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In a paper I have the following statement:

The integral $$\phi_{\infty}= \frac{1}{\sqrt{2M}}\int_{0}^{\frac{1}{P}}\frac{dt}{\sqrt{G(t)}}$$ with $$G(t)= t^3 - \frac{1}{2M}t^2 + \frac{P-2M}{2MP^3}$$ can be transformed into a classsical Jacobian elliptic integral $$\phi_{\infty}= 2\left(\frac{P}{Q}\right)^{\frac{1}{2}}\int_{\zeta_{\infty}}^{\frac{\pi}{2}}\frac{dt}{\sqrt{1-k^2\sin^2{t}}}$$

(the constants $M,P,Q$ and $k=k(M,P,Q)$ are given, but are not relevant, I think, for the problem I'm facing).
Question: I tried hard to go from the first form to the second but did not find a way to achieve this (the term $t^2$ get's in the way).\ Can someone give any clue how to perform this transformation or conclude that this is not possible?

Bufo Viridis
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1 Answers1

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Given fixed but arbitrary $M>0\land P>0$, define the function $G:\mathbb{R}\rightarrow\mathbb{R}$ by the cubic polynomial

$$G{\left(t\right)}:=t^{3}-\frac{1}{2M}t^{2}+\frac{P-2M}{2MP^{3}},$$

and let $\mathcal{I}{\left(M,P\right)}$ denote the value of the following integral for all $(M,P)$ where it converges and is real-valued:

$$\mathcal{I}{\left(M,P\right)}:=\frac{1}{\sqrt{2M}}\int_{0}^{\frac{1}{P}}\mathrm{d}t\,\frac{1}{\sqrt{G{\left(t\right)}}}.$$

Factoring the cubic:

We require $G{(t)}$ be nonnegative over the interval of integration $\left[0,\frac{1}{P}\right]$. Examining the values of the function at the endpoints, we quickly find $G{\left(\frac{1}{P}\right)}=0$ and

$$G{\left(0\right)}=\frac{P-2M}{2MP^{3}}\ge0\implies P\ge 2M.$$

It's not difficult to see that if $P=2M$, we would then have $G{\left(t\right)}<0$ for all $t\in\left(0,\frac{1}{P}\right)$. Hence, we require the strict inequality $P>2M$.

Setting $\Delta:=\left(P-2M\right)\left(P+6M\right)>0$ for convenience, the cubic $G$ can be factored as follows:

$$\begin{align} G{\left(t\right)} &=t^{3}-\frac{1}{2M}t^{2}+\frac{P-2M}{2MP^{3}}\\ &=\frac{2MP^{3}t^{3}-P^{3}t^{2}+P-2M}{2MP^{3}}\\ &=\frac{\left(Pt-1\right)\left[2MP^{2}t^{2}+\left(2M-P\right)Pt+2M-P\right]}{2MP^{3}}\\ &=\frac{\left(Pt-1\right)\left[4^{2}M^{2}P^{2}t^{2}+2\left(2M-P\right)4MPt+8M\left(2M-P\right)\right]}{4^{2}M^{2}P^{3}}\\ &=\frac{\left(Pt-1\right)\left[\left(4MPt\right)^{2}+2\left(4MPt\right)\left(2M-P\right)+8M\left(2M-P\right)\right]}{4^{2}M^{2}P^{3}}\\ &=\frac{\left(Pt-1\right)\left[\left(4MPt+2M-P\right)^{2}-\left(P-2M\right)\left(P+6M\right)\right]}{4^{2}M^{2}P^{3}}\\ &=\frac{\left(1-Pt\right)\left[\left(P-2M\right)\left(P+6M\right)-\left(P-2M-4MPt\right)^{2}\right]}{4^{2}M^{2}P^{3}}\\ &=\frac{\left(1-Pt\right)\left[\Delta-\left(P-2M-4MPt\right)^{2}\right]}{4^{2}M^{2}P^{3}}\\ &=\frac{\left(1-Pt\right)\left[\sqrt{\Delta}+\left(P-2M-4MPt\right)\right]\left[\sqrt{\Delta}-\left(P-2M-4MPt\right)\right]}{4^{2}M^{2}P^{3}}\\ &=\frac{\left(1-Pt\right)\left[\left(P-2M\right)+\sqrt{\Delta}-4MPt\right]\left[4MPt-\left(P-2M\right)+\sqrt{\Delta}\right]}{4^{2}M^{2}P^{3}}\\ &=\left(\frac{P-2M+\sqrt{\Delta}}{4MP}-t\right)\left(\frac{1}{P}-t\right)\left(t-\frac{P-2M-\sqrt{\Delta}}{4MP}\right).\\ \end{align}$$

