Evaluating the elliptic integral $\int^1_0 \frac{K(k)}{\sqrt{1-k^2}}\,dk$

Question

Find the elliptic integral $$\int\limits^1_0 \frac{K(k)}{\sqrt{1-k^2}}\,\mathrm{d}k,$$ where $$K(k) = \int\limits_{0}^{1} \frac {\mathrm dt}{\sqrt{(1 - t^2)(1 - k^2t^2)}}=\int\limits_{0}^{\pi/2} \frac{\text{d}\vartheta}{\sqrt{1 - k^2 \sin^2\vartheta}}$$

My approach: Let $k=\dfrac{\sin\theta}{\sin\phi}$

Now after some calculation, I have reached the following: $$\int\limits_{0}^{\pi/2}\int\limits_{0}^{\phi}\dfrac{d\theta d\phi}{\sqrt{\sin^2\phi-\sin^2\theta}}$$ Now how to proceed from this point..please give some idea/hint.

Definition of elliptic integral: https://en.wikipedia.org/wiki/Elliptic_integral

Answer 1

1. Replace $K$ with its $_2F_1$ representation (DLMF 19.5.1).
2. Evaluate using the Euler integral representation of $_3F_2$ (DLMF 16.5.2).
3. Simplify using Dixon's well-poised $_3F_2$ sum (DLMF 16.4.4).
4. Simplify using the functional relations for $\Gamma$ (DLMF 5.5).

The result is

$$\frac{\Gamma(1/4)^4}{16\pi}\text{.}$$

In detail:

$\newcommand{\d}{\mathrm{d}}$

$$\begin{split} \int_0^1 \frac{K(k)}{\sqrt{1-k^2}}\d k &= \frac{\pi}{2} \int_0^1 \frac{_2F_1(\tfrac{1}{2},\tfrac{1}{2};1;k^2)} {\sqrt{1-k^2}}\d k \\ &= \frac{\pi}{4} \int_0^1 \frac{_2F_1(\tfrac{1}{2},\tfrac{1}{2};1;x)} {\sqrt{x(1-x)}}\d x \\ &= \frac{\pi^2}{4}\,_3F_2(\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2};1,1;1) \\ &= \frac{\pi^2}{4}\left(\frac{\Gamma(\tfrac{5}{4})\Gamma(1)\Gamma(1)\Gamma(\tfrac{1}{4})}{\Gamma(\tfrac{3}{2})\Gamma(\tfrac{3}{4})\Gamma(\tfrac{3}{4})\Gamma(\tfrac{1}{2})}\right) \\ &= \frac{\Gamma(\tfrac{1}{4})^4}{16\pi}\text{.} \end{split}$$