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I'm having difficulty re-deriving a result a calculation from a paper. The integral is $$\int_0^{2\pi} \int_0^{2\pi} \frac{\sinh\eta}{(\cosh\eta-\cos\theta)^2}\left(1-c\sinh^2\eta\sin\phi\right)^\frac12d\theta d\phi,$$ where $\eta$ and $c$ are parameters such that $\sinh^2\eta = 2$ and $c\sinh^2\eta < 1.$

From what I've read about hypergeometric functions, one could potentially evaluate this using elliptic integrals of the first and second kind, \begin{align*} K(z) &= \int_0^{\pi/2} (1-z^2\sin^2 t)^{-1/2} dt \\ E(z) &= \int_0^{\pi/2} (1-z^2\sin^2 t)^{1/2} dt. \end{align*} These can be written in hypergeometric form as \begin{align*} K(z) &= \frac{\pi}{2}F(\frac{1}{2},\frac{1}{2};1;z^2)\\ E(z) &= \frac{\pi}{2}F(-\frac{1}{2},\frac{1}{2};1;z^2). \end{align*} The expected result of the integral is $$S = 8\pi^2 \frac{b^2}{a} G_1(c/a).$$ where $a = \sqrt{c^2+b^2}$ and $$G_1(x) = F(3/2,1/2,1;x^2)+x^2/2F(3/2,3/2,2;x^2).$$

I've tried doing this, but I am inexperienced in these calculations and it's going to take me a while. I'll keep working at it, but I was wondering if anybody could take a look and tell me whether I am even going about this in the right way.

Best,

Michael

  • 2
    Why don't you "solve" $$\int_0^{2\pi } {\frac{{\sqrt 2 }}{{{{\left( {\sqrt 3 - \cos \theta } \right)}^2}}}d\theta \int_0^{2\pi } {{{\left( {1 - 2c\sin \phi } \right)}^{\frac{1}{2}}}d\phi ,} } $$ then?2012-08-19
  • 0
    $$\int_1^\infty\frac{\log(x)}{a+bx}\,\mathrm{d}x$$ In Mathematica: `Integrate[Log(x)/(a+b*x),{x,1,Infinity}]`2012-12-24

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