I have two positive, independent, absolutely continuous random variables $X$ and $Y$ with their moment generating functions $M_X$ and $M_Y$. In Foata Fuchs, there is an exercise stating $$\mathbb E\Big[\frac{X}{X+Y}\Big]=-\int_{\mathbb R}M_X'(t) M_Y(t)\,dt.$$
Trying this for $X,Y\sim Exp(1)$, implies $M_X(t)=M_Y(t)=(1-t)^{-1}$. These functions are only defined for $|t|<1$. Furthermore, $M'_X(t)=(1-t)^{-2}$. And finally $$-\int_{\mathbb R}\frac{1}{(1-t)^3}\,dt.$$ This integral, however, does not converge.
However, $$\mathbb E\Big[\frac{X}{X+Y}\Big]=\int_0^\infty\int_0^\infty\frac{x}{x+y}e^{-(x+y)}\,dxdy$$
I do know $$\frac{1}{x+y}=-\int_0^\infty e^{-(x+y)t}dt$$
Provided, I can exchange the integrals, that should give me the answer above... What is wrong with this formula?
Hmmm, I looked through the book and there seems to be confusion in the notation and functions involved. The book suggests in particular that $$ \mathbb{E}\left( \frac{X}{X+Y}\right) = -\int_{\mathbb{R}^{+}} g_1'(u) g_2(u)du $$ where $$ g_1'(u) = \frac{d}{du} \mathbb{E}(e^{-uX}), \quad g_2(u) = \mathbb{E}(e^{-uY}) $$ let's calculate all this \begin{align*} \mathbb{E}(e^{-uX}) & = \int_{0}^{\infty} e^{-ux} e^{-x} dx = \frac{1}{1+u} \\ \frac{d}{du} \mathbb{E}(e^{-uX}) & = - \frac{1}{(1+u)^2} \end{align*} we use the formula \begin{align*} \mathbb{E}\left( \frac{X}{X+Y}\right)& = -\int_{\mathbb{R}^{+}} g_1'(u) g_2(u)du \\ & = -\int_{0}^{\infty} \frac{-1}{(1+u)^2} \frac{1}{1+u} du \\ & = \int_{0}^{\infty} \frac{1}{(1+u)^3} du \\ & = -\frac{1}{2(1+x)^2} \Big|_{0}^{\infty} \\ & = 1/2 \end{align*} Using what you suggest we can do the following \begin{align*} \int_{0}^{\infty} \int_{0}^{\infty} \frac{x}{x+y} e^{-x-y} dydx & = \int_{0}^{\infty} \int_{0}^{\infty} \int_{0}^{\infty} x e^{-x-y} e^{-t(x+y)} dtdydx \\ & = \int_{0}^{\infty} \int_{0}^{\infty} \int_{0}^{\infty} x e^{-(x+y)(t+1)} dtdydx \\ & = \int_{0}^{\infty} \int_{0}^{\infty} \int_{0}^{\infty} x e^{-(x+y)(t+1)} dydxdt \\ & = \int_{0}^{\infty} \int_{0}^{\infty} \frac{x e^{-x(t+1)}}{t+1} dxdt \\ & = \int_{0}^{\infty} \frac{1}{(t+1)^3} dt \\ & = 1/2 \end{align*}