_0^\infty e^{-st} G(t) \ dt = \int_{-b/a}^\infty ae^{(u - b)s/a} F(u) \ du,$$
where we have made the substitution $t = au + b$, so that $G(t) = F(u)$. In terms of $\overline{f}(as)$, this is
$$ae^{-sb} \overline{f}(as) + ae^{-sb} \int_{-b/a}^0 e^{-ast} F(t) \ dt.$$
This solution seems like nonsense to me. My work proceeds as follows:
$$\mathcal{L}\{ G(t) \} = \int_0^\infty e^{-st} G(t) \ dt = \int_{-b/a}^\infty e^{-s(au + b) F(u)},$$
where $t = au + b \Rightarrow u = \dfrac{t - b}{a}$.
How can the author's $\int_{-b/a}^\infty ae^{(u - b)s/a} F(u) \ du$ be correct? And what is $\overline{f}$ supposed to be?
I would greatly appreciate it if people would please take the time to review this.
With $a \ne 0$
$$ G(t)=F\left(\frac{t-b}{a}\right)\Rightarrow \mathcal{L}\left[G(t)\right]=\mathcal{L}\left[F\left(\frac{t-b}{a}\right)\right] $$
but
$$ \mathcal{L}\left[F\left(\frac{t-b}{a}\right)\right]=\int_0^{\infty}e^{-st}F\left(\frac{t-b}{a}\right)dt $$
now making $\tau = \frac{t-b}{a}$ we have
$$ \int_0^{\infty}e^{-st}F\left(\frac{t-b}{a}\right)dt = a e^{-sb}\int_{-\frac ba}^{\infty}e^{-sa\tau}F(\tau)d\tau = a e^{-s b}\left(\int_0^{\infty}e^{-sa\tau}F(\tau)d\tau+\int_{-\frac ba}^0e^{-sa\tau}F(\tau)d\tau\right)=a e^{-sb}\left(\bar f(as)+\int_{-\frac ba}^0e^{-sa\tau}F(\tau)d\tau\right) $$
hence
$$ \mathcal{L}\left[G(t)\right]=a e^{-sb}\left(\bar f(as)+\int_{-\frac ba}^0e^{-sa\tau}F(\tau)d\tau\right) $$