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$\int_0^1 \sqrt{\frac{2-x^4}{1-x^4}}\, dx$ in terms of the gamma function or elliptic integrals

Too long for comment and not really an answer to the central question, but here is how we can recover the hypergeometric result: $$\begin{align*} I &= \int_0^1 \sqrt{\frac{2-x^4}{1-x^4}} \, dx \\ &...
user170231's user avatar
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1 vote

Exact solution to integral (Keplerian free fall time)

We have the ODE with $k=G M$ $$r''(t)=-\frac{k}{r(t)^2}$$ We multiply the factor $2r'(t)$ on both sides. $$2r'(t)r''(t)=-\frac{2k}{r(t)^2}r'(t)$$ No we integrate the equation over time $$r'(t)^2=\int -...
gpmath's user avatar
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6 votes
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Is this a 'standard' elliptic integral?

The first integral is a third type elliptic one, of the form $$\frac1{(1+\mu)^{3/2}}\int\dfrac{d\alpha}{\left(1-\frac{2\mu}{1+\mu}\cos^2\frac\alpha2\right)\sqrt{(1-\frac{2\mu}{1+\mu}\cos^2\frac\alpha2)...
Yves Daoust's user avatar
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10 votes

Is this a 'standard' elliptic integral?

Maple says $$ \int_{0}^{\pi}\! \left( 1-\mu\,\cos \left( \alpha \right) \right) ^{- {\frac{3}{2}}}\,{\rm d}\alpha=-2\,{\frac {1}{\sqrt {\mu+1} \left( -1+ \mu \right) }{E} \left( {\frac {\sqrt {2}\...
GEdgar's user avatar
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0 votes

An Elliptic Curve has 3 real roots if and only if $\frac{\omega_2}{\omega_1}$ is purely imaginary

I think that for the other directon OP's approach is good. For the other direction, WLOG, we may assume that $e_3<e_2<e_1$. Then, by direct calculation, we can find that $$\omega_1 = \int_{\...
Bob Dobbs's user avatar
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