# Double integral - failure of Fubini's theorem

I want to calculate the integral $$I = \int_{-\infty}^{+\infty} dy \int_1^{\infty}dx x^2 \frac{x^4+10x^2 y^2 -15 y^4}{(x^2 + y^2)^4}.$$ If I perform the $x$ integration first, I obtain $$I = \int_{-\infty}^{+\infty} dy \frac{1+5y^2}{(1+y^2)^3} = \pi,$$ but when I reverse the order of integration, I find that the first integration over $y$ vanishes, $$I = \int_1^{\infty}dx x^2 \int_{-\infty}^{+\infty} dy \frac{x^4+10x^2 y^2 -15 y^4}{(x^2 + y^2)^4} = \int_1^{\infty}dx x^2 \times 0 = 0.$$ Why doesn't the Fubini theorem hold in this case? The function looks pretty well behaved.

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Fubini's theorem requires the integrand to be integrable, the integral of the absolute value must be finite. This is not the case here, hence the dependence on the order of integration. –  Daniel Fischer Mar 31 '14 at 14:46