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I am being asked to find $\int_1^\infty\int_1^\infty \frac{x^2-y^2}{(x^2+y^2)^2}\,dx\,dy$ and $\int_1^\infty\int_1^\infty \frac{x^2-y^2}{(x^2+y^2)^2}\,dy\,dx$. I am even told to "notice that" $\int_1^\infty \frac{x^2-y^2}{(x^2+y^2)^2}\,dy=\frac{1}{x^2+1}$.

I can put both of these in my calculator to find the respective values are $\frac{\pi}{4}$ and $-\frac{\pi}{4}$. I have two problems...

  1. I believe the "notice that" integral should be negative. If it were taken with respect to $x$, then it would be positive. Am I correct?
  2. If I assume this piece of information, I know how to compute the rest using the an inverse trigonometric function. The problem is I do not know why this is true. I have tried factoring the difference of two squares and PFD, but that does not seem to help me.

Suggestions? Thank you!

fullyhip
  • 436

1 Answers1

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Probably the easiest proof here is found by noting that $\frac{x^2-y^2}{(x^2+y^2)^2} = -\frac{\partial^2}{\partial x \partial y}\arctan(y/x)$ $$\int_1^\infty\frac{x^2-y^2}{(x^2+y^2)^2}\,dy$$ $$\int_1^\infty \frac{\partial^2}{\partial x \partial y}\arctan(y/x)\,dy$$ $$=\lim_{t \to \infty}\left.\left(\frac{\partial}{\partial x}\arctan(y/x) \,dy\right)\right|_{y=1}^{y=t}$$ $$=\lim_{t \to \infty}\left.\left(\frac{-y}{x^2 + y^2}\right)\right|_{y=1}^{y=t}$$ $$=\frac{-1}{x^2 + 1}$$
We can now easily integrate with respect to $x$ to get the result as claimed... note that the OP is correct though, the provided hint has the wrong sign. Note that this is an example of Fubini's theorem (and a function that violates it) and further info can be found here

Brevan Ellefsen
  • 14,793