TITLE: Need help evaluating this improper double integral
QUESTION [5 upvotes]: Need help evaluating this improper integral
  $$\iint_D (1-x+y)e^{(-2x+y)}\,dx\,dy$$
  where $D$ is the region: $0 \le y  \le x $.

I've tried doing the substitution:
$u = -2x +y \\ v = 1-x+y$
which gives us the integral
$\iint_E ve^{u}\,du\,dv$ over the new region $E$.
To find the bounds for the intregral, I tried plugging in the values of $x$ and $y$ expressed in terms of  $u $ and $v$ so I get the inequality:
$$0 \le 2v-u-2\le -u-1+v  \le n$$ where $n$ tends to $\infty$.
But I'm not really sure how to interpret that inequality and find the bounds for the integral.

REPLY [4 votes]: A good method : looking for a domain which is a product $A \times B$.
The idea here is to make a change of variable keeping $y \in A=[0,+\infty)$ and introducing $z=x-y$. 
$z$ is "the offset" from $x$ to $y$, naturally $\geq 0$ ; $z$ varies in the whole domain $B=[0,+\infty)$.
The integral becomes (due to a unit jacobian) :
$$\iint_{(y,z)\in A \times B} (1-z)e^{(-y-2z)}dydz=\iint_{A \times B} e^{(-y-2z)}dydz-\iint_{A \times B} ze^{(-y-2z)}dydz$$
that you can split into one-dimensional easily computable integrals.