# complete elliptic integral of the 1st kind with infinitely large argument

I am trying to evaluate this integral. $$\int_0^1 dz\frac{1}{\sqrt{(a_2-a_1z^2)(z^2-1)}} = \frac{K(\sqrt{a_1/a_2})}{\sqrt{-a_2}}$$ I know when $a_2$ is not $0$. It is a normal complete elliptic integral of the 1st kind. However, the problem requires to evaluate it when $a_2 = 0$. It seems to diverge in that case. Will it converge when $a_2 = 0$ and how to evaluate it if it is convergent?

Thanks

• I don't know anything about evaluating elliptic integrals, but when $a_2=0$ the associated curve is $y^2 = -a_1 z^2(z^2-1)$, which is singular. – André 3000 Mar 9 '17 at 23:18

## 1 Answer

All right, maybe I do know a something about evaluating this integral. As I commented, the curve $C: y^2 = a_1 x^2(1-x^2)$ is singular, and in fact, you can show that its genus is $0$. This means that it is birational to a line, so we can find a parametrization of it. Using Magma, I found the parametrization $$(x(t),y(t)) = \left(\frac{t^{4} - a_{1}^{2}}{{\left(t^{2} + a_{1}\right)}^{2}}, \frac{2 a_{1} t {\left(a_1 - t^2 \right)}}{{\left(t^{2} + a_{1}\right)}^{2}} \right) \, .$$ Then $$x'(t) = \frac{4 \, t^{3}}{{\left(t^{2} + a_{1}\right)}^{2}} - \frac{4 \, {\left(t^{4} - a_{1}^{2}\right)} t}{{\left(t^{2} + a_{1}\right)}^{3}}$$ and substituting these expressions into the integral yields \begin{align*} \int_0^1 \frac{1}{\sqrt{(-a_1 x^2)(x^2-1)}} \, dx = -2 \int_{\sqrt{a_1}}^\infty \frac{t^2+a_1}{(t^2 - a_1)^2} \, dt. \end{align*} We compute the antiderivative $$-2 \int \frac{t^2+a_1}{(t^2 - a_1)^2} \, dt = \frac{2t}{t^2-a_1} \, .$$ which tends to $\infty$ as $t \to \sqrt{a_1}$ from the right.

• Does the derivation prove this integral is singular when a2 = 0? – Owls Mar 13 '17 at 19:07
• Yes, I think it shows that the integral diverges when $a_2 = 0$. The fancy geometry terms aside, I just made a change of variable. – André 3000 Mar 13 '17 at 21:32