Reduction of elliptic integrals to Legendre's normal form

Question

Let $P(x)$ be a real polynomial of degree four without multiple roots and suppose that all roots are real. According to my textbook (Königsberger, Analysis 1, first printing, p. 208), there is a rational transformation $x=T(t)=\frac{at+b}{ct+d}$ such that $$ \frac{dx}{\sqrt{P(x)}}=\frac{1}{\sqrt{P(T(t))}}\cdot\frac{ad-bc}{(ct+d)^2}dt=\text{const.}\cdot\frac{dt}{\sqrt{Q(t)}} $$ where $Q(t)=(1-t^2)(1-k^2t^2)$ and $k=\frac{x_1-x_2}{x_1-x_4}\colon\!\frac{x_3-x_2}{x_3-x_4}$ with $x_1<x_2<x_3<x_4$ - the real roots in order.

By making the Ansatz $P(T(t))(ct+d)^4=Q(t)$ I was able to determined $a,b,c,d$, namely $a=\frac{2x_1}{x_1-x_2}-1$, $b=1$, $c=\frac{2}{x_1-x_2}$, $d=0$, so that $P(T(t))(ct+d)^4$ becomes $$ (1-t^2)(1+(2(x_1-x_3)/(x_1-x_2)-1)t))(1+(2(x_1-x_4)/(x_1-x_2)-1)t) $$ but the last two factors in the last displayed equation do not summarize nicely as $(1-k^2t^2)$ and I suspect that I am doing something wrong although I get something close to the desired Doppelverhältnis expression for $k$.

What is the right fractional linear transformation in terms of $a,b,c,d\in\mathbb{R}$ and especially what is the constant $\text{const.}$ in the first displayed equation? Is it $ad-bc$ as I expected?

Any help is appreciated!

---Edit---

You can write $P(x)=a_0(x-x_1)(x-x_2)(x-x_3)(x-x_4)$, then, because $\sqrt{a_0}$ merges with $\text{const.}$, we get $$ P(T(t))(ct+d)^4=(at+b-(ct+d)x_1)(at+b-(ct+d)x_2)(at+b-(ct+d)x_3)(at+b-(ct+d)x_4)\,. $$ By comparing coefficients and still changing the constant, we get a system of equations as follows: $$ a-cx_1=-q\,,\,a-cx_2=p\,,\,b-dx_1=q\,,\,b-x_2d=p\,,\,a-cx_3=-kr\,,a-cx_4=ks,b-x_3d=r\,,\,b-x_4d=s $$ with yet to be determined constants $p,q,r,s$ and where $k$ is the Doppelverhältnis of $x_1,x_2,x_3,x_4$ in a, yet to be determined, order. Solving these euqations should be possible and $k$ should be as desired. I tried but I did not come to an end. The values of $a,b,c,d$ above are obviously wrong and arise from a special choice of constants $p,q,r,s$ which is too restrictive.

Answer 1

Assume that

$$P(x) = \pm (x - x_1) (x - x_2) (x - x_3) (x - x_4)$$

with the order of the roots as above. Consider the substitution $x=T(t^2)$ with the transformation $T$ as defined above.

Note that $T'(t) = \frac{ad-bc}{(ct+d)^2}$ and the inverse transformation is $T^{-1}(x)=-\frac{b-dx}{a-cx}$. We want to have

$$\frac{1}{\sqrt{P(x)}}\,dx = \frac{2t}{\sqrt{P(T(t^2))}}T'(t^2)\,dt = \text{const.} \cdot \frac{1}{\sqrt{Q(t)}}\,dt,$$

that is,

$$\pm \Big( \frac{at^2+b}{ct^2+d} - x_1 \Big) \Big( \frac{at^2+b}{ct^2+d} - x_2 \Big) \Big( \frac{at^2+b}{ct^2+d} - x_3 \Big) \Big( \frac{at^2+b}{ct^2+d} - x_4 \Big) \frac{(ct^2+d)^4}{4(ad-bc)^2t^2} = \mu \cdot Q(t)$$

with some positive constant $\mu$. This leads to

$$\pm \frac{1}{t^2} ((a-cx_1)t^2 + b-dx_1) \cdot ((a-cx_2)t^2 + b-dx_2) \cdot ((a-cx_3)t^2 + b-dx_3) \cdot ((a-cx_4)t^2 + b-dx_4) = 4(ad-bc)^2\mu \cdot Q(t).$$

By expanding, we see that the left-hand side contains the powers $t^6,t^4,t^2,t^0$, and $t^{-2}$. By comparing with the powers in $Q$, we see that the coefficients in front of $t^6$ and $t^{-2}$ have to vanish. To achieve this, we need $a=cx_i$ for some $i$ and $b=dx_j$ for some $j \neq i$.

