# Writing complete elliptic integral of first kind as a hypergeometric function

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.

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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.

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Yes. Indeed that was the missing link. Thank you. –  Frank Nov 3 '11 at 0:00

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. $b=\frac{1}{2}$, $c=1$, $a=\frac{1}{2}$: $$K(k) = \frac{1}{2} B\left(\frac{1}{2}, \frac{1}{2}\right) {}_2 F_1\left( \frac{1}{2}, \frac{1}{2}; 1; k^2 \right) = \frac{\pi}{2} {}_2 F_1\left( \frac{1}{2}, \frac{1}{2}; 1; k^2 \right)$$

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