# Complete Elliptic Integral of the First Kind as Power Series

## Theorem

$\displaystyle \map K k = \int_0^{\pi / 2} \frac {\rd \phi} {\sqrt {1 - k^2 \sin^2 \phi} } = \int_0^1 \frac {\rd v} {\sqrt {\paren {1 - v^2} \paren {1 - k^2 v^2} } }$

can be expressed as the power series:

 $\displaystyle \map K k$ $=$ $\displaystyle \frac \pi 2 \sum_{i \mathop \ge 0} \paren {\prod_{j \mathop = 1}^i \frac {2 j - 1} {2 j} }^2 k^{2 i}$ $\displaystyle$ $=$ $\displaystyle \frac \pi 2 \paren {1 + \paren {\frac 1 2}^2 k^2 + \paren {\frac {1 \cdot 3} {2 \cdot 4} }^2 k^4 + \paren {\frac {1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6} }^2 k^6 + \cdots}$

## Proof

From Reduction Formula for Integral of Power of Sine, $\forall i \in \N$:

 $\displaystyle \int_0^{\pi / 2} \sin^{2 i} \phi \rd \phi$ $=$ $\displaystyle \frac {2 i - 1} {2 i} \int_0^{\pi / 2} \sin^{2 i - 2} \phi \rd \phi - \intlimits {\frac {\sin^{2 i - 1} x \cos x} i} {x = 0} {x = \frac \pi 2}$ $\displaystyle$ $=$ $\displaystyle \frac {2 i - 1} {2 i} \int_0^{\pi / 2} \sin^{2 i - 2} \phi \rd \phi$ $\sin 0 = \cos \dfrac \pi 2 = 0$ $\displaystyle$ $=$ $\displaystyle \cdots$ $\displaystyle$ $=$ $\displaystyle \prod_{j \mathop = 1}^i \frac {2 j - 1} {2 j} \int_0^{\pi / 2} 1 \rd \phi$ $\displaystyle$ $=$ $\displaystyle \frac \pi 2 \prod_{j \mathop = 1}^i \frac {2 j - 1} {2 j}$

Hence:

 $\displaystyle \map K k$ $=$ $\displaystyle \int_0^{\pi / 2} \frac {\rd \phi} {\sqrt{1 - k^2 \sin^2 \phi} }$ $\displaystyle$ $=$ $\displaystyle \int_0^{\pi / 2} \sum_{i \mathop \ge 0} \binom {- \frac 1 2} i \paren {- k^2 \sin^2 \phi}^i \rd \phi$ General Binomial Theorem $\displaystyle$ $=$ $\displaystyle \sum_{i \mathop \ge 0} k^{2 i} \paren {\prod_{j \mathop = 1}^i \frac {\frac 1 2 - j} j} \paren {-1}^i \int_0^{\pi / 2} \sin^{2 i} \phi \rd \phi$ Definition of Binomial Coefficient $\displaystyle$ $=$ $\displaystyle \sum_{i \mathop \ge 0} k^{2 i} \paren {\prod_{j \mathop = 1}^i \frac {1 - 2 j} {2 j} } \paren {\prod_{j \mathop = 1}^i \paren {-1} }\paren {\frac \pi 2 \prod_{j \mathop = 1}^i \frac {2 j - 1} {2 j} }$ from above $\displaystyle$ $=$ $\displaystyle \frac \pi 2 \sum_{i \mathop \ge 0} \paren {\prod_{j \mathop = 1}^i \frac {2 j - 1} {2 j} }^2 k^{2 i}$

$\blacksquare$