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 ?

By definition we have

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


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}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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