Assuming $P>3M$, it can be shown that

$$\frac{P-2M+\sqrt{\Delta}}{4MP}>\frac{1}{P}>0>\frac{P-2M-\sqrt{\Delta}}{4MP}.$$

This also guarantees that the integral $\mathcal{I}{\left(M,P\right)}$ is real-valued and convergent.

Reduction of the elliptic integral to Legendre form:

For $(a,b,c,z)\in\mathbb{R}^{4}$ such that $a>b>z>c$, define the function $\mathcal{E}{\left(a,b,c,z\right)}$ by the elliptic integral

$$\mathcal{E}{\left(a,b,c,z\right)}:=\int_{z}^{b}\mathrm{d}t\,\frac{1}{\sqrt{\left(a-t\right)\left(b-t\right)\left(t-c\right)}},$$

and set

$$\varphi:=\arcsin{\left(\sqrt{\frac{z-c}{b-c}}\right)}\in\left(0,\frac{\pi}{2}\right),$$

$$\kappa:=\sqrt{\frac{b-c}{a-c}}\in(0,1).$$

Consider the transformation given by the substitution $\sqrt{\frac{t-c}{b-c}}=x$. Observing that

$$\sqrt{\frac{t-c}{b-c}}=x\implies t=c+\left(b-c\right)x^{2}\implies \frac{dt}{dx}=2\left(b-c\right)x,$$

we find after applying the substitution that $\mathcal{E}$ can be rewritten as

$$\begin{align} \mathcal{E}{\left(a,b,c,z\right)} &=\int_{z}^{b}\mathrm{d}t\,\frac{1}{\sqrt{\left(a-t\right)\left(b-t\right)\left(t-c\right)}}\\ &=\int_{\sqrt{\frac{z-c}{b-c}}}^{1}\mathrm{d}x\,\frac{2\left(b-c\right)x}{\sqrt{\left[a-c-\left(b-c\right)x^{2}\right]\left[b-c-\left(b-c\right)x^{2}\right]\left(b-c\right)x^{2}}}\\ &=\frac{2}{\sqrt{a-c}}\int_{\sqrt{\frac{z-c}{b-c}}}^{1}\mathrm{d}x\,\frac{1}{\sqrt{\left(1-x^{2}\right)\left[1-\left(\frac{b-c}{a-c}\right)x^{2}\right]}}\\ &=\frac{2}{\sqrt{a-c}}\int_{\sin{\left(\varphi\right)}}^{1}\mathrm{d}x\,\frac{1}{\sqrt{\left(1-x^{2}\right)\left(1-\kappa^{2}x^{2}\right)}}\\ &=\frac{2}{\sqrt{a-c}}\int_{\varphi}^{\frac{\pi}{2}}\mathrm{d}\tau\,\frac{1}{\sqrt{1-\kappa^{2}\sin^{2}{\left(\tau\right)}}};~~~\small{\left[x=\sin{\left(\tau\right)}\right]}\\ &=\frac{2}{\sqrt{a-c}}\left[K{\left(\kappa\right)}-F{\left(\varphi,\kappa\right)}\right],\\ \end{align}$$

where here the incomplete and complete elliptic integrals of the first kind are defined with the following argument convention:

$$F{\left(\theta,k\right)}:=\int_{0}^{\theta}\mathrm{d}\tau\,\frac{1}{\sqrt{1-k^{2}\sin^{2}{\left(\tau\right)}}};~~~\small{\theta\in\mathbb{R}\land k\in(-1,1)},$$

$$K{\left(k\right)}:=F{\left(\frac{\pi}{2},k\right)};~~~\small{k\in(-1,1)}.$$

From here, obtaining a final expression for $\mathcal{I}{\left(M,P\right)}$ is simply a matter of plugging-and-chugging!

As a parting gift, I refer you to this amazing set of lecture notes in case you want to know more about systematically reducing elliptic integrals. Cheers! ;)

David H
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