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I have been reading Zaid Alyafeai's book Advanced Integration Techniques, and I came across the following identity: $$K(k)=\frac1{1+k}K\left(\frac{2\sqrt k}{1+k}\right),$$ where $ K(k)$ is the Complete Elliptic Integral of first kind: $$K(k) = \int_0^1 \frac{dx}{\sqrt{1-x^2}\sqrt{{1-(kx)^2}}} $$ In the book, he proved it by applying Quadratic transformation in the hypergeometric representation of $K(k)$.

My Question: Is there any other way (that does not involve Quadratic Transformation) to prove this identity?

I am unfamiliar with the transformation and hence would like to know an alternative. Thanks in advance.

Anne Bauval
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Quphine
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2 Answers2

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You may prove that both sides fulfill the same differential equation by performing change of variables, or you may exploit the fact that the arithmetic-geometric mean fulfills

$$\text{AGM}(1-\sqrt{x},1+\sqrt{x}) = \text{AGM}(1,\sqrt{1-x})\tag{1} $$

and that (according to Mathematica's notation)

$$ K(x) = \int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1-x\sin^2\theta}}=\frac{\pi}{2}\sum_{n\geq 0}\left(\frac{\binom{2n}{n}}{4^n}\right)^2 x^n =\frac{\pi}{2\,\text{AGM}(1,\sqrt{1-x})}\tag{2}$$

for any $|x|<1$. In particular, since the $\text{AGM}$ is homogeneous,

$$\begin{eqnarray*} K(x)&=&\frac{\pi}{2\,\text{AGM}(1,\sqrt{1-x})}=\frac{\pi}{2\,\text{AGM}(1-\sqrt{x},1+\sqrt{x})}\\&=&\frac{1}{1+\sqrt{x}}\cdot\frac{\pi}{2\,\text{AGM}\left(1,\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)}\\&=&\frac{1}{1+\sqrt{x}}\cdot\frac{\pi}{2\,\text{AGM}\left(\sqrt{1-\frac{4\sqrt{x}}{(1+\sqrt{x})^2}}\right)}\\&=&\frac{1}{1+\sqrt{x}}\,K\left(\frac{4\sqrt{x}}{(1+\sqrt{x})^2}\right)\tag{3}.\end{eqnarray*}$$

Jack D'Aurizio
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Using the Binomial Theorem: $$\sum_{n=0}^{\infty}\frac{(2n)!x^{n}}{2^{2n}n!^2}=\frac{1}{\sqrt{1-x}} \tag{1}$$ The expansion for $K(k)$ wich follows after replacing $x=k^2\sin^2{x}$ in $(1)$, integrating in $(0,\pi/2)$ and aplying Wallis' formula $\int_{0}^{\pi/2}\sin^{2n}{(x)}dx=\frac{\pi}{2}\frac{(2n)!}{2^{2n}n!^2}$: $$\frac{\pi}{2}\sum_{n=0}^{\infty}\frac{(2n)!^2k^{2n}}{2^{4n}n!^4}=\int_{0}^{\pi/2}\frac{dx}{\sqrt{1-k^2\sin^2x}}=K(k) \tag{2}$$ Then replacing in $(1)$ with $x=-ke^{ix}$ using $(2)$ and Parseval's Theorem: $$\frac{2}{\pi}K(k)=\sum_{n=0}^{\infty}\left(\frac{(2n)!(-k)^n}{2^{2n}n!^2}\right)^2=\frac{1}{\pi}\int_{-\pi}^{\pi}|{\frac{1}{\sqrt{1+ke^{ix}}}}|^2dx \\= \frac{1}{\pi}\int_{0}^{\pi}\frac{dx}{\sqrt{2k\cos{x}+k^2+1}}=\frac{2}{\pi}\int_{0}^{\pi/2}\frac{dx}{\sqrt{-4k\sin^2{x}+(k+1)^2}} \\=\frac{2}{\pi}\frac{1}{1+k}\int_{0}^{\pi/2}\frac{dx}{\sqrt{1-\left(\frac{2\sqrt{k}}{1+k}\right)^2\sin^2{x}}}=\frac{2}{\pi}\frac{K(\frac{2\sqrt{k}}{1+k})}{1+k}.$$ Where we have used $\cos{2x}=1-2\sin^2{x}$ in the fifth step. And we are done. $$K(k)=\frac{1}{1+k}K(\frac{2\sqrt{k}}{1+k}).$$

User-Refolio
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