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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?

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

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  • $\begingroup$ 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. $\endgroup$
    – Bryan Clair
    Commented Jun 11, 2013 at 14:15

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