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{{Short description|Special function defined by an integral}}
In integral calculus, an '''elliptic integral''' is one of a number of related functions defined as the value of certain integrals, which were first studied by Giulio Fagnano and [[Biography:Leonhard Euler|Leonhard Euler]] ({{Circa|1750}}). Their name originates from their originally arising in connection with the problem of finding the [[Arc length|arc length]] of an [[Ellipse|ellipse]].

Modern mathematics defines an "elliptic integral" as any [[Function (mathematics)|function]] {{math|''f''}} which can be expressed in the form

<math display="block"> f(x) = \int_{c}^{x} R{\left({\textstyle t, \sqrt{ P(t)} }\right)} \, dt,</math>

where {{math|''R''}} is a [[Rational function|rational function]] of its two arguments, {{math|''P''}} is a [[Polynomial|polynomial]] of degree 3 or 4 with no repeated roots, and {{math|''c''}} is a constant.

In general, integrals in this form cannot be expressed in terms of [[Elementary function|elementary function]]s. Exceptions to this general rule are when {{math|''P''}} has repeated roots, or when {{math|''R''(''x'', ''y'')}} contains no odd powers of {{math|''y''}} or if the integral is pseudo-elliptic. However, with the appropriate [[Integration by reduction formulae|reduction formula]], every elliptic integral can be brought into a form that involves integrals over rational functions and the three [[Legendre form|Legendre canonical form]]s (i.e. the elliptic integrals of the first, second and third kind).

Besides the Legendre form given below, the elliptic integrals may also be expressed in [[Carlson symmetric form]]. Additional insight into the theory of the elliptic integral may be gained through the study of the [[Schwarz–Christoffel mapping]]. Historically, elliptic functions were discovered as inverse functions of elliptic integrals.

==Argument notation==
''Incomplete elliptic integrals'' are functions of two arguments; ''complete elliptic integrals'' are functions of a single argument. These arguments are expressed in a variety of different but equivalent ways (they give the same elliptic integral). Most texts adhere to a canonical naming scheme, using the following naming conventions.

For expressing one argument:
* {{math|''α''}}, the ''modular angle''
* {{math|1=''k'' = sin ''α''}}, the ''elliptic modulus'' or ''[[Eccentricity (mathematics)|eccentricity]]''
* {{math|1=''m'' = ''k''<sup>2</sup> = sin<sup>2</sup> ''α''}}, the ''parameter''

Each of the above three quantities is completely determined by any of the others (given that they are non-negative). Thus, they can be used interchangeably.

The other argument can likewise be expressed as {{math|''φ''}}, the ''amplitude'', or as {{math|''x''}} or {{math|''u''}}, where {{math|1=''x'' = sin ''φ'' = sn ''u''}} and {{math|sn}} is one of the Jacobian elliptic functions.

Specifying the value of any one of these quantities determines the others. Note that {{math|''u''}} also depends on {{math|''m''}}. Some additional relationships involving {{math|''u''}} include
<math display="block">\cos \varphi = \operatorname{cn} u, \quad \textrm{and} \quad \sqrt{1 - m \sin^2 \varphi} = \operatorname{dn} u.</math>

The latter is sometimes called the ''delta amplitude'' and written as {{math|1=Δ(''φ'') = dn ''u''}}. Sometimes the literature also refers to the ''complementary parameter'', the ''complementary modulus,'' or the ''complementary modular angle''. These are further defined in the article on [[Quarter period|quarter period]]s.

In this notation, the use of a vertical bar as delimiter indicates that the argument following it is the "parameter" (as defined above), while the backslash indicates that it is the modular angle. The use of a semicolon implies that the argument preceding it is the sine of the amplitude:
<math display="block"> F(\varphi, \sin \alpha) = F\left(\varphi \mid \sin^2 \alpha\right) = F(\varphi \setminus \alpha) = F(\sin \varphi ; \sin \alpha).</math>
This potentially confusing use of different argument delimiters is traditional in elliptic integrals and much of the notation is compatible with that used in the reference book by [[Abramowitz and Stegun]] and that used in the integral tables by [[Gradshteyn and Ryzhik]].

