I have seen questions on how to integrate double integrals but I do not know enough maths to be able to understand it. I have come across the need to integrate this function but don't know where to start.


I mostly want to know what the integral is but would also appreciate some pointers on how to go about doing this myself. I tries WolframAlpha but it kept timing out on me.

Thanks for any help.

  • $\begingroup$ I suppose one could use the transformation $t \mapsto y - t$ and $s \mapsto x - s$ and then use the Jacobian for the transformations effect on $dt ds$. Once you do this, maybe use polar coordinates? $\endgroup$ – ADF Dec 15 '11 at 17:55
  • $\begingroup$ It may help to know which part you don't understand. If you understand single integrals, I believe this is the same as $\int_{-S}^S(\int_{-T}^T\frac1{(x-s)^2+(y-t)^2}dt)ds$. In other words, the first integration is in terms of t and s is treated as a constant. The first integration looks as if it will yield some sort of inverse tangent function. $\endgroup$ – Mike Dec 15 '11 at 19:29
  • $\begingroup$ I think you should impose some conditions on the integration domain to give a meaning to such an integral. What gets stuck Mathematica is exactly this. $\endgroup$ – Jon Dec 15 '11 at 19:45
  • $\begingroup$ Do you know any residue calculus? Do you have any condition on $x,y$? Note that $\int\int(x^2+y^2)^{-1}dxdy=\int x^{-1}\int(1+(y/x)^2)^{-1}dydx=\int \arctan(y/x)/x dx\sim \int \arctan(ye^u)du$, which seems difficult. $\endgroup$ – AD. Dec 15 '11 at 20:18
  • $\begingroup$ What is meant by a condition on x or y? In this x, y, S and T are constants. @Mike is it correct to just integrate as you suggested, I have spoken to other people and they give me the impression that this is the wrong thing to do. $\endgroup$ – DrYap Dec 16 '11 at 16:08

You can try a change of variables in the form: $s - x = rcos(\theta), y - t = rsin(\theta) $, and you'll get the integral of $\frac{1}{r^2}$ in other similar domain.

  • 2
    $\begingroup$ That seems like it would be difficult with a square domain though. $\endgroup$ – W. Cadegan-Schlieper Dec 15 '11 at 18:54
  • $\begingroup$ You should improve your answer providing a proper integration domain. $\endgroup$ – Jon Dec 15 '11 at 19:44
  • $\begingroup$ Yes, I think you are right. It would be difficult with a square domain. $\endgroup$ – ofer Dec 15 '11 at 19:55

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