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The theory of elliptic functions tells us that any elliptic curve defined by a cubic $y^2 = 4(x - e_1(x - e_2)(x - e_3)$ with distinct roots is isomorphic to the quotient $\mathbb{C}$ by a lattice. The periods of this lattice can be found by evaluating the period integrals $$\int_{\infty}^{e_1}[(x - e_1)(x - e_2)(x - e_3)]^{-1/2}dx$$ and $$\int_{e_1}^{e_3}[(x - e_1)(x - e_2)(x - e_3)]^{-1/2}dx.$$ My understanding is that to actually calculate these integrals, one performs a transformation which turns them into complete elliptic integrals of the first kind $$K(k) = \int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1 - k^2\sin^2\theta}},$$ which are given by the Gaussian hypergeometric series. Does anyone have a reference for the change of variables required to convert the period integrals into the elliptic integral? Thanks!

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Not sure this is want you want, but

$\sin \theta = t, d \theta = \frac{dt}{ \sqrt{1-t^2}}$ so you get $$\int_0^{\pi / 2} \frac{d \theta}{\sqrt{1-k^2 \sin^2 \theta}} = \int_0^1 \frac{dt}{\sqrt{(1-k^2 t^2)(1-t^2)}}$$ Then $t = \sqrt{u}, dt = \frac{du}{2 \sqrt{u}}$ $$\int_0^1 \frac{dt}{\sqrt{(1-k^2 t^2)(1-t^2)}}=\int_0^1 \frac{du}{2\sqrt{u(1-k^2 u)(1-u)}}$$ With the correct change of variable you should be able to reduce $(x-e_1)(x-e_2)(x-e^3)$ to $u(1-k^2 u)(1-u)$.

Also take a look at "the Pentagonal Number Theorem and Modular Forms" for a larger picture.

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  • $\begingroup$ Thanks, but what I wanted was the change in coordinates needed to go from the integral of the cubic to the trigonometric integral. $\endgroup$ – Ethan Alwaise Oct 22 '16 at 1:46
  • $\begingroup$ @EthanAlwaise $u = \frac{x-e_1}{e_3-e_1}$ so that $(e_3-e_1)^3 u (1-k^2 u)(1-u) = (x-e_1)(x-e_3)(e_3-e_1+k^2e_1-k^2 x)$ $\endgroup$ – reuns Oct 22 '16 at 1:58
  • $\begingroup$ I'm not seeing how that helps me transform $(x - e_1)(x - e_2)(x - e_3)$. $\endgroup$ – Ethan Alwaise Oct 22 '16 at 5:49
  • $\begingroup$ @EthanAlwaise come on.. $e_2 = e_3+(k^2-1)e_1$, $k = \sqrt{(e_2+e_1-e_3)/e_1}$ $\endgroup$ – reuns Oct 22 '16 at 6:03
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$K(k)=K(-k)$, $$K(1)=\int_{0}^{\pi/2}\frac{1}{\sqrt{1-\sin^2\theta}}d\theta=\int_{0}^{\pi/2}\frac{1}{\cos\theta}d\theta=\ln|\tan\theta+\sec\theta| \;\bigg|_{0}^{\pi/2}=\infty.$$

if $k>1 \Rightarrow\;k^2>1^2,\;-k^2<-1^2,\;\sqrt{1-k^2\sin^2\theta}<\sqrt{1-1^2\sin^2\theta},$ $$K(k)=\int_{0}^{\pi/2}\frac{1}{\sqrt{1-k^2\sin^2\theta}}d\theta>\int_{0}^{\pi/2}\frac{1}{\sqrt{1-\sin^2\theta}}d\theta=K(1)=\infty.$$

So $$ K(k) = \begin{cases} \infty, & |k|\geq1 \\ \int_{0}^{\pi/2}\frac{1}{\sqrt{1-k^2\sin^2\theta}}d\theta, & |k|<1\end{cases}$$

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