# Can elliptic integral singular values generate cubic polynomials with integer coefficients?

For the elliptic integral of first kind, $K(m)=\int_0^{\pi/2}\frac{d\theta}{\sqrt{1-m^2sin^2\theta}}$, it is well-known that $K(m)$ can be expressed in what Chowla and Selberg call "finite terms" (i.e. algebraic numbers and a finite product of Gamma functions of rational values) whenever $i\frac{K(\sqrt{1-m^2})}{K(m)}$ belongs to an imaginary quadratic field $\mathbb Q(\sqrt{d})$ (see Theorem 7 in S. Chowla and A. Selberg, On Epstein's zeta function, J. reine angew. Math. 227, 86-110, 196).

Examples for these so called elliptic integral singular values are given on this Wolfram page (with some small typos) and in the note of J.M. Borwein and I.J. Zucker, "Elliptic integral evaluation of the Gamma function at rational values of small denominator," IMA Journal on Numerical Analysis, 12 (1992), 519- 526.

The following question arises:

• For these singular values, is there (always, or, if not always: when?) a polynomial $P(t)$ of degree 3 with integer coefficients such that $K(m)=c\int\limits_{t_0}^\infty\dfrac{dt}{\sqrt{P(t)}}$ with $c\in\mathbb Q$?
(EDIT: After Noam Elkies' remark, introduced $t_0$, the biggest real zero of $P$, instead of $0$ as the lower limit. Only "complete" integrals make sense here.)

In particular for $d=-7$, we have by the Carlson symmetric form $$\frac12\int\limits_0^\infty\dfrac{dt}{\sqrt{t(t+1)(t+\frac{8+3\sqrt 7}{16})}} =K(k_7)=\dfrac1{7^{1/4}4\pi}\Gamma\left(\dfrac17\right)\Gamma\left(\dfrac27\right)\Gamma\left(\dfrac47\right),$$ on the other hand I have seen somewhere (I can't remember the reference) $$\int\limits_0^\infty\dfrac{dt}{\sqrt{t(t^2+21t+112)}} =\dfrac1{4\pi\sqrt{7}}\Gamma\left(\dfrac17\right)\Gamma\left(\dfrac27\right)\Gamma\left(\dfrac47\right)=\frac{K(k_7)}{7^{1/4}}.$$

I would like to get it straight at least for this example:

• Can the polynomial in the first integral be transformed into one with integer coefficients? And is there any sort of relationship between both above polynomials?
Note that the ratio of the discriminants of the two above polynomials is $-2^{24}\cdot7^3$, and both of them do not yield affirmative answers, as the second one would have to be divided by $\sqrt7$ to obtain $K(k_7)$ directly!

• EDIT: Follow-up question: If $P(t)$ is an integer cubic polynomial such that $\int\limits_{t_0}^\infty\dfrac{dt}{\sqrt{P(t)}}$ (with $t_0$ its biggest real zero) can be written in "finite terms", is this value always an algebraic multiple of an elliptic integral singular value $K(m)$?

• Is this really about elliptic curves with complex multiplication? If so, then the answer is going to be yes if and only if the class number of $\mathbb Q(\sqrt{d})=1$. – Will Sawin Feb 4 '12 at 22:42
• I didn't necessarily think of that restriction. So if we restrict them to that, which polynomial(s) would that be for the $d=-7$ case? And how to find them? Unique? – Wolfgang Feb 5 '12 at 0:03
• If I understand what is going on correctly, any polynomial with the same $j$-invariant should do: en.wikipedia.org/wiki/J-invariant#Algebraic_definition – Will Sawin Feb 5 '12 at 4:27
• spanferkel's demand that the integral be over $(0,\infty)$ means that the curve must also have a rational 2-torsion point, which cuts the usual list of 13 CM discriminants down to seven: $$-3, -4, -7, -8, -12, -16, -28.$$ For example, the disc.-11 curve yields the complete elliptic integral of $dt / \sqrt{4t^3-4t^2-28t+41}$ which is a rational multiple of $$\pi^{-2} \phantom.\Gamma(1/11)\phantom.\Gamma(3/11)\phantom.\Gamma(4/11) \Gamma(5/11)\phantom.\Gamma(9/11),$$ but the cubic is irreducible so the integral cannot be written as $\int_0^\infty dt/\sqrt P(t)$. – Noam D. Elkies Feb 5 '12 at 7:12
• Thank you very much. I have edited the lower limit, as it is exactly examples like yours I'm interested in. I suppose the curve is uniquely determined by $d$, up to translation? So again the question: What about $n=-7$? – Wolfgang Feb 5 '12 at 13:16

