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:
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 $2$.
$\blacksquare$