There are still other conventions for the notation of elliptic integrals employed in the literature. The notation with interchanged arguments, {{math|''F''(''k'', ''φ'')}}, is often encountered; and similarly {{math|''E''(''k'', ''φ'')}} for the integral of the second kind. [[Abramowitz and Stegun]] substitute the integral of the first kind, {{math|''F''(''φ'', ''k'')}}, for the argument {{mvar|φ}} in their definition of the integrals of the second and third kinds, unless this argument is followed by a vertical bar: i.e. {{math|''E''(''F''(''φ'', ''k'') {{!}} ''k''<sup>2</sup>)}} for {{math|''E''(''φ'' {{!}} ''k''<sup>2</sup>)}}. Moreover, their complete integrals employ the ''parameter'' {{math|''k''<sup>2</sup>}} as argument in place of the modulus {{math|''k''}}, i.e. {{math|''K''(''k''<sup>2</sup>)}} rather than {{math|''K''(''k'')}}. And the integral of the third kind defined by [[Gradshteyn and Ryzhik]], {{math|Π(''φ'', ''n'', ''k'')}}, puts the amplitude {{mvar|φ}} first and not the "characteristic" {{mvar|n}}.

Thus one must be careful with the notation when using these functions, because various reputable references and software packages use different conventions in the definitions of the elliptic functions. For example, [[Company:Wolfram Research|Wolfram]]'s [[Software:Mathematica|Mathematica]] software and [[Physics:Wolfram Alpha|Wolfram Alpha]] define the complete elliptic integral of the first kind in terms of the parameter {{math|''m''}}, instead of the elliptic modulus {{math|''k''}}.

==Incomplete elliptic integral of the first kind==
The '''incomplete elliptic integral of the first kind''' {{mvar|F}} is defined as

<math display="block"> F(\varphi,k) = F\left(\varphi \mid k^2\right) = F(\sin \varphi ; k) = \int_0^\varphi \frac {d\theta}{\sqrt{1 - k^2 \sin^2 \theta}}.</math>

This is the trigonometric form of the integral; substituting {{math|1=''t'' = sin ''θ''}} and {{math|1=''x'' = sin ''φ''}}, one obtains the Legendre normal form:

<math display="block"> F(x ; k) = \int_{0}^{x} \frac{dt}{\sqrt{\left(1 - t^2\right)\left(1 - k^2 t^2\right)}}.</math>

Equivalently, in terms of the amplitude and modular angle one has:
<math display="block"> F(\varphi \setminus \alpha) = F(\varphi, \sin \alpha) = \int_0^\varphi \frac{d\theta}{\sqrt{1-\left(\sin \theta \sin \alpha\right)^2}}.</math>

With {{math|1=''x'' = sn(''u'', ''k'')}} one has:
<math display="block">F(x;k) = u;</math>
demonstrating that this Jacobian elliptic function is a simple inverse of the incomplete elliptic integral of the first kind.

The incomplete elliptic integral of the first kind has following addition theorem:
<math display="block">F\bigl[\arctan(x),k\bigr] + F\bigl[\arctan(y),k\bigr] = F\left[\arctan\left(\frac{x\sqrt{k'^2y^2+1}}{\sqrt{y^2+1}}\right) + \arctan\left(\frac{y\sqrt{k'^2x^2+1}}{\sqrt{x^2+1}}\right),k\right] </math>

The elliptic modulus can be transformed that way:
<math display="block">F\bigl[\arcsin(x),k\bigr] = \frac{2}{1+\sqrt{1-k^2}}F\left[\arcsin\left(\frac{\left(1+\sqrt{1-k^2}\right)x}{1+\sqrt{1-k^2x^2}}\right),\frac{1-\sqrt{1-k^2}}{1+\sqrt{1-k^2}}\right] </math>

==Incomplete elliptic integral of the second kind==
The '''incomplete elliptic integral of the second kind''' {{math|''E''}} in trigonometric form is

<math display="block"> E(\varphi,k) = E\left(\varphi \,|\,k^2\right) = E(\sin\varphi;k) = \int_0^\varphi \sqrt{1-k^2 \sin^2\theta}\, d\theta.</math>

Substituting {{math|1=''t'' = sin ''θ''}} and {{math|1=''x'' = sin ''φ''}}, one obtains the Legendre normal form:

<math display="block"> E(x;k) = \int_0^x \frac{\sqrt{1-k^2 t^2} }{\sqrt{1-t^2}}\,dt.</math>

Equivalently, in terms of the amplitude and modular angle:
<math display="block"> E(\varphi \setminus \alpha) = E(\varphi, \sin \alpha) = \int_0^\varphi \sqrt{1-\left(\sin \theta \sin \alpha\right)^2} \, d\theta.</math>

