Evaluate $\int_{0}^{\pi/2} \ln\left[ \tan\left ( \frac{\theta}{2}\right) \right ]^2 K\left ( \sin\theta \right )\text{d}\theta$

Question

Let us define $K(x)$ as complete elliptic integral of the first kind, where $x$ is elliptic modulus. A possible closed-form is ($G$ denotes Catalan's constant.) $$ \int_{0}^{\pi/2} \ln\left[ \tan\left ( \frac{\theta}{2}\right) \right ]^2 K\left ( \sin\theta \right )\text{d}\theta =\frac{\Gamma\left ( \frac14 \right )^4G }{8\pi}. $$ It looks like a "product" of two solvable integrals (both are elementary): $$\int_{0}^{\pi/2} \ln\left[ \tan\left ( \frac{\theta}{2}\right) \right ]^2\text{d}\theta =\frac{\pi^3 }{8}$$ and $$ \int_{0}^{\pi/2} K\left ( \sin\theta \right )\text{d}\theta =\frac{\Gamma\left ( \frac14 \right )^4 }{16\pi}. $$

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Question: How can we evaluate the integral? I try to utilize the Fourier series $K(\sin\theta) =\pi\sum_{n\ge0} \frac{\left ( \frac12 \right )_n^2 }{(n!)^2} \sin\left ( \left ( 4n+1 \right )\theta \right )$ to prove, but seems not to go well. I appreciate for your help.

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An Interesting Observation: We find $$ \int_{0}^{\pi/2} \ln\left[ \tan\left ( \frac{\theta}{2}\right) \right ]^4 K\left ( \sin\theta \right )\text{d}\theta =\frac{3\,\Gamma\left ( \frac14 \right )^4}{4\pi}(G^2+\beta(4)) $$ where $\beta(.)$ is Dirichlet's $\beta$ function.

Answer 1

We have $$\int_0^{\pi/2} \log^{2n} (\tan \frac{x}{2}) K(\sin x) dx = \int_0^1 \frac{2 \log^{2n} t}{1+t^2} K(\frac{2t}{1+t^2}) dt$$ Their values can be extracted by differentiating $$\tag{*}\int_0^1 \frac{t^{4a} + t^{-4a}}{1+t^2} K(\frac{2t}{1+t^2}) dt = \frac{\pi}{8}\cot(\pi(\frac{1}{4}-a))\frac{\Gamma \left(a+\frac{1}{4}\right)^2}{\Gamma \left(a+\frac{3}{4}\right)^2} \quad -1/4<\Re(a)<1/4$$

I give two different proofs of $(*)$, I come up with the longer proof first.

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First proof of $(*)$: using quadratic transformation $(1+t^2)K(t^2) = K(2t/(1+t^2))$, we have $$ \begin{aligned}\int_0^1 \frac{t^{4a}}{1+t^2} K(\frac{2t}{1+t^2}) dt &= \frac{1}{2} \int_0^1 t^{2a-1/2} K(t) dt \\ &= \frac{1}{2} \frac{\pi}{1+4a}{_3F_2}(\frac{1}{2}, \frac{1}{2}, \frac{1}{4} + a; 1, \frac{5}{4} + a; 1) \end{aligned}$$ here we noted that $K(t)$ is a $_2F_1$. Using third formulas here gives $${_3F_2}(\frac{1}{2}, \frac{1}{2}, \frac{1}{4} + a; 1, \frac{5}{4} + a; 1) = \frac{4 a+1}{4 a-1} {_3F_2(\frac{1}{2}, \frac{1}{2}, \frac{1}{4} - a; 1, \frac{5}{4} - a; 1)} +\frac{\Gamma \left(\frac{1}{4}-a\right) \Gamma \left(a+\frac{5}{4}\right)}{\Gamma \left(\frac{3}{4}-a\right) \Gamma \left(a+\frac{3}{4}\right)}$$ we see the $a$ in $_3F_2$ becomes $-a$, giving $(*)$. Q.E.D.

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Second proof of $(*)$: It's much longer. See edit history.

Answer 2

By expanding $K(k)$, we deduce $$ \,_5F_4\left ( \frac14,\frac14,\frac14,\frac12,\frac12; 1,\frac54,\frac54,\frac54;1 \right )=\frac{\Gamma\left ( \frac14 \right)^4}{16\pi^2}G, $$ $$ \,_7F_6\left ( \frac14,\frac14,\frac14,\frac14,\frac14,\frac12,\frac12; 1,\frac54,\frac54,\frac54,\frac54,\frac54;1 \right )=\frac{\Gamma\left ( \frac14 \right)^4}{32\pi^2}(G^2+\beta(4)). $$