The flurry of comments did not yet produce an answer to the question concerning the complete elliptic integrals $$I_1 := \frac12 \int\limits_0^\infty\dfrac{dt}{\sqrt{t(t+1)(t+\frac{8+3\sqrt{7}}{16})}}$$ and $$I_2 := \int\limits_0^\infty\dfrac{dt}{\sqrt{t(t^2+21t+112)}}.$$ It turns out that (i) Yes, $I_1$ can be transformed to a complete elliptic integral associated to a cubic with integer coefficients, and (ii) The identity $I_1 = 7^{1/4} I_2$ can then be recovered via a form of Landen's transformation.
This could be surmised by calculating the $j$-invariants of the corresponding elliptic curves $$y^2 = x(x+1)(x+\frac{8+3\sqrt{7}}{16}) \phantom{\cong}\text{and}\phantom{\cong} y^2 = x(x^2+21x+112).$$
The first $j$-invariant is $j_1 = 16581375 = 255^3$, which is rational, so there's a linear change of variable that transforms $x(x+1)(x+\frac{8+3\sqrt{7}}{16})$ to a polynomial with integer coefficients. The second $j$-invariant is $j_2 = -3375 = -15^3 \neq j_1$, so we can't get immediately from $j_1$ to $j_2$. But $j_1$ and $j_2$ are still related by a $2$-isogeny [indeed $j_1 = j(\sqrt{-7})$ and $j_2 = j((1+\sqrt{-7})/2)$], so $I_1$ and $I_2$ are related by a Landen transformation.
For (i), first translate $t$ by $(8+3\sqrt{7})/16$ to get $$I_1 := \frac12 \int\limits_{\frac{8+3\sqrt{7}}{16}}^\infty \dfrac{dt}{\sqrt{t \bigl(t-\frac{8+3\sqrt{7}}{16}\bigr) \bigl(t-\frac{-8+3\sqrt{7}}{16}\bigr) }} \phantom{0}.$$ Then observe that $$\bigl(t-\frac{8+3\sqrt{7}}{16}\bigr) \bigl(t-\frac{-8+3\sqrt{7}}{16}\bigr) = t^2 - \frac{3\sqrt{7}}{8} + \frac{3^2 7 - 8^2}{16^2} = t^2 - \frac{6\sqrt{7}}{16} - \frac1{16^2}.$$ Thus the change of variable $x = 16 \sqrt{7} \cdot t$ yields $$I_1 = 2 \cdot 7^{1/4} \int\limits_{21+8\sqrt{7}}^\infty \dfrac{dx}{\sqrt{x (x^2-42x-7)}}$$ with the integrand having integer coefficients.
For (ii), we apply Landen's change of variable for elliptic integrals $\int dt/\sqrt{t(t^2+at+b)}$ with $a,b>0$: if $x = (t^2+at+b)/t$ then $dt/\sqrt{t(t^2+at+b)} = dx/\sqrt{x(x^2+Ax+B)}$ where $(A,B) = (-2a, a^2-4b)$. Moreover the map $t \mapsto (t^2+at+b)/t$ maps the interval $0 < t < \infty$ to $a + 2\sqrt{b} \leq x < \infty$, with each $x$ value arising twice except for the left endpoint $x = a+2\sqrt{b}$ which has a single preimage $t = \sqrt{b}$ of multiplicity $2$. Therefore $$\int\limits_0^\infty\dfrac{dt}{\sqrt{t(t^2+at+b)}} = 2 \int\limits_{a+2\sqrt{b}}^\infty\dfrac{dx}{\sqrt{x(x^2-AX-B)}}.$$ The $I_2$ coefficients $(a,b) = (21,112)$ yield $(A,B) = (-42,-7)$ and $a + 2 \sqrt{b} = 21 + 8 \sqrt{7}$. Therefore $I_1 = 7^{1/4} I_2$ as desired.