Equivalence of Definitions of Incomplete Elliptic Integral of the First Kind

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Theorem

The following definitions of the concept of Incomplete Elliptic Integral of the First Kind are equivalent:

Definition 1

$\ds \map F {k, \phi} = \int \limits_0^\phi \frac {\d \phi} {\sqrt {1 - k^2 \sin^2 \phi} }$

is the incomplete elliptic integral of the first kind, and is a function of the variables:

$k$, defined on the interval $0 < k < 1$
$\phi$, defined on the interval $0 \le \phi \le \pi / 2$.

Definition 2

$\ds \map F {k, \phi} = \int \limits_0^x \frac {\d v} {\sqrt {\paren {1 - v^2} \paren {1 - k^2 v^2} } }$

is the incomplete elliptic integral of the first kind, and is a function of the variables:

$k$, defined on the interval $0 < k < 1$
$x = \sin \phi$, where $\phi$ is defined on the interval $0 \le \phi \le \pi / 2$.


Proof

Let $\map K {k, \phi}$ be the incomplete elliptic integral of the first kind by definition $1$.


Let $v := \sin \phi$.

Then we have:

\(\ds \dfrac {\d v} {\d \phi}\) \(=\) \(\ds \cos \phi\) Derivative of Sine Function
\(\ds \phi\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds 0\)
\(\ds \phi\) \(=\) \(\ds \dfrac \pi 2\)
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds 1\)

So as $0 \le \phi \le \dfrac \pi 2$, we have that $0 \le v \le 1$.


Hence:

\(\ds \map K {k, \phi}\) \(=\) \(\ds \int \limits_0^\phi \frac {\d \phi} {\sqrt {1 - k^2 \sin^2 \phi} }\) Definition 1 of Incomplete Elliptic Integral of the First Kind
\(\ds \) \(=\) \(\ds \int \limits_0^x \frac {\d v} {\cos \phi \sqrt {1 - k^2 \sin^2 \phi} }\) Integration by Substitution: $x = \sin \phi$
\(\ds \) \(=\) \(\ds \int \limits_0^x \frac {\d v} {\sqrt {1 - \sin^2 \phi} \sqrt {1 - k^2 \sin^2 \phi} }\) Sum of Squares of Sine and Cosine
\(\ds \) \(=\) \(\ds \int \limits_0^x \frac {\d v} {\sqrt {1 - v^2} \sqrt {1 - k^2 v^2} }\) substituting for $v$
\(\ds \) \(=\) \(\ds \int \limits_0^x \frac {\d v} {\sqrt {\paren {1 - v^2} \paren {1 - k^2 v^2} } }\) making the equivalence explicit


Thus $\map K {k, \phi}$ is the incomplete elliptic integral of the first kind by definition $2$.

$\blacksquare$