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Consider the integral

$$\displaystyle \int_{0}^{2\pi} \int_{0}^{2\pi} (x^2 + y^2)^{-3/4}((b-x)^2 + (c-y)^2)^{-3/4}\mathrm{d}x\mathrm{d}y,$$

where $b, c \in \mathbb{Z}$ are non-zero constants. It seems like there should be an obvious substitution to make here, but I can't spot one that would turn the integral into something nicer. In particular, I'm having trouble finding a substitution that works with the constants. If this were an integral in one dimension, a substitution like $x = bt$ would work rather easily, given that we could just pull the constant out. In this case, however, the fact that we have two constants here makes it tricky. Converting to polar co-ordinates would certainly simplify the $(x^2 + y^2)^{-3/4}$ term but then dealing with the second term is still an issue, as are the constants. Can anyone recommend a substitution to use here?

For anyone wondering why I need this result, see here. The NIntegrate function in Mathematica quickly gives results if we plug in values for $b,c$, but I need a general result for $b,c$. I am also willing to accept a bound for the integral provided the resulting expression in $b,c$ is square-summable (see the link).

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user363087
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  • Should this be an integral over a rectangle? The limits make me think you might have some kind of typo. – Ben Grossmann Oct 10 '16 at 18:30
  • No, the limits are correct. – user363087 Oct 10 '16 at 18:31
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    If you use polar coordinates, we can apply the law of cosines in order to rewrite the second term as $$ \left[r^2 + b^2 + c^2 - 2\sqrt{b^2 + c^2}r\cos\left(\theta - \arctan(c/b)\right)\right]^{-3/4} $$ but then you're taking a polar integral over a rectangular region – Ben Grossmann Oct 10 '16 at 18:38
  • looks hopeless! – tired Oct 10 '16 at 19:39
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    Mathematica does not know the answer, either the indefinite or the definite integral. So it seems likely that no simple substitution will work. – Greg Martin Oct 10 '16 at 19:58
  • That seems to be the case. However, NIntegrate seems to work if we put in values for $b,c$, but of course does not give a symbolic answer for general $b,c$. – user363087 Oct 10 '16 at 20:03

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