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How can I compute this integral?

$$\int _0 ^{\infty} \int _0 ^{\infty }\frac { 4xy-x^2-y^2}{(x+y)^4}dx dy$$

I tried polar coordinate, but it didn't work.

Would anyone please give me a hint?

Thanks for your help and advice.

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  • $\begingroup$ Go polar coordinates. The region of integration is in the first quadrant. So you can imagine a circle with radius from 0 to infinity, and angle from 0 to pi/2 $\endgroup$ – cgo Nov 17 '15 at 6:08
  • $\begingroup$ @cgo yes I tried that, but it was still hard to compute. Would you please help me more? $\endgroup$ – user115608 Nov 17 '15 at 6:14
  • $\begingroup$ As a Lebesgue double integral, it diverges because $\int 1/r\; dr$ diverges. But in polar coordinates the integral over $\theta$ is $0$. I'm not sure what happens as an iterated improper integral in $x$ and $y$. $\endgroup$ – Robert Israel Nov 17 '15 at 6:19
  • $\begingroup$ What have you done? $\endgroup$ – Euler....IS_ALIVE Nov 17 '15 at 6:19
  • $\begingroup$ @RobertIsrael Do you think this user knows about Lebesgue? $\endgroup$ – Euler....IS_ALIVE Nov 17 '15 at 6:19
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Let $u=x+y$, so $x = u-y$.

$$ \begin{align*} &\int _0 ^{\infty} \int _y ^{\infty}\frac { 4(u-y)y-(u-y)^2-y^2}{u^4}du\ dy\\ &=\int _0 ^{\infty} \int _y ^{\infty }4yu^{-3}-4y^2u^{-4}-u^{-2}+2yu^{-3}-y^2u^{-4}-y^2u^{-4} \ du\ dy\\ &=\int _0 ^{\infty} \int _y ^{\infty }6yu^{-3}-6y^2u^{-4}-u^{-2} \ du\ dy\\ &=\int _0 ^{\infty} -3yu^{-2}+2y^2u^{-3}+u^{-1} \bigg|_y^\infty\ dy \\ &=\int _0 ^{\infty} 3y^{-1}-2y^{-1}-y^{-1} \ dy \\ &=\int _0 ^{\infty} 0\ dy\\ &= 0 \end{align*} $$

This matches Wolfram's answer.

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  • $\begingroup$ Oops... just saw the comments about how a change of variable is not justified. Perhaps someone can point out the error for posterity's sake. $\endgroup$ – zahbaz Nov 17 '15 at 6:25
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    $\begingroup$ Change of variables for the inner ($x$) integral is OK, and does give $0$. Change to polar (or other) coordinates would be problematic. $\endgroup$ – Robert Israel Nov 17 '15 at 6:27
  • $\begingroup$ Interesting... Thanks for the comment. $\endgroup$ – zahbaz Nov 17 '15 at 6:39
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Try the partial fraction expansion $$\begin{align} \frac{4xy-x^2-y^2}{(x+y)^4} & = \frac{6xy-(x+y)^2}{(x+y)^4} \\[1ex] & = \frac{-6y^2}{(x+y)^4}+\frac{6y(x+y)}{(x+y)^4} - \frac{(x+y)^2}{(x+y)^4} \\[1ex]& = \frac{-6y^2}{(x+y)^4}+\frac{6y}{(x+y)^3} - \frac{1}{(x+y)^2} \end{align}$$

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