0
$\begingroup$

I've run into the following integral:

$\int \frac{K(k)}{k} dk$

where $K$ is the complete elliptic integral of the first kind

$K(k) = \int_0^{\pi/2} \frac{d\theta}{\sqrt{1-k^2 \sin\theta}}$.

I've looked in Byrd, Friedman, "Handbook of Elliptic Integrals...", and found that $\int K/k dk = \int E/k dk - E$ (where $E$ is the complete elliptic integral of the second kind). There are some other similar formulas, too, such as $\int K/k^2 dk = -E/k$.

This leads me to suspect that there is no "nice" formula for $\int \frac{K(k)}{k} dk$. Is there a sense in which I can make this precise?

$\endgroup$

1 Answer 1

1
$\begingroup$

Wolfram is your friend. Mathematica comes up with $${\pi x\over 8} {}_4F_3(1,1,3/2,3/2;2,2,2;x)+{\pi\over 2}\log(x).$$

$\endgroup$
1
  • $\begingroup$ Wolfram's EllipticK satisfies $K(k) = EllipticK[k^2]$, and what you've posted is $\int \frac{EllipticK[x]}{x} dx$. But thanks, because I've now found [ \int \frac{K(k)}{k} dk = -\frac{1}{4} G_{3,3}^{2,2}\left(-k^2| \begin{array}{c} \frac{1}{2},\frac{1}{2},1 \\ 0,0,0 \\ \end{array} \right) ] This uses the Meijer $G$ function, which isn't exactly what I meant by nice. In fact, it doesn't seem to offer any advantage over term-by-term integration of a series approximation. But maybe it does - I suppose I want to know it's not expressible in elementary and elliptic functions. $\endgroup$ Jun 11, 2013 at 14:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.