Relations with the [[Jacobi elliptic functions]] include
<math display="block">\begin{align}
E{\left(\operatorname{sn}(u ; k) ; k\right)}
= \int_0^u \operatorname{dn}^2 (w ; k) \, dw
&= u - k^2 \int_0^u \operatorname{sn}^2 (w ; k) \, dw \\[1ex]
&= \left(1-k^2\right) u + k^2 \int_0^u \operatorname{cn}^2 (w ; k) \,dw.
\end{align}</math>

The [[Earth:Meridian arc|meridian arc]] length from the [[Astronomy:Equator|equator]] to [[Earth:Latitude|latitude]] {{math|''φ''}} is written in terms of {{math|''E''}}:
<math display="block">m(\varphi) = a\left(E(\varphi,e)+\frac{d^2}{d\varphi^2}E(\varphi,e)\right),</math>
where {{math|''a''}} is the semi-major axis, and {{math|''e''}} is the [[Eccentricity (mathematics)|eccentricity]].

The incomplete elliptic integral of the second kind has following addition theorem:
<math display="block">E{\left[\arctan(x), k\right]}
 + E{\left[\arctan(y), k\right]}
= E{\left[\arctan\left(\frac{x\sqrt{k'^2y^2+1}}{\sqrt{y^2+1}}\right) + \arctan\left(\frac{y\sqrt{k'^2x^2+1}}{\sqrt{x^2+1}}\right),k\right]}
 + \frac{k^2xy}{k'^2x^2y^2+x^2+y^2+1}\left(\frac{x\sqrt{k'^2y^2+1}}{\sqrt{y^2+1}}+\frac{y\sqrt{k'^2x^2+1}}{\sqrt{x^2+1}}\right) </math>

The elliptic modulus can be transformed that way:
<math display="block">E{\left[\arcsin(x),k\right]}
= \left(1+\sqrt{1-k^2}\right) E{\left[\arcsin\left(\frac{\left(1+\sqrt{1-k^2}\right)x}{1+\sqrt{1-k^2x^2}}\right),\frac{1-\sqrt{1-k^2}}{1+\sqrt{1-k^2}}\right]}
 - \sqrt{1-k^2} F{\left[\arcsin(x),k\right]}
  + \frac{k^2x\sqrt{1-x^2}}{1+\sqrt{1-k^2x^2}} </math>

==Incomplete elliptic integral of the third kind==
The '''incomplete elliptic integral of the third kind''' {{math|Π}} is 
<math display="block"> \Pi(n ; \varphi \setminus \alpha) = \int_0^\varphi \frac{1}{1-n\sin^2 \theta} \frac{d\theta}{\sqrt{1-\left(\sin\theta\sin \alpha\right)^2}}</math>

or

<math display="block"> \Pi(n ; \varphi \,|\,m) = \int_{0}^{\sin \varphi} \frac{1}{1-nt^2} \frac{dt}{\sqrt{\left(1-m t^2\right)\left(1-t^2\right) }}.</math>

The number {{math|''n''}} is called the '''characteristic''' and can take on any value, independently of the other arguments. Note though that the value {{math|Π(1; {{sfrac|π|2}} {{!}} ''m'')}} is infinite, for any {{math|''m''}}.

A relation with the Jacobian elliptic functions is
<math display="block"> \Pi\bigl(n; \,\operatorname{am}(u;k); \,k\bigr) = \int_0^u \frac{dw} {1 - n \,\operatorname{sn}^2 (w;k)}.</math>

The meridian arc length from the equator to latitude {{math|''φ''}} is also related to a special case of {{math|Π}}:

<math display="block">m(\varphi)=a\left(1-e^2\right)\Pi\left(e^2 ; \varphi \,|\,e^2\right).</math>

==Complete elliptic integral of the first kind==
<!-- This section is redirected from Complete elliptic integral of the first kind -->
[[Image:Mplwp complete ellipticKk.svg|thumb|300px|Plot of the complete elliptic integral of the first kind {{math|''K''(''k'')}}]]
Elliptic Integrals are said to be 'complete' when the amplitude {{math|1=''φ'' = {{sfrac|π|2}}}} and therefore {{math|1=''x'' = 1}}. The '''complete elliptic integral of the first kind''' {{math|''K''}} may thus be defined as
<math display="block">K(k) = \int_0^\tfrac{\pi}{2} \frac{d\theta}{\sqrt{1-k^2 \sin^2\theta}} = \int_0^1 \frac{dt}{\sqrt{\left(1-t^2\right)\left(1-k^2 t^2\right)}},</math>
or more compactly in terms of the incomplete integral of the first kind as
<math display="block">K(k) = F\left(\tfrac{\pi}{2},k\right) = F\left(\tfrac{\pi}{2} \,|\, k^2\right) = F(1;k).</math>

