This post discusses Example 3.5.4 in Resnick's book on page 203. However, when I try to understand the next example 3.5.5, I still have some doubts.
Example 3.5.5
If $F(dx) = xe^{-x}dx$, then we have the following: $$\sum_{n=1}^{\infty}F^{n*}(dx)=\frac{1}{2}-\frac{1}{2}e^{-2y}$$ In this case, $$U*z(t)=z(t)+\int_0^tz(t-y)\{\frac{1}{2}-\frac{1}{2}e^{-2y}\}dy$$
My Doubts
We know that \begin{align} U(t) &= \sum_{n=0}^{\infty}F^{n*}(dx)\\ &= F^{0*}+\sum_{n=1}^{\infty}F^{n*}(dx)\\ &= 1+\Bigg(\frac{1}{2}-\frac{1}{2}e^{-2x}\Bigg)\\ &= \frac{3}{2}+\frac{1}{2}e^{-2x} \end{align}
By using the explanation from previous post: \begin{align} U(t) *z(t)&=\Bigg(\frac{3}{2}+\frac{1}{2}e^{-2x}\Bigg)*z(t) \\ &= \frac{3}{2}*z(t) + \frac{1}{2}e^{-2x}*z(t) \\ &= \frac{3}{2}z(t) + \int_0^tz(t-y)\{-\frac{1}{2} d(e^{-2y}) \}\\ &= \frac{3}{2}z(t) + \int_0^tz(t-y)e^{-2y}dy \end{align} Which is different from the textbook solution. What's wrong here?
Secondly, I am very confuse about the meaning of $F(dx)$ For example, in the definition of convolution, we have $$F_1*F_2=\int_0^tF_2(t-x)F_1(dx)$$.
So $F(dx) = dF(x)$?