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Let be three real numbers, $\chi_*$, $\chi\in [-a,+a]\ (a>0)$, $\xi_*> 0$ and let be $\phi\in[0,2\pi]$.

Q1: I seek some analytical expression for the following integral,

$\int_{0}^{2\pi} d\phi \frac{(\xi_*\cos\phi -1)(\xi_* - \cos\phi)}{[\xi_*^2-2\xi_*\cos\phi + 1 +(\chi_*-\chi)^2]^2}$

Q2: As a step further I also seek for some formula after integration over the $\chi$ variable i.e.,

$\int_{-a}^{a} d\chi \int_{0}^{2\pi} d\phi \frac{(\xi_*\cos\phi -1)(\xi_* - \cos\phi)}{[\xi_*^2-2\xi_*\cos\phi + 1 +(\chi_*-\chi)^2]^2}$

Thanks for your help, Pete

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    $\begingroup$ For Q1, the integrand is a rational function of $\cos\phi$, and can thus be solved analytically by the substitution $u = \tan\frac{phi}{2}$. Alternatively, write it as the contour integral around the unit circle of the rational function obtained by substituting $\cos\phi\mapsto \frac{z+ z^{-1}}{2},\mathrm d\phi\mapsto \frac{\mathrm dz}{2\pi i z}$, which can be calculated analytically using the residue theorem. Elliptic integrals are not needed. $\endgroup$
    – Bertram Arnold
    Commented Mar 14, 2018 at 0:40
  • $\begingroup$ Thank you Arnold for your precious comment. I was not sure of a possible link with elliptic integrals neither, that's why I only suggested their usage in the body of the question. I consider Q1 solved and re-edit the question. $\endgroup$
    – Pete
    Commented Mar 14, 2018 at 1:51
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    $\begingroup$ "I only suggested their usage in the body of the question." Actually, you required their usage in the title of the question, which was "Integration by means of elliptic integrals." $\endgroup$
    – Gerry Myerson
    Commented Mar 14, 2018 at 4:44
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    $\begingroup$ You're right Gerry, the initial title was really inappropriate. It did not reflect all the hesitation I've had about a possible link with Elliptic integrals at the time of writing it. I decided to reformulate it in order to prevent mathoverflow audience who might be interested in expressing formulas from being misdriven. All my apologies ... $\endgroup$
    – Pete
    Commented Mar 14, 2018 at 8:58

1 Answer 1

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Q1: $$I(\chi)\equiv\int_{0}^{2\pi} d\phi \frac{(\xi_*\cos\phi -1)(\xi_* - \cos\phi)}{[\xi_*^2-2\xi_*\cos\phi + 1 +(\chi_*-\chi)^2]^2}=$$ $$-\frac{\pi}{2\xi_\ast}+\frac{\pi}{2\xi_\ast}\frac{\delta\chi^6+3 \delta\chi^4 \left(\xi_\ast^2+1\right)+3 \delta\chi^2 \left(\xi_\ast^2-1\right)^2+\left(\xi_\ast^2-1\right)^2 \left(\xi_\ast^2+1\right)}{\left(2 \left(\delta\chi^2-1\right)\xi_\ast^2+\left(\delta\chi^2+1\right)^2+\xi_\ast^4\right)^{3/2}}$$ with the abbreviation $\delta\chi=\chi_\ast-\chi$.

Q2:

$$\int I(\chi)\,d\chi=\frac{\pi\delta\chi}{2\xi_\ast}\frac{\delta\chi^2+\xi_\ast^2+1-\sqrt{2 \left(\delta\chi^2-1\right) \xi_\ast^2+\left(\delta\chi^2+1\right)^2+\xi_\ast^4}}{\sqrt{\delta\chi^4+2 \delta\chi^2 \left(\xi_\ast^2+1\right)+\left(\xi_\ast^2-1\right)^2}}$$

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  • $\begingroup$ Thank you very much Carlo for your formulation. I have to derive it on my own, now. I've only checked that for $\xi_*\rightarrow 0+$ your solution $I(\chi)$ was trivially zero. As soon as I have time to confirm I'll upvote your answer. Many thanks again. $\endgroup$
    – Pete
    Commented Mar 17, 2018 at 7:38

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