Equivalence of Definitions of Complete Elliptic Integral of the Third Kind

Theorem

The following definitions of the concept of Complete Elliptic Integral of the Third Kind are equivalent:

Definition 1

$\ds \map \Pi {k, n} = \int \limits_0^{\pi / 2} \frac {\d \phi} {\paren {1 + n \sin^2 \phi} \sqrt {1 - k^2 \sin^2 \phi} }$

is the complete elliptic integral of the third kind, and is a function of the variables:

$k$, defined on the interval $0 < k < 1$

$n \in \Z$

Definition 2

$\ds \map \Pi {k, n} = \int \limits_0^1 \frac {\d v} {\paren {1 + n v^2} \sqrt {\paren {1 - v^2} \paren {1 - k^2 v^2} } }$

is the complete elliptic integral of the third kind, and is a function of the variables:

$k$, defined on the interval $0 < k < 1$

$n \in \Z$

Proof

Let $\map \Pi {k, n}$ be the complete elliptic integral of the third kind by definition $1$.

Let $v := \sin \phi$.

Then we have:

\(\ds \dfrac {\d v} {\d \phi}\)

\(=\)

\(\ds \cos \phi\)

Derivative of Sine Function

\(\ds \phi\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds v\)

\(=\)

\(\ds 0\)

\(\ds \phi\)

\(=\)

\(\ds \dfrac \pi 2\)

\(\ds \leadsto \ \ \)

\(\ds v\)

\(=\)

\(\ds 1\)

Hence:

\(\ds \map \Pi {k, n}\)

\(=\)

\(\ds \int \limits_0^{\pi / 2} \frac {\d \phi} {\paren {1 + n \sin^2 \phi} \sqrt {1 - k^2 \sin^2 \phi} }\)

Definition 1 of Complete Elliptic Integral of the Third Kind

\(\ds \)

\(=\)

\(\ds \int \limits_0^1 \frac {\d v} {\paren {1 + n \sin^2 \phi} \cos \phi \sqrt {1 - k^2 \sin^2 \phi} }\)

Integration by Substitution

\(\ds \)

\(=\)

\(\ds \int \limits_0^1 \frac {\d v} {\paren {1 + n \sin^2 \phi} \sqrt {1 - \sin^2 \phi} \sqrt {1 - k^2 \sin^2 \phi} }\)

Sum of Squares of Sine and Cosine

\(\ds \)

\(=\)

\(\ds \int \limits_0^1 \frac {\d v} {\paren {1 + n v^2} \sqrt {\paren {1 - v^2} \paren {1 - k^2 v^2} } }\)

substituting for $v$

Thus $\map \Pi {k, n}$ is the complete elliptic integral of the third kind by definition $2$.

$\blacksquare$