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](http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1063041/pdf/pnas01544-0031.pdf)). 


Examples for these so called **elliptic integral singular values** are given on [this Wolfram page](http://mathworld.wolfram.com/EllipticIntegralSingularValue.html) (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](http://carma.newcastle.edu.au/jon/Preprints/Books/EMA/Exercises/For%20others/K-beta.pdf). 


See also what Tito Piezas has to say about this in his pleasant-to-read [Collection of Algebraic Identities](https://www.sites.google.com/site/tpiezas/0026).


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](http://en.wikipedia.org/wiki/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 $K(m)$?**



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


