# Integral of inverse of square root of quartic function with real roots

I was doing a physics problem and in order to finish it, I need to prove that:

$$\int_{x1}^{x2}\frac{dx}{{((x1 - x)(x - x2)(x - x3)(x - x4))}^{1/2}} = \int_{x3}^{x4}\frac{dx}{{((x1 - x)(x - x2)(x - x3)(x - x4))}^{1/2}}$$

Where all the roots of the polynomial are real and $x1 < x2 < x3 < x4.$

I don't know the range of validity of my claim, however I do have tested this using mathematica and it was true for all the cases. Also, I know that I can change the integrand using some transformations to get an elliptic integral of first kind. However, I think it will be a lot of work and will not pay the price (maybe I'm wrong), so I'm looking to a more elegant and direct proof.

• Is it true that $x_4-x_3=x_2-x_1$? In such a case, you can simply exploit the fact that the graphics of the integrand function is symmetric with respect to the $x=\frac{x_2+x_3}{2}$ line. – Jack D'Aurizio Sep 1 '14 at 18:51
• No, this is not always true. I've checked using different values of x1,x2,x3 and x4 and my statement seems to be always true (at least up to 5 decimal places lol). – Stephen Dedalus Sep 1 '14 at 18:56
• Yes, I fixed that. My expression is less symmetrical now. :( But I think is better doing this way instead of talking about branches and stuff. – Stephen Dedalus Sep 1 '14 at 19:09
• Also, I really appreciate the effort to make this question more precise. As I've said, this was just an integral that appearead after some mathematical manipulation. – Stephen Dedalus Sep 1 '14 at 19:13
• @JackD'Aurizio, there is some evidence supporting this, see my answer – Will Jagy Sep 1 '14 at 20:11

## 3 Answers

EVEN MORE: given $0,1 < C < D$ we get an interchange from the Moebius transformation $$m(z) = \frac{-Cz \; + \; CD}{(D-C-1) \; z \; + \; C}$$ Here $$m(0) = D, m(D) = 0, m(1) = C, m(C) = 1.$$ Note that if $D-C = 1,$ we have a linear map and Jack's observation applies directly, the two intervals would then be the same length.

Alright, that actually worked, proof by Moebius transformation. Take $$W = D - C - 1,$$ which could be positive, negative, or zero, we don't know. With $$x = \frac{-Cz \; + \; CD}{W \; z \; + \; C},$$ $$dx = \frac{-C (C + DW)}{ (Wz+C)^2} dz.$$ But $C + DW = (D-C)(D-1),$ so $$dx = \frac{-C (D-C)(D-1)}{ (Wz+C)^2} dz.$$ Also use $W+C = D-1, W+1=D-C.$

We need four items, $$(Wz+C) x = C(D-z),$$ $$(Wz+C) (1- x) = (W+C) z + (C-CD) = (D-1)z + C (1 -D) = (D-1)(z-C),$$ $$(Wz+C)(C-x) = C(D-C)(z-1),$$ $$(Wz+C)(D-x) = (D-C)(D-1)z.$$

Together $$\color{magenta}{ -C (D-C)(D-1) \int_D^C \frac{(Wz+C)^2}{\sqrt{C^2 (D-1)^2(D-C)^2 z (z-1)(z-C)(D-z)}} \frac{dz}{(Wz+C)^2}}$$

$$\color{magenta}{ C (D-C)(D-1) \int_C^D \frac{dz}{ C (D-C)(D-1)\sqrt{ z (z-1)(z-C)(D-z)}} }$$

$$\color{magenta}{ \int_C^D \frac{dz}{\sqrt{ z (z-1)(z-C)(D-z)}} }$$

EXTRA comment: from @Jack's observation, it works for $0,1,C, C+1.$ A fair amount of work to get right, you can then find the derivative by $D$ of the two integrals with roots $0,1,C,D.$ The first one is not so hard, differentiate under the integral sign, the second has the extra problem that $D$ is both a limit and in the integrand, might be better to transform first so that it is no longer one of the integration endpoints; need to think about the extent to which that can be done.

