# Relation between full elliptic integrals of the first and third kind

I am working on a calculation involving the Ronkin function of a hyperplane in 3-space. I get a horrible matrix with full elliptic integrals as entries. A priori I know that the matrix is symmetrical and that give me a relation between full elliptic integrals of the first and third kind. I can not find transformations in the literature that explain the relation and I think I need one in order to simplify my matrix.

## The relation

With the notation

$\operatorname{K}(k) = \int_0^{\frac{\pi}{2}}\frac{d\varphi}{\sqrt{1-k^2\sin^2\varphi}},$ $\qquad$ $\Pi(\alpha^2,k)=\int_0^{\frac{\pi}{2}}\frac{d\varphi}{(1-\alpha^2\sin^2\varphi)\sqrt{1-k^2\sin^2\varphi}}$

$k^2 = \frac{(1+a+b-c)(1+a-b+c)(1-a+b+c)(-1+a+b+c)}{16abc},\quad a,b,c > 0$

the following is true:

$2\frac{(1+a+b-c)(1-a-b+c)(a-b)}{(a-c)(b-c)}\operatorname{K}(k)+$

$(1-a-b+c)(1+a-b-c)\Pi\left( \frac{(1+a-b+c)(-1+a+b+c)}{4ac},k\right) +$

$\frac{(a+c)(1+b)(1-a-b+c)(-1-a+b+c)}{(a-c)(-1+b)}\Pi\left( \frac{(1+a-b+c)(-1+a+b+c)(a-c)^2}{4ac(1-b)^2},k\right)+$

$(1-a-b+c)(-1+a+b+c)\Pi\left( \frac{(1-a+b+c)(-1+a+b+c)}{4ac},k\right)+$

$\frac{(1+a)(b+c)(-1-a+b+c)(-1+a+b-c)}{(1-a)(c-b)}\Pi\left( \frac{(1-a+b+c)(-1+a+b+c)(b-c)^2}{4ac(1-a)^2},k\right)$

$==0$.

Is there some addition formula or transformation between elliptic integrals of the first and third kind that will explain this?

Are you sure that $a$, $b$, and $c$ have no relations among them?
I ask this because I attempted to (painstakingly!) copy your proposed equation over to Mathematica (taking care of the fact that the functions in it take the modulus $m=k^2$ as argument), and I keep getting complex results instead of 0.