It can be expressed as a [[Power series|power series]]
<math display="block">K(k) = \frac{\pi}{2}\sum_{n=0}^\infty \left(\frac{(2n)!}{2^{2 n} (n!)^2}\right)^2 k^{2n} = \frac{\pi}{2} \sum_{n=0}^\infty \bigl(P_{2 n}(0)\bigr)^2 k^{2n},</math>

where {{math|''P''<sub>''n''</sub>}} is the [[Legendre polynomials]], which is equivalent to

<math display="block">K(k) = \frac{\pi}{2}\left(1+\left(\frac{1}{2}\right)^2 k^2+\left(\frac{1\cdot 3}{2\cdot 4}\right)^2 k^4+\cdots+\left(\frac{\left(2n-1\right)!!}{\left(2n\right)!!}\right)^2 k^{2n}+\cdots\right),</math>

where {{math|''n''!!}} denotes the [[Double factorial|double factorial]]. In terms of the Gauss [[Hypergeometric function|hypergeometric function]], the complete elliptic integral of the first kind can be expressed as

<math display="block">K(k) = \tfrac{\pi}{2} \,{}_2F_1 \left(\tfrac{1}{2}, \tfrac{1}{2}; 1; k^2\right).</math>

The complete elliptic integral of the first kind is sometimes called the [[Quarter period|quarter period]]. It can be computed very efficiently in terms of the [[Arithmetic–geometric mean|arithmetic–geometric mean]]:{{sfn|Carlson|2010|loc=19.8}}
<math display="block">K(k) = \frac{\pi}{2\operatorname{agm}\left(1,\sqrt{1-k^2}\right)}.</math>

Therefore the modulus can be transformed as:

<math display="block">\begin{align}
K(k) &= \frac{\pi}{2\operatorname{agm}\left(1,\sqrt{1-k^2}\right)} \\[4pt]
& = \frac{\pi}{2\operatorname{agm}\left(\frac12+\frac\sqrt{1-k^2}{2},\sqrt[4]{1-k^2}\right)} \\[4pt]
&= \frac{\pi}{\left(1+\sqrt{1-k^2}\right)\operatorname{agm}\left(1,\frac{2\sqrt[4]{1-k^2}}{\left(1+\sqrt{1-k^2}\right)}\right)} \\[4pt]
& = \frac{2}{1+\sqrt{1-k^2}}K\left(\frac{1-\sqrt{1-k^2}}{1+\sqrt{1-k^2}}\right)
\end{align}</math>

This expression is valid for all <math>n \isin \mathbb{N}</math> and {{math|0 ≤ ''k'' ≤ 1}}:

<math display="block">K(k) = n\left[\sum_{a = 1}^{n} \operatorname{dn}\left(\frac{2a}{n}K(k);k\right)\right]^{-1}K\left[k^n\prod_{a=1}^{n}\operatorname{sn}\left(\frac{2a-1}{n}K(k);k\right)^2\right] </math>

===Relation to the gamma function===
If {{math|1=''k''<sup>2</sup> = ''λ''(''i''{{sqrt|''r''}})}} and <math>r \isin \mathbb{Q}^+</math> (where {{mvar|λ}} is the [[Modular lambda function|modular lambda function]]), then {{math|''K''(''k'')}} is expressible in closed form in terms of the [[Gamma function|gamma function]].<ref>{{Cite book |last1=Borwein |first1=Jonathan M. |last2=Borwein| first2=Peter B. |title=Pi and the AGM: A Study in Analytic Number Theory and Computational Complexity |publisher=Wiley-Interscience |year=1987 |edition=First |isbn=0-471-83138-7}} p. 296</ref> For example, {{math|1=''r'' = 2}}, {{math|1=''r'' = 3}} and {{math|1=''r'' = 7}} give, respectively,<ref>{{Cite book |last1=Borwein |first1=Jonathan M. |last2=Borwein| first2=Peter B. |title=Pi and the AGM: A Study in Analytic Number Theory and Computational Complexity |publisher=Wiley-Interscience |year=1987 |edition=First |isbn=0-471-83138-7}} p. 298</ref>

