$$\iint_D \frac{(x+y) e^{y-x}}{x^2 y^2}dx \, dy$$
$$D= \{(x,y) ; 0\leq y+1\leq x , xy\geq 1 \}$$
Iv been stuck on this for past two hours , I need some hint .
My bounds are : $\frac{1+\sqrt{5}}{2}\leq X<\infty $
$\frac 1 x \leq Y\leq x-1$ are the bounds correct ?
I need some hints, Thanks in advance
On
I think this integral converges. Let $a= (1+\sqrt 5)/2.$ Note that $(x+y)/(x^2y^2) = 1/(xy^2) + 1/(x^2y).$ Let's look at the integral that involves the first of these terms. That equals
$$\tag 1 \int_a^\infty \frac{e^{-x}}{x} \int_{1/x}^{x-1}\frac{e^y}{y^2}\,dy \, dx.$$
The inner integral is bounded above by
$$\tag 2 \int_{1/x}^{1}\frac{e^y}{y^2}\,dy + \int_{1}^{x}\frac{e^y}{y^2}\,dy.$$
The first integral in $(2)$ is bounded above $e(x-1).$ If we insert that into the outer integral in $(1),$ we get a convergent integral. For the second integral in $(2)$ note that by L'Hopital,
$$\frac{\int_1^x (e^y/y^2)\, dy}{e^x/x^2} \to 1$$
as $x\to \infty.$ So this integral is $\le 2e^x/x^2$ for large $x.$ Inserting that term into the outer integral in $(1)$ also gives a convergent integral.
That takes care of the part of the original integral involving $1/(xy^2).$ The part involving $1/(x^2y)$ can be handled the same way.
HINT
Try $u = xy$
$v=y-x$
Everything will simplfy with the Jacobian.