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Let $(y_1,y_2,y_3,y_4)$ be an arbitrary permutation of $(x_1,x_2,x_3,x_4)$ and assume that $a=cy_1$ and $b=dy_2$. Then

$$\pm (b-dy_1) \cdot (a-cy_2) \cdot ((a-cy_3)t^2 + b-dy_3) \cdot ((a-cy_4)t^2 + b-dy_4) = 4(ad-bc)^2\mu \cdot Q(t).$$

By comparing coefficients, we get

$$\begin{align*} \pm (b-dy_1) \cdot (a-cy_2) \cdot (a-cy_3) \cdot (a-cy_4) &= 4(ad-bc)^2\mu \cdot k^2, \\ \pm (b-dy_1) \cdot (a-cy_2) \cdot ((a-cy_3)(b-dy_4) + (b-dy_3)(a-cy_4)) &= 4(ad-bc)^2\mu \cdot (-1-k^2), \\ \pm (b-dy_1) \cdot (a-cy_2) \cdot (b-dy_3) \cdot (b-dy_4) &= 4(ad-bc)^2\mu. \end{align*}$$

Since $ad-bc=cd(y_1-y_2)$, the third equation implies $\mu = \mp \frac{d}{4c}(y_2-y_3)(y_2-y_4)$.

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Let $S(y) = \frac{a-cy}{b-dy}$ and $s_i = S(y_i)$. By dividing the first and the third equation, we obtain $s_3 s_4 = k^2$. By dividing the second and the third equation, we obtain $s_3 + s_4 = -1 - k^2$. So $s_3$ and $s_4$ solve the following equation in $s$:

$$s + \frac{k^2}{s} = - 1 - k^2.$$

The solutions of this equation are $-1$ and $-k^2$. Assume that $s_3 = -1$ and $s_4 = -k^2$. Then

$$\begin{align*} c(y_1-y_3) &= -d(y_2-y_3), \\ c(y_1-y_4) &= -k^2 d(y_2-y_4) \end{align*}$$

From the first equation we deduce $-\frac{d}{c}=\frac{y_1-y_3}{y_2-y_3}$, so

$$T^{-1}(x) = -\frac{d}{c}\frac{y_2-x}{y_1-x} = \frac{y_1-y_3}{y_2-y_3}\frac{y_2-x}{y_1-x}, \quad k^2 = \frac{y_2-y_3}{y_1-y_3}\frac{y_1-y_4}{y_2-y_4} =: (y_1,y_2;y_3,y_4), \quad \mu = \pm \frac{(y_1-y_3)(y_2-y_4)}{4}.$$

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Now we consider domains.

• If the leading coefficient $c$ of $P$ is $+1$, the domains where $\sqrt{P}$ is defined are $(-\infty,x_1]$, $[x_2,x_3]$, and $[x_4,\infty)$.
• If the leading coefficient $c$ of $P$ is $-1$, the domains where $\sqrt{P}$ is defined are $[x_1,x_2]$ and $[x_3,x_4]$.

Depending on the domain, we have to choose a suitable permutation of $(x_1,x_2,x_3,x_4)$ such that

• the transformation $t=\sqrt{T^{-1}(x)}$ is defined on the domain,
• the cross-ratio $k^2$ is in $(0,1)$,
• $\mu$ is a positive number.

$c$	domain of $x$	permutation	transformation $t$	$t$ maps to	cross-ratio $k^2$
+1	$[x_2,x_3]$	$x_1,x_2,x_3,x_4$	$\sqrt{\frac{x_1-x_3}{x_2-x_3}\frac{x_2-x}{x_1-x}}$	$[0,1]$	$(x_1,x_2;x_3,x_4)$
+1	$(-\infty,x_1]$	$x_3,x_4,x_1,x_2$	$\sqrt{\frac{x_3-x_1}{x_4-x_1}\frac{x_4-x}{x_3-x}}$	$(\sqrt{\frac{x_3-x_1}{x_4-x_1}},1]$	$(x_3,x_4;x_1,x_2)$
+1	$[x_4,\infty)$	$x_3,x_4,x_1,x_2$	$\sqrt{\frac{x_3-x_1}{x_4-x_1}\frac{x_4-x}{x_3-x}}$	$[0,\sqrt{\frac{x_3-x_1}{x_4-x_1}})$	$(x_3,x_4;x_1,x_2)$
-1	$[x_1,x_2]$	$x_4,x_1,x_2,x_3$	$\sqrt{\frac{x_4-x_2}{x_1-x_2}\frac{x_1-x}{x_4-x}}$	$[0,1]$	$(x_4,x_1;x_2,x_3)$
-1	$[x_3,x_4]$	$x_2,x_3,x_4,x_1$	$\sqrt{\frac{x_2-x_4}{x_3-x_4}\frac{x_3-x}{x_2-x}}$	$[0,1]$	$(x_2,x_3;x_4,x_1)$

Does this make sense? :)