<math display="block">K\left(\sqrt{2}-1\right)=\frac{\Gamma \left(\frac18\right)\Gamma \left(\frac38\right)\sqrt{\sqrt{2}+1}}{8\sqrt[4]{2}\sqrt{\pi}},</math>

and

<math display="block">K\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)=\frac{1}{8\pi}\sqrt[4]{3}\,\sqrt[3]{4}\,\Gamma\biggl(\frac{1}{3}\biggr)^3</math>

and

<math display="block">K\left(\frac{3-\sqrt{7}}{4\sqrt{2}}\right)=\frac{\Gamma \left(\frac17\right)\Gamma \left(\frac27\right)\Gamma \left(\frac47\right)}{4\sqrt[4]{7}\pi}.</math>

More generally, the condition that
<math display="block">\frac{iK'}{K}=\frac{iK\left(\sqrt{1-k^2}\right)}{K(k)}</math>
be in an [[Quadratic field|imaginary quadratic field]]<ref group="note">{{mvar|K}} can be [[Analytic continuation|analytically extended]] to the [[Complex plane|complex plane]].</ref> is sufficient.<ref>{{Cite journal|title=On Epstein's Zeta Function (I).|last1=Chowla|first1=S.|last2=Selberg|first2=A.|journal=Proceedings of the National Academy of Sciences|year=1949|volume=35|issue=7|page=373|doi=10.1073/PNAS.35.7.371|pmid=16588908|pmc=1063041|bibcode=1949PNAS...35..371C|s2cid=45071481}}</ref><ref>{{Cite journal|url=https://eudml.org/doc/150803|title=On Epstein's Zeta-Function|last1=Chowla|first1=S.|last2=Selberg|first2=A.|journal=Journal für die Reine und Angewandte Mathematik|year=1967|volume=227|pages=86–110}}</ref> For instance, if {{math|1=''k'' = ''e''<sup>5''πi''/6</sup>}}, then {{math|1={{sfrac|''iK''{{prime}}|''K''}} = ''e''<sup>2''πi''/3</sup>}} and<ref>{{Cite web|url=https://fungrim.org/topic/Legendre_elliptic_integrals/|title = Legendre elliptic integrals (Entry 175b7a)}}</ref>

<math display="block">K\left(e^{5\pi i/6}\right)=\frac{e^{-\pi i/12}\Gamma ^3\left(\frac13\right)\sqrt[4]{3}}{4\sqrt[3]{2}\pi}.</math>

===Relation to Jacobi theta function===
The relation to [[Theta function|Jacobi's theta function]] is given by 
<math display="block">K(k)=\frac{\pi}{2}\theta_3^2(q),</math>
where the [[Nome (mathematics)|nome]] {{math|''q''}} is
<math display="block">q(k) = \exp\left(-\pi \frac{K\left(\sqrt{1-k^2}\right)}{K(k)}\right).</math>

===Asymptotic expressions===
<math display="block">K\left(k\right)\approx\frac{\pi}{2}+\frac{\pi}{8}\frac{k^2}{1-k^2}-\frac{\pi}{16}\frac{k^4}{1-k^2}</math>
This approximation has a relative precision better than {{val|3|e=−4}} for {{math|''k'' < {{sfrac|1|2}}}}. Keeping only the first two terms is correct to 0.01 precision for {{math|''k'' < {{sfrac|1|2}}}}.{{citation needed|date=January 2017}}

===Differential equation===
The differential equation for the elliptic integral of the first kind is
<math display="block">\frac{d}{dk}\left(k\left(1-k^2\right)\frac{dK(k)}{dk}\right) = k \, K(k)</math>

A second solution to this equation is <math>K\left(\sqrt{1-k^2}\right)</math>. This solution satisfies the relation
<math display="block">\frac{d}{dk}K(k) = \frac{E(k)}{k\left(1-k^2\right)}-\frac{K(k)}{k}.</math>

===Continued fraction===
A [[Continued fraction|continued fraction]] expansion is:<ref>N.Bagis,L.Glasser.(2015)"Evaluations of a Continued fraction of Ramanujan". Rend.Sem.Mat.Univ.Padova, Vol.133 pp 1-10</ref> 
<math display="block">\frac{K(k)}{2\pi} = -\frac{1}{4} + \sum^{\infty}_{n=0} \frac{q^n}{1+q^{2n}} = -\frac{1}{4} + \cfrac{1}{1-q+ \cfrac{\left(1-q\right)^2}{1-q^3+ \cfrac{q\left(1-q^2\right)^2}{1-q^5+ \cfrac{q^2\left(1-q^3\right)^2}{1-q^7+\cfrac{q^3\left(1-q^4\right)^2}{1-q^9+\cdots}}}}},</math>
where the [[Nome (mathematics)|nome]] is <math> q = q(k) = \exp[-\pi K'(k)/K(k)] </math> in its definition.

