Writing complete elliptic integral of first kind as a hypergeometric function

Question

I am trying to show $K=\int_{0}^{\frac{\pi}{2}}\frac{dz}{\sqrt{1-k^{2}\sin^{2}(z)}}$ can be written as $\frac{\pi}{2} \mathstrut_{2}F_{1}(\frac{1}{2},\frac{1}{2};1;k^{2})$. First I used $(1+z)^{n}=\mathstrut_{2}F_{1}(-n,b;b;-z)$ to expand the integrand to get $\frac{dz}{\sqrt{1-k^{2}\sin^{2}(z)}} = 1+\frac{1}{2}k^{2}\sin^{2}(z) + \ldots +\frac{(2n-1)!!}{2^{n}n!}k^{2n}\sin^{2n}(z)$ but when I go to integrate with respect to $z$ I have to integrate $\sin^{2n}(z)$. I'm not sure where to go since even integrating the first few terms I get a $\cos(z)$ and plugging in $\frac{\pi}{2}$ kills the term which it shouldn't. Thanks for any help or tips you can give.

Answer 1

It looks to me you'll want one of the Wallis formulae:

$$\int_0^{\frac{\pi }{2}} \sin^{2n}u \,\mathrm du=\frac{\pi(2n-1)!}{4^n n! (n-1)!}$$

Just plug where needed, and use duplication formulae/conversion to Pochhammer symbols whenever necessary.

Answer 2

As an alternative way of deriving the representation: $$ K(k) = \int_0^\frac{\pi}{2} \frac{\mathrm{d} z}{\sqrt{1-k^2 \sin^2 (z)}} \stackrel{u = \sin^2(z)}{=} \int_0^1 \frac{1}{2}\frac{\mathrm{d} u}{\sqrt{\left(1-k^2 u\right) \left( 1-u\right) u }} $$

The latter is exactly the Euler's integral representation of the Gauss's hypergeometric function with $b=\frac{1}{2}$, $c-b=\frac{1}{2}$ and $a=\frac{1}{2}$, $z=k^2$, i.e. $b=\frac{1}{2}$, $c=1$, $a=\frac{1}{2}$: $$ K(k) = \frac{1}{2} B\left(\frac{1}{2}, \frac{1}{2}\right) {}_2 F_1\left( \frac{1}{2}, \frac{1}{2}; 1; k^2 \right) = \frac{\pi}{2} {}_2 F_1\left( \frac{1}{2}, \frac{1}{2}; 1; k^2 \right) $$

Answer 3

$$\left|(1-x^2)^{-\frac 1{2}}{\cos(2n\sin^{-1}x)} = {_2F_1}(\frac1{2} + n, \frac 1{2} -n; \frac 1{2};x^2) \right|_{n =0, x = k\sin(z) }$$ Integrate w.r.t $z$ over $[0, \frac {\pi}{2}]$ $$\begin{align*} \int_{0}^{\frac{\pi}{2}}\frac{dz}{\sqrt{1-k^{2}\sin^{2}(z)}} & = \int_0^{\frac {\pi}{2}}{{_2F_1}(\frac1{2}, \frac 1{2}; \frac 1{2};(k\sin(z))^2)}dz\\ & = \sum_{u\ge0}{\frac {(\frac 1{2})_u(\frac 1{2})_u}{(\frac 1{2})_uu!}}k^{2u}\int_0^{\frac {\pi}{2}}\sin^{2u}(z)dx\\ & = \frac {\pi}{2}{_2F_1}(\frac 1{2}, \frac 1{2};\frac 1{2};k^2) = P_{-\frac 1{2}}(1-2k^2) \end{align*}$$