Çinlar's Probability and Stochastic, Examples 2.11 and 2.12

Question

I have two questions about a few steps in Çinlar's derivations of Examples 2.11 and 2.12.

1. (Blue highlight) How did he go from $\mathbb{E}f(X+Y, \frac{X}{X+Y})$ to $\int_{0}^{\infty}dx\frac{x^{a-1}e^{-x}}{\Gamma(a)}\int_{0}^{\infty}dy\frac{x^{b-1}e^{-y}}{\Gamma(b)}f(x+y, \frac{x}{x+y})$? I can understand that from Theorem 2.4, we have $\mathbb{E}f\circ X = \mu f$, so the integral will look something like $\int_{0}^{\infty}\mu(dx)\int_{0}^{1}K(x,dy)f(x,y)$, but where did the transitional kernel go in his derivation?

2. (Green highlight) He said that $\mathbb{P}\{R > r\} = (\frac{c}{c+r})^a$ according to the Laplace transform computation above in 2.11, but I don't really get it... What does this have anything to do with 2.11 (the previous example)? It doesn't seem like the previous example has anything to do with the Laplace transform.

Thanks to everyone in advance for helping out!

Here are the screenshots of the above examples and Theorem 2.4:

Example 2.11:

Example 2.12:

Theorem 2.4:

Answer 1

1. The blue highlight is just using the fact that if $X$ and $Y$ are independent with respective densities $f_X(x),f_Y(y),$ then by LOTUS,

$$E[g(X,Y)]=\int_\mathbb{R_+}\int_\mathbb{R_+}g(x,y)f_X(x)f_Y(y)dydx\\ =\int_\mathbb{R_+}f_X(x)\int_\mathbb{R_+}g(x,y)f_Y(y)dydx.$$

Yours is the case where $g(X,Y)=f(X+Y,\frac{X}{X+Y})$ for a positive Borel function $f$ on $\mathbb{R_+}\times [0,1],$ and $f_X,f_Y$ are standard gamma densities.

2. As for the green highlight, not sure exactly how the author meant to tie it into Example 2.11 (is there an equation also labelled 2.11?). I guess one way to tie it in is by using $f(u,v)=e^{-ruv}$ since the integral in Example 2.11 would then be computing $E[e^{-rX}]$ as desired (at least for standard gamma $X$). But you can also get their result by using the MGF of a gamma random variable.