==Complete elliptic integral of the second kind==
<!-- This section was copied from [[Ellipse]] -->
<!-- This section is redirected from Complete elliptic integral of the second kind -->
[[Image:Mplwp complete ellipticEk.svg|thumb|300px|Plot of the complete elliptic integral of the second kind {{math|''E''(''k'')}}]]

The '''complete elliptic integral of the second kind''' {{math|''E''}} is defined as

<math display="block">E(k) = \int_0^\tfrac{\pi}{2} \sqrt{1-k^2 \sin^2\theta} \, d\theta = \int_0^1 \frac{\sqrt{1-k^2 t^2}}{\sqrt{1-t^2}} \, dt,</math>

or more compactly in terms of the incomplete integral of the second kind {{math|''E''(''φ'',''k'')}} as

<math display="block">E(k) = E\left(\tfrac{\pi}{2},k\right) = E(1;k).</math>

For an ellipse with semi-major axis {{math|''a''}} and semi-minor axis {{math|''b''}} and eccentricity {{math|1=''e'' = {{sqrt|1 − ''b''<sup>2</sup>/''a''<sup>2</sup>}}}}, the complete elliptic integral of the second kind {{math|''E''(''e'')}} is equal to one quarter of the [[Ellipse#Circumference|circumference]] {{math|''C''}} of the ellipse measured in units of the semi-major axis {{math|''a''}}. In other words:

<math display="block">C = 4 a E(e).</math>

The complete elliptic integral of the second kind can be expressed as a [[Power series|power series]]<ref>{{Cite web|url=https://functions.wolfram.com/EllipticIntegrals/EllipticE/06/01/03/01/0003/|title=Complete elliptic integral of the second kind: Series representations (Formula 08.01.06.0002)}}</ref>

<math display="block">E(k) = \frac{\pi}{2}\sum_{n=0}^\infty \left(\frac{(2n)!}{2^{2n} \left(n!\right)^2}\right)^2 \frac{k^{2n}}{1-2n},</math>

which is equivalent to

<math display="block">E(k) = \frac{\pi}{2}\left(1-\left(\frac12\right)^2 \frac{k^2}{1}-\left(\frac{1\cdot 3}{2\cdot 4}\right)^2 \frac{k^4}{3}-\cdots-\left(\frac{(2n-1)!!}{(2n)!!}\right)^2 \frac{k^{2n}}{2n-1}-\cdots\right).</math>

In terms of the Gauss hypergeometric function, the complete elliptic integral of the second kind can be expressed as

<math display="block">E(k) = \tfrac{\pi}{2} \,{}_2F_1 \left(\tfrac12, -\tfrac12; 1; k^2 \right).</math>

The modulus can be transformed that way:
<math display="block">E(k) = \left(1+\sqrt{1-k^2}\right)\,E\left(\frac{1-\sqrt{1-k^2}}{1+\sqrt{1-k^2}}\right) - \sqrt{1-k^2}\,K(k) </math>

===Computation===

Like the integral of the first kind, the complete elliptic integral of the second kind can be computed very efficiently using the arithmetic–geometric mean.{{sfn|Carlson|2010|loc=19.8}}

Define sequences {{mvar|a<sub>n</sub>}} and {{mvar|g<sub>n</sub>}}, where {{math|1=''a''<sub>0</sub> = 1}}, {{math|1=''g''<sub>0</sub> = {{sqrt|1 − ''k''<sup>2</sup>}} = ''k''{{prime}}}} and the recurrence relations {{math|1=''a''<sub>''n'' + 1</sub> = {{sfrac|''a<sub>n</sub>'' + ''g<sub>n</sub>''|2}}}}, {{math|1=''g''<sub>''n'' + 1</sub> = {{sqrt|''a<sub>n</sub> g<sub>n</sub>''}}}} hold. Furthermore, define
<math display="block">c_n=\sqrt{\left|a_n^2-g_n^2\right|}.</math>

By definition,

<math display="block">a_\infty = \lim_{n\to\infty} a_n = \lim_{n\to\infty} g_n = \operatorname{agm}\left(1, \sqrt