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

Question

Question:

Prove that

$$\int^1_0 \frac{K(k)}{\sqrt{1-k^2}}\,dk=\frac{1}{16\pi}\Gamma^4\left( \frac{1}{4}\right)$$

My attempt

Start by the transformation

$$k \to \frac{2\sqrt{k}}{1+k}$$

Hence we have

$$\int^{1}_0 K\left(\frac{2\sqrt{k}}{1+k}\right)\,\frac{1}{\sqrt{k}(1+k)}dk$$

Now we use that

$$K(k)=\frac{1}{k+1}K\left( \frac{2\sqrt{k}}{1+k} \right)$$

So we have

$$\int^1_0 \frac{K(k)}{\sqrt{k}}\,dk=2\int^1_0 K(k^2)\,dk$$

[1] I have no idea how to solve the last integral?

[2] Should I use another approach to solve the integral ?

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By definition we have

$$K(k) = \int^1_0 \frac{dx}{\sqrt{1-x^2}\sqrt{1-k^2\,x^2}}\,$$

Answer 1

Consider the integral \begin{align} I = \int_{0}^{1} \frac{K(x)}{\sqrt{1-x^2}} \ dx \end{align} where $K(x)$ is the complete elliptic integral of the first kind. It can be shown that the hypergeometric form is \begin{align} K(x) = \frac{\pi}{2} \ {}_{2}F_{1}(\frac{1}{2}, \frac{1}{2}; 1; x^{2}). \end{align} By using the series form the integral becomes \begin{align} I &= \frac{\pi}{2} \ \sum_{r=0}^{\infty} \frac{(1/2)_{r} (1/2)_{r}}{r! (1)_{r}} \ \int_{0}^{1} x^{2r} (1-x^{2})^{-1/2} \ dx. \end{align} By making the substitution $x = \sqrt{t}$ the integral can be cast into Beta function form and yields \begin{align} I &= \frac{\pi}{4} \ \sum_{r=0}^{\infty} \frac{(1/2)_{r} (1/2)_{r}}{r! (1)_{r}} \ B(r+1/2, 1/2) \\ &= \left( \frac{\pi}{2} \right)^{2} \ {}_{3}F_{2}(a,a,a; 1,1; 1) \end{align} where $a = 1/2$. Now, by using the identity \begin{align} {}_{3}F_{2}(a,a,a; 1,1; 1) = \frac{\Gamma\left(1 - \frac{3a}{2}\right) \ \cos\left( \frac{a \pi}{2} \right)}{\Gamma^{3}\left( 1 - \frac{a}{2}\right)} \end{align} which is valid for $Re(a) < 2/3$, the integral value is seen to be \begin{align} I = \left( \frac{\pi}{2} \right)^{2} \ \frac{\Gamma\left(\frac{1}{4}\right) \ \cos\left( \frac{\pi}{4} \right)}{\Gamma^{3}\left(\frac{3}{4}\right)}. \end{align} Now using the relation \begin{align} \Gamma\left(\frac{3}{4}\right) = \frac{\sqrt{2} \pi}{\Gamma\left(\frac{1}{4}\right)} \end{align} the final value is obtained, namely, \begin{align} I = \frac{1}{16 \pi} \Gamma^{4}\left(\frac{1}{4}\right). \end{align} Hence \begin{align} \int_{0}^{1} \frac{K(x)}{\sqrt{1-x^2}} \ dx = \frac{1}{16 \pi} \Gamma^{4}\left(\frac{1}{4}\right). \end{align}

Answer 2

A simple method for the last integral: \begin{align} \int \limits_0^1 \frac{\operatorname{K}(k)}{\sqrt{k}} \, \mathrm{d} k &= \int \limits_0^1 \int \limits_0^{\pi/2} \frac{\mathrm{d} \phi}{\sqrt{k(1-k^2 \sin^2(\phi))}} \, \mathrm{d} k \stackrel{\text{Tonelli}}{=} \int \limits_0^{\pi/2} \int \limits_0^1 \frac{\mathrm{d} k}{\sqrt{k(1-k^2 \sin^2(\phi))}} \, \mathrm{d} \phi \\ &\!\!\!\!\!\!\stackrel{k = \frac{\sin(\theta)}{\sin(\phi)}}{=} \int \limits_0^{\pi/2} \int \limits_0^\phi \frac{\mathrm{d} \theta \, \mathrm{d} \phi}{\sqrt{\sin(\theta)\sin(\phi)}} = \frac{1}{2} \int \limits_0^{\pi/2} \int \limits_0^{\pi/2} \frac{\mathrm{d} \theta \, \mathrm{d} \phi}{\sqrt{\sin(\theta)\sin(\phi)}} = \frac{1}{2} \left[\int \limits_0^{\pi/2} \frac{\mathrm{d} t}{\sqrt{\sin(t)}}\right]^2 \\ &= \frac{1}{8} \operatorname{B}^2\left(\frac{1}{4},\frac{1}{2}\right) = \frac{1}{8} \left(\frac{\operatorname{\Gamma}\left(\frac{1}{4}\right) \operatorname{\Gamma}\left(\frac{1}{2}\right)}{\operatorname{\Gamma}\left(\frac{3}{4}\right)}\right)^2 \stackrel{\Gamma\text{-reflection}}{=} \frac{1}{8} \left(\frac{\operatorname{\Gamma}^2\left(\frac{1}{4}\right) \operatorname{\Gamma}\left(\frac{1}{2}\right) \sin\left(\frac{\pi}{4}\right)}{\pi}\right)^2 \\ &= \frac{\operatorname{\Gamma}^4\left(\frac{1}{4}\right)}{16 \pi} \, . \end{align}