Inverse Laplace transform of $\frac{s}{as+b}F(s)$

Question

I'm trying to find $\mathcal{L}^{-1}\left\{\frac{s}{as+b}F(s)\right\}$, where $F(s)=\mathcal{L}\{f(t)\}$. I have found an answer but applying the Laplace transform to check my answer doesn't take me back there, and I don't see where I'm wrong: $$\text{let }G(s)=\mathcal{L}\{g(t)\}=\frac{1}{as+b}F(s)$$ $$\Rightarrow \mathcal{L}^{-1}\left\{\frac{s}{as+b}F(s)\right\}=\mathcal{L}^{-1}\{sG(s)\}$$ $$\mathcal{L}\left\{\frac{dg}{dt}\right\}=sG(s)-g(0)\Rightarrow \mathcal{L}^{-1}\{sG(s)\}=\frac{dg}{dt}+\delta(t)g(0)$$ So all we have to do is find $g(t)$. $$\mathcal{L}\left\{e^{\frac{b}{a}t}g(t)\right\}=G\left(s-\frac{b}{a}\right)=\frac{1}{as}F\left(s-\frac{b}{a}\right)=\frac{1}{as}\mathcal{L}\left\{e^{\frac{b}{a}t}f(t)\right\}$$ $$\Rightarrow e^{\frac{b}{a}t}g(t)=\frac{1}{a}\int_0^t{e^{\frac{b}{a}t}f(t)dt}$$ $$\Rightarrow g(t)=\frac{1}{a}e^{-\frac{b}{a}t}\int_0^t{e^{\frac{b}{a}t}f(t)dt}$$ $$g(0)=\frac{1}{a}e^{-\frac{b}{a}(0)}\int_0^0{e^{\frac{b}{a}t}f(t)dt}=0$$ $$\text{so }\mathcal{L}^{-1}\left\{\frac{s}{as+b}F(s)\right\}=\frac{dg}{dt}=\frac{1}{a}\left(f(t)-\frac{b}{a}e^{-\frac{b}{a}t}\int_0^t{e^{\frac{b}{a}t}f(t)dt}\right)$$ This should mean that applying the Laplace transform should take me back to the original function: $$\mathcal{L}\left\{\frac{1}{a}\left(f(t)-\frac{b}{a}e^{-\frac{b}{a}t}\int_0^t{e^{\frac{b}{a}t}f(t)dt}\right)\right\}=\frac{1}{a}\left(F(s)-\frac{b}{a}\mathcal{L}\left\{e^{-\frac{b}{a}t}\int_0^t{e^{\frac{b}{a}t}f(t)dt}\right\}\right)$$ $$\mathcal{L}\left\{e^{\frac{b}{a}t}f(t)\right\}=F\left(s-\frac{b}{a}\right)$$ $$\Rightarrow \mathcal{L}\left\{\int_0^t{e^{\frac{b}{a}t}f(t)dt}\right\}=\frac{1}{s}F\left(s-\frac{b}{a}\right)$$ $$\Rightarrow \mathcal{L}\left\{e^{-\frac{b}{a}t}\int_0^t{e^{\frac{b}{a}t}f(t)dt}\right\}=\frac{1}{s+\frac{b}{a}}F(s)$$ And the Laplace transform is equal to: $$\frac{1}{a}\left(F(s)-\frac{b}{a}\left(\frac{1}{s+\frac{b}{a}}F(s)\right)\right)$$ $$=\frac{1}{a}\left(1-\frac{b}{as+b}\right)F(s)$$ Which does not look like anything that boils down to $\frac{s}{as+b}F(s)$.

Answer 1

Oh well, it was my mistake to assume that $\frac{1}{a}\left(1-\frac{b}{as+b}\right)F(s)$ doesn't equal $\frac{s}{as+b}F(s)$. $$\frac{1}{a}\left(1-\frac{b}{as+b}\right)=\frac{1}{a}\left(\frac{as+b-b}{as+b}\right)=\frac{1}{a}\left(\frac{as}{as+b}\right)=\frac{s}{as+b}$$