Finding the difference of two random variables Give that $X_1$ and $X_2$ are random variables and the joint of $$f_{x_1x_2}(x_1,x_2) = e^{-x_1},\quad 0\le x_2\le x_1\le \infty$$
Given that $Y=X_1-X_2$
Find the pdf for Y.
So far I have.
$$P(Y \le y)=\iint_D e^{-x_1}\,dx_1\,dx_2.$$
$$P(Y \le y)=\int_{0} ^{\infty}\int_{0} ^{y+x_2} e^{-x_1}\,dx_1\,dx_2.$$
After I evaluate the integral I get $$e^{-y-x_2}+x_2$$ which if I try to evaluate from 0 to infinity it diverges and that doesn't seem right.
 A: We will use $X$ instead of $X_1$, and $Y$ instead of $X_2$. So we need a new letter for what you called $Y$. Let $W=X-Y$.   
Take $w\ge 0$. We have $W\le w$ if $X\le Y+w$. Thus
$$\Pr(W\le w)=\iint_D e^{-x}\,dx\,dy,$$
where $D$ is the part of the first quadrant between the lines  $y=x-w$ and $y=x$.
This strip is slightly unpleasant to integrate over. It is easier to note that
$$\Pr(W\le w)=1-\Pr(X\gt Y+w)=1-\iint_E  e^{-x}\,dx\,dy,$$
where $E$ is the part of the first quadrant below the line $y=x-w$.
We have
$$\iint_E  e^{-x}\,dx\,dy=\int_{x=w}^\infty \left(\int_{y=0}^{x-w} 1\,dy\right)e^{-x}\,dx.$$
Integrate. The inner integral is $x-w$, so we end up with $\int_w^\infty (x-w)e^{-x}\,dx$, which is $e^{-w}$.
It follows that (for $w\gt 0$), we have $F_W(w)=1-e^{-w}$. 
Added: There is a poor choice of order of integration above, which resulted in an integration by parts for the outer integral. It is better to note that
$$\iint_E  e^{-x}\,dx\,dy=\int_{y=0}^\infty \left(\int_{x=y+w}^{\infty} e^{-x}\,dx\right)\,dy.$$
