Laplace transform of $f(at+b)$ in terms of the Laplace transform of $f(t)$

Question

If the Laplace transform of $f(t)$ is known, is there a way to calculate the Laplace transform of $f(at+b)$?

In particular, I am interested in the Laplace transform of $\operatorname{erf}(at+b)$.

Answer 1

For $b=0,a>0$ there is a familiar identity, the "change of scale property" $L(f(at))=\frac{1}{a} F(s/a)$ which is easy to prove by a simple change of variable.

For $b<0,a>0$ you can use the familiar translation property provided that $f(x)$ is zero whenever $x<0$. Otherwise you are adding information about negative values of $x$ that the regular Laplace transform of $f$ cannot see, and so there can be no simple identity only involving $F$.

I think otherwise there is no way to make this work in full generality, i.e. without some relationship between $f(x)$ for $x>0$ and $f(x)$ for $x<0$. For erf, you have symmetry: $\operatorname{erf}(-x)=-\operatorname{erf}(x)$, so you might be able to use this symmetry to simplify the Laplace transform.

Answer 2

$$\mathcal{L}_t[\text{erf}(a t+b)](s)=\frac{e^{\frac{s (4 b+s)}{4 a^2}} \left(1-\text{erf}\left(\frac{2 b+s}{2 a}\right)\right)+\text{erf}\left(\frac{b}{a}\right)}{a s}$$ In general if $F(s)=\mathcal{L}_t[f(t)](s)$ then $$\mathcal{L}_t[f(a t+b)](s)=\frac{e^{-b s} F\left(\frac{s}{a}\right)}{a}$$