Suppose $u$ satisfies the equation $$u(x)=\beta\int_0^x u(x-y)dF(y)+v(x)$$ where $0<\beta<1$ and $F(y)$ a distribution function with $F(0)=0$ and $v(x)\geq 0$ continuous for $x\geq 0$. The Laplace Stieltjes transform is $\tilde{f}(s)=\int_0^\infty e^{-sy}dF(s)$. Let $F^{*n}$ be the $n$ times convolution of $F$ with itself. Then the Laplace transform is $\int_0^\infty e^{-sy}F^{*n}(y)=\frac{\tilde{f}(s)}{s}$. Define the compound geometric density function $G(y)=\sum_{n=0}^\infty(1-\beta)\beta^nF^{*n}(y)$. Show that $u$ can be written as $$u(x)=\frac{1}{1-\beta}\int_0^x v(x-y)dG(y)+v(x)$$.
My ideas: I want to take the Laplace transform of both equations and show that they are the same. The first equation gives me $\tilde{u}(s)=\beta \tilde{u}(s)\tilde{f}(s)+\tilde{v}(s)$. This implies $\tilde{u}(s)=\frac{\tilde{v}(s)}{1-\beta\tilde{f}(s)}$. So I have to show that the second equation gives the same result. How can I go on?
Let's recall that the the Laplace-Stieltjes transform maps the convolution product to the usual product. In consequence, the second equation is transformed into $$ \tilde{u}(s) = \frac{\tilde{v}(s)\tilde{g}(s)}{1-\beta}, $$ where $$ \frac{\tilde{g}(s)}{1-\beta} = \sum_{n\ge0} \beta^n \tilde{f}(s)^n = \frac{1}{1-\beta\tilde{f}(s)} $$ thanks to the geometric series, hence in the end $$ \tilde{u}(s) = \frac{\tilde{v}(s)}{1-\beta\tilde{f}(s)}, $$ which coincides with your first result. N.B. : your second equation has got one term too many, namely the final $\tilde{v}(s)$, probably due to copying and pasting.