In my paper on geodesics on an ellipsoid, I express the area between a geodesic segment and the equator in terms of an indefinite integral $$\int \frac{t(e'^2) - t(k^2\sin^2\sigma)}{e'^2-k^2\sin^2\sigma} \frac{\sin\sigma}2 \,d\sigma,$$ where $$t(x) = x + \sqrt{x^{-1} + 1}\,\sinh^{-1}\!\sqrt x,$$ $e'$ is the second eccentricity, $k = e'\cos\alpha_0$, and $\alpha_0$ is the azimuth of the geodesic when crossing the equator. For oblate ellipsoids, we have $0 < k \le e'$.

In the paper, I evaluate this integral by Taylor expanding the integrand in the limit that $e' \rightarrow 0$. I would like to relax this assumption. I have made some unsystematic (and unsuccessful) stabs at expressing the integral in terms of elliptic integrals.

I would appreciate help with

  • expressing the integral in terms of elliptic integrals,
  • pointing me to a systematic procedure for doing this,
  • proving that the integral can't be expressed in terms of elliptic integrals, or
  • expressing the integral in terms of other special functions (especially those which can be numerical evaluated easily).

I have seen your paper. You have worked out a series with $e'\rightarrow 0$. You can do the same on the other side $e'\rightarrow\infty$. Also for this series the integrals can be evaluated. In this way, you will be able to get a satisfactory numerical evaluation of the integral for a wide range of values of $e'$. That this can be done can be easily seen by noting the asymptotic series $$ t(x)=x+\ln 2+\frac{1}{2}\ln x+\left(\frac{1}{2}\ln 2+\ln x+\frac{1}{4}\right)\frac{1}{x}+\ldots $$ that entails both logarithm and power terms. For example, for your integral you will get $$ \int\frac{t(e'^2)-t(k^2\sin^2\sigma)}{e'^2-k^2\sin^2\sigma}\frac{\sin\sigma}{2}d\sigma= -\frac{1}{2}\cos\sigma+O\left(\frac{1}{e'^2}\right) $$ and you can evaluate the higher order terms by any computer based program you prefer.

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    $\begingroup$ Thanks for looking into this! The expressions for geodesics in terms of elliptic integrals allows geodesic problems to be solved to arbitrary precision for arbitrary eccentricity rather easily. I'd like to be able to do the same with the computation of areas. It's rather difficult to do this with series expansions. The expansions (with two small parameters) quickly get big; Visual Studio barfed (with an internal compiler error) on the 30th-order expansions, until I rearranged the code. $\endgroup$ – cffk Jun 15 '17 at 18:58
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    $\begingroup$ I just checked... the number of coefficients in the nth order expansion of the area integral is about n^3/6. $\endgroup$ – cffk Jun 15 '17 at 19:31

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