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Let $\Delta=\sqrt{1-k^2\sin^2x}$, $E(k)=\int_0^{\frac\pi 2}\Delta dx$ and $K(k)=\int\frac{dx}{\Delta}.$

Start wearing purple, gives a nice answer for the interesting identity $\int_0^{\frac\pi 2}\frac{dx}{\Delta^3}=\frac{E(k)}{1-k^2}.$

It is stuck in my head: Is $\int_0^{\frac\pi 2}\frac{dx}{\Delta^2}$ an elliptic integral? If so or not, can it be represented in terms of $k, E(k), K(k)$?

Thanks for your reading.

Bob Dobbs
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1 Answers1

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By the standard substution $\sin x=y, dx =du/\sqrt{1-u^2}$ yields the integral

$$\int \frac{du}{\sqrt{(1-u)(1+u)(1-k u)^2(1+k u)^2)}}$$

that is not an elliptic integral by the definition, the polynomial to be of degree 4 at most. So by complex topology, it cannot be expressed by elliptic integrals, just by the different structure of the Riemann surfaces.

The indefinite integral is trival of a form $\frac{\tan ^{-1}\left(\sqrt{1-k^2} \tan (x)\right)}{\sqrt{1-k^2}}$, whose inversion is not doubly periodic in the complex domain, the central property of elliptic functions. That happens for elliptic integrals for the special cases $k=0, k=1$ only.

The reason of algebraic triviality is the fact, that the denominator has the simple trigonometric form

$$ \int \frac{dx}{a + b \cos(2x)}$$

Roland F
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  • Wikipedia says "... of degree 3 or 4" somewhere in the definition. https://en.m.wikipedia.org/wiki/Elliptic_integral – Bob Dobbs Feb 28 '24 at 10:35
  • Yes, but the standard projection for elliptic integrands onto the real line here yields degree 6, with 6 possible branch points . But that is topologically not true, instead the integrand has an infinity of simple poles at $a+\cos(2x) =0$. – Roland F Feb 28 '24 at 11:56
  • I didnt understand your answer. :( – Bob Dobbs Mar 01 '24 at 07:42