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.

  • 2
    $\begingroup$ 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? $\endgroup$ – Pedro Tamaroff Aug 19 '12 at 0:59
  • $\begingroup$ $$\int_1^\infty\frac{\log(x)}{a+bx}\,\mathrm{d}x$$ In Mathematica: Integrate[Log(x)/(a+b*x),{x,1,Infinity}] $\endgroup$ – user54110 Dec 24 '12 at 8:36

Going off Peter's comment, note that your integral is separable, and can thus be factored into a product of two one-dimensional integrals:

$$\begin{split}&\int_0^{2\pi} \int_0^{2\pi} \frac{\sinh\eta}{(\cosh\eta-\cos\theta)^2}\left(1-c\sinh^2\eta\sin\phi\right)^\frac12\mathrm d\theta\mathrm d\phi=\\&\quad\left(\color{green}{\sinh\eta\int_0^{2\pi} \frac{\mathrm d\theta}{(\cosh\eta-\cos\theta)^2}}\right)\left(\color{blue}{\int_0^{2\pi} \sqrt{1-c\sinh^2\eta\sin\phi} \mathrm d\phi}\right)\end{split}$$

The first integral is elementary:

$$\color{green}{\sinh\eta\int_0^{2\pi} \frac{\mathrm d\theta}{(\cosh\eta-\cos\theta)^2}}=\frac{2\pi\cosh\,\eta}{\sinh^2\eta}$$

while the second requires the services of the incomplete elliptic integral of the second kind $E(\phi\mid m)$ (see Byrd and Friedman, formula 288.01 for the required identity):

$\displaystyle\begin{split}&\color{blue}{\int_0^{2\pi} \sqrt{1-c\sinh^2\eta\sin\phi} \mathrm d\phi}=\\&\Tiny 2\left(\sqrt{\frac{1+c\sinh^2\eta}{1-c\sinh^2\eta}}\;E\left(\arcsin\left(\sqrt{\frac{1+c\sinh^2\eta}{2}}\right)\mid\frac{2c\sinh^2\eta}{1+c\sinh^2\eta}\right)-\frac{c\sinh^2\eta}{\sqrt{1-c\sinh^2\eta}}+2\sqrt{1+c\sinh^2\eta}\;E\left(\frac{\pi}{4}\mid\frac{2c\sinh^2\eta}{1+c\sinh^2\eta}\right)\right)\end{split}$

  • $\begingroup$ J.M., thanks so much! I forgot to write this in the OP, but yes, of course let's split the integrals. I didn't know how to integrate $\int_0^{2\pi} (\cosh\eta-\cos\theta)^{-2} d\theta$ -- did you evaluate it or check with Wolfram? Going to see if the answers agree, will accept if they do. Thanks again! $\endgroup$ – snar Aug 19 '12 at 16:23
  • $\begingroup$ "I didn't know how to integrate..." - the Weierstrass substitution $u=\tan(\theta/2)$ is the first thing to try... $\endgroup$ – J. M. isn't a mathematician Aug 19 '12 at 17:18

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