long for a comment; I think you are onto something. $$\int_0^1 \frac{1}{\sqrt{x - x^2}} = \pi,$$ antiderivative is $\; \; \; - \arcsin (1-2x), \; \;$ so $$\int_A^B \frac{1}{\sqrt{(x-A)(B-x)}} = \pi.$$

I found $-1,0 < C< D$ a little more convenient, pull out $(C-x)(D-x)$ and bound without integrating those,... $$\frac{\pi}{\sqrt{(C+1)(D+1)}} \leq \int_{-1}^0 \frac{dx}{\sqrt{-(x+1)x(C-x)(D-x)}} \leq \frac{\pi}{\sqrt{CD}}$$ while $$\frac{\pi}{\sqrt{(C+1)(D+1)}} \leq \int_C^D \frac{dx}{\sqrt{(x+1)x(x-C)(D-x)}} \leq \frac{\pi}{\sqrt{CD}}$$

I do not yet have the explicit transformation, if indeed your two integrals are always equal, but I suspect they are. This is too much of a coincidence otherwise. Here my idea is $C,D$ large, maybe close together, maybe far apart, maybe very very far apart. If you are worried about $C,D$ very small, then just take a linear change so that those are $-1,0$ instead

• There is a point $z$ in the $(x_2,x_3)$ interval such that $\frac{z-x_2}{x_3-z}=\frac{x_2-x_1}{x_4-x_3}$ and a point $w$ in the same interval such that $\frac{w-x_2}{x_3-w}=\frac{x_4-x_3}{x_2-x_1}$. I bet that by by considering $z$ or $w$ as the origin we can exhibit an explicit transformation between the two integrals (that are complete elliptic integrals of the second kind). – Jack D'Aurizio Sep 1 '14 at 20:11
• Yes, probably it works. We just need to switch $x_1,x_2$ and $x_4,x_3$ with a birational map. – Jack D'Aurizio Sep 1 '14 at 20:19
• We must take $u\in(x_2,x_3)$ such that $\frac{u-x_1}{x_4-u}=\frac{x_3-u}{u-x_2}$. – Jack D'Aurizio Sep 1 '14 at 20:21
• @StephenDedalus, added to answer. Loved your work in Ulysses... note that for proof of equality, it suffices to take $A=0, B=1.$ You called them $x_1,x_2,x_3,x_4$ I guess. – Will Jagy Sep 1 '14 at 20:36
• @JackD'Aurizio, it is a little different, I was never quite sure what you wanted with the special value for $u,$ so that is yours in any case. Strange problem, I would have ignored it if the extreme behavior gave an obvious inequality, quite surprised when the estimates were identical. – Will Jagy Sep 1 '14 at 21:35

We can assume that $x_1=0,x_2=1,x_3=C,x_4=D$ with $D>C>1$.

There is point $z$ for which: $$(z-x_2)(x_3-z)=(z-x_1)(x_4-z),\tag{1}$$ and by solving $(1)$ we get: $$z = \frac{C}{1+C-D}\tag{2}$$ (if $D-C=1$, the observation I made in the comments settles the question) from which we have: $$z-x_2=\frac{D-1}{1+C-D},\quad x_3-z =\frac{C(C-D)}{1+C-D},\quad x_4-z=\frac{(C-D)(D-1)}{1+C-D}.\tag{3}$$ Now the first integral can be written as: $$I_L = \int_{-\frac{C}{1+C-D}}^{-\frac{D-1}{1+C-D}}\frac{du}{\sqrt{-(u+\frac{C}{1+C-D})(u+\frac{D-1}{C+D-1})(u-\frac{C(C-D)}{C+D-1})(u-\frac{(D-1)(C-D)}{1+C-D})}}\tag{4}$$ while the second integral can be written as: $$I_R = \int_{\frac{C(C-D)}{C+D-1}}^{\frac{(D-1)(C-D)}{1+C-D}}\frac{dv}{\sqrt{-(v+\frac{C}{1+C-D})(v+\frac{D-1}{C+D-1})(v-\frac{C(C-D)}{C+D-1})(v-\frac{(D-1)(C-D)}{1+C-D})}}\tag{5}$$ and the equivalence between $(4)$ and $(5)$ follows from the substitution $$v = -\frac{C(C-D)(D-1)}{(1+C-D)^2 u}.\tag{6}$$

So, if we perform a translation first, we just need to apply an inversion to map the first integral into the second one, and calculations become a little easier to perform. The content of this answer, however, is more or less the same as Will Jagy's answer: Moebius map works.

With the help of Mathematica, I also got a closed form expression:

$$I_L = I_R = \frac{2}{\sqrt{CD-C}}\cdot K\left(\frac{D-C}{CD-C}\right),\tag{7}$$

where $K(\cdot)$ it the complete elliptic integral of the first kind (with the Mathematica convention): $$K(k)=\int_{0}^{1}\frac{dt}{\sqrt{(1-t^2)(1-kt^2)}}=\frac{\pi}{2}\sum_{n=0}^{+\infty}\left(\frac{1}{4^n}\binom{2n}{n}\right)^2 k^n.$$ Notice that the argument of the complete elliptic integral in $(7)$ is deeply related with the cross-ratio of $x_1,x_2,x_3,x_4$.

• Great minds think alike. – Will Jagy Sep 1 '14 at 21:31

The projective curve $X$ given by $Y^2Z^2=(X-x_1Z)(X-x_2Z)(X-x_3Z)(X-x_4Z)$ is smooth and has genus $1$. Let $\omega = dx/y$. Then $\mathbb C\cdot\omega = \Omega^1(X/\mathbb C)$. Let $\Lambda \subseteq \mathbb C$ be the period lattice of $\omega$. Then $\Lambda$ is a free $\mathbb Z$-module of rank $2$. Since $X$ and $\omega$ are defined over $\mathbb R$, $\Lambda = (i\mathbb R \cap \Lambda) \oplus (\mathbb R \cap \Lambda)$. Let $\sigma \in \mathbb R \cap \Lambda$ be the smallest positive real period. I claim that both of your integrals are equal to $\sigma/2$.

Consider in $H_1(X(\mathbb C), \mathbb Z)$ the homology classes of the cycles $\gamma_1$ and $\gamma_2$ given by:

$\gamma_1$ is the cycle which goes from $x_1$ to $x_2$ along the first sheet and comes back to $x_1$ through the second sheet;

$\gamma_2$ is the cycle which goes from $x_3$ to $x_4$ along the first sheet and comes back to $x_3$ through the second sheet.

(Here I am viewing $X$ as a double-sheeted covering of $\mathbb P^1$ given by $(x,y) \mapsto x$.)

It is not hard to see that the classes of $\gamma_1$ and $\gamma_2$ are equal (draw a picture of $X$ lying over $\mathbb P^1$). Hence $\int_{\gamma_1} \omega = \int_{\gamma_2}\omega = \sigma$, and your integral is half of that.

Alternatively, make $X$ into an elliptic curve by choosing the point $P_1=[X_1 : 0 : 1]$ as a base point. Then the points $P_i = [X_i : 0 : 1]$ $(1 \leq i \leq 4)$ are the $2$-torsion points on $X$. Let $f: X \to X$ be $P \mapsto P+P_3$. The differential $\omega$ is translation-invariant, hence

$$\int_{P_1}^{P_2} \omega = \int_{P_1}^{P_2} f^*\omega = \int_{P_1+P_3}^{P_2+P_3} \omega = \int_{P_3}^{P_4}\omega.$$