Equivalence of Definitions of Incomplete Elliptic Integral of the Second Kind
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Theorem
The following definitions of the concept of Incomplete Elliptic Integral of the Second Kind are equivalent:
Definition 1
- $\ds \map E {k, \phi} = \int \limits_0^\phi \sqrt {1 - k^2 \sin^2 \phi} \rd \phi$
is the incomplete elliptic integral of the second kind, and is a function of the variables:
Definition 2
- $\ds \map E {k, \phi} = \int \limits_0^x \dfrac {\sqrt {1 - k^2 v^2} } {\sqrt {1 - v^2}} \rd v$
is the incomplete elliptic integral of the second 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 $\map E {k, \phi}$ be the incomplete elliptic integral of the second 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:
\(\ds \map E {k, \phi}\) | \(=\) | \(\ds \int \limits_0^\phi \sqrt {1 - k^2 \sin^2 \phi} \rd \phi\) | Definition 1 of Incomplete Elliptic Integral of the Second Kind | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \limits_0^x \sqrt {1 - k^2 \sin^2 \phi} \frac {\d v} {\cos \phi}\) | Integration by Substitution: $x = \sin \phi$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \limits_0^x \sqrt {1 - k^2 \sin^2 \phi} \frac {\d v} {\sqrt {1 - \sin^2 \phi} }\) | Sum of Squares of Sine and Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \limits_0^x \dfrac {\sqrt {1 - k^2 v^2} } {\sqrt {1 - v^2} } \rd v\) | substituting for $v$ |
Thus $\map E {k, \phi}$ is the incomplete elliptic integral of the second kind by definition $2$.
$\blacksquare$