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$$\iint_D \frac{(x+y) e^{y-x}}{x^2 y^2}dx \, dy$$

$$D= \{(x,y) ; 0\leq y+1\leq x , xy\geq 1 \}$$

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

Kasmir Khaan
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    So far I can only confirm that the bound of your domain are indeed correct – b00n heT Nov 07 '16 at 22:20
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    Thanks for taking your time , we just started on this topic and I don't know how to proceed , integrating with respect to x and y both fails. – Kasmir Khaan Nov 07 '16 at 22:25
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    My next idea is to try comparison test but what should I campare it with ? – Kasmir Khaan Nov 07 '16 at 22:26
  • Maybe I'm wrong but you can take for comparison, $y=x-1$, while $x-1>2$, if you suppose that it converges. Or take $y=2$, if you suppose that it will diverge. $y=2$ is the minimum of $\frac{e^{y}}{y^{2}}$. So you can compare the double itegral to one integral, with either $(x-1)$ or with the $2$ for $y$, but you should always multiply it with the length $x-1-\frac{1}{x}$. – kolobokish Nov 07 '16 at 22:46
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    I've deleted my incomplete answer, but I'm leaning toward applying a comparison test after massaging the integral a bit. $\qquad$ – Michael Hardy Nov 08 '16 at 01:26
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    That was my idea too but I could not find any function to compare with that diverges . – Kasmir Khaan Nov 08 '16 at 16:15
  • Huh. Doing some legerdemain in Mathematica and asking it to fully simplify the answer, I get a final result of $1/e$ which is far simpler than I'd have expected. – Semiclassical Nov 09 '16 at 22:01

2 Answers2

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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.

zhw.
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    I will take a look right now at your solution , but am very sure that the i wrote the question right , ie diverges , i looked it up again , that been being said my teacher could made a mistake. Thanks for your answer . – Kasmir Khaan Nov 08 '16 at 17:31
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    If i can show that $e^{y-x}$ diverges and what we multiply it by converges , is that considered to be a legal step ? is the product of divergent and convergent = divergent ? – Kasmir Khaan Nov 08 '16 at 17:55
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    Of course not. In fact divergent times divergent can be convergent ($1/x$ on $[1,\infty)$ for example) – zhw. Nov 08 '16 at 18:00
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    do you know where i can find the rules such as , sum and product of convergent and divergent series ,and what the aritmatic on them? thanks – Kasmir Khaan Nov 08 '16 at 18:14
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    Not sure why you're looking to me for a proof of divergence, when I think the integral converges. – zhw. Nov 08 '16 at 19:44
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    you were right ! it does converge , am sorry and thank you ! – Kasmir Khaan Nov 09 '16 at 23:10
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HINT

Try $u = xy$

$v=y-x$

Everything will simplfy with the Jacobian.