I have studied that changing the order of double integration will not change the answer if both the limits of integration are constants. But this function is not agreeing with what I have studied: $$1)\int_0^1\left(\int_0^1{\frac{x-y}{(x+y)^3}dy}\right)dx$$ $$2)\int_0^1\left(\int_0^1{\frac{x-y}{(x+y)^3}dx}\right)dy$$ The answer to the first integral is 0.5 and that of the second integral is -0.5 respectively.

Can anyone please explain why is this so?

  • 1
    $\begingroup$ Have you checked that the integral of the absolute value is finite? Otherwise you wouldn't be in the conditions of the Fubini theorem. $\endgroup$ – Javi Dec 7 '17 at 15:06
  • $\begingroup$ $(0,0)$ is a problematic point for $\frac{x-y}{(x+y)^3}$. $\endgroup$ – Malcolm Dec 7 '17 at 15:07

In general you cannot switch the order of integration without additional constraints. These are typically given by Fubini's theorem. In particular the example you've given does not converge absolutely so switching the order changes the answer.

  • $\begingroup$ What do you mean by absolute convergence? Can you explain it? $\endgroup$ – Jigar Faria Dec 9 '17 at 10:45
  • $\begingroup$ @jigarfaria1995 Sure. Absolute convergence in this case means the convergence of the absolute value of the function. This is analogous to absolute convergence of a series which is defined similarly using the sum of the absolute values of the terms. en.wikipedia.org/wiki/Absolute_convergence $\endgroup$ – CyclotomicField Dec 9 '17 at 13:53
  • $\begingroup$ Thank you for your reply. It has helped me to understand some other integrals too. $\endgroup$ – Jigar Faria Dec 10 '17 at 16:08

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