Transforming Limits of a Double Integral I´m working on trasnforming a double integral: $\int_0^\infty \int_0^x e^{-(x+y)} \,dy\,dx$ using the following identities identities. I need to get the limit to be 0 and1  in order to integrate it by using Monte Carlo´s integration

$$\theta=\int_0^\infty g(x)\,\mathrm dx,$$ we could apply the substitution $y=1/(x+1),\mathrm dy=-\mathrm dx/(x+1)^2=-y^2\mathrm dx,$ to obtain the identity $$\theta=\int_0^1h(y)\,\mathrm dy,$$ where #1 $$h(y)=\dfrac{g\left(\tfrac1y-1\right)}{y^2}.$$

and #2 $$h(y)=g(a+(b-a)y)(b-a).$$
when

$$\theta=\int_a^b g(x)\,\mathrm dx,$$
  So far this is what I´ve done:

applying  #1 $h(y)$:
$$\int_0^1 \int_0^x \dfrac{e^{-((1/y) -1)+y}}{y^2} \,dy\,dx$$
and now applying #2 :
$$\int_0^1 \int_0^1 \dfrac{e^{-((1/yx) -1)+xy}}{(xy)^2}(x) \,dy\,dx$$
taking $a=0$ and $b=x$
Can anyone help to check my answers?
Edit: I found a hint reading several books, but I still can´t transform both limits to 0 and 1.

I should use the function above equate the integralto one in which both terms go from 0 to ∞.]
Thank you very much

After serveral attempts I was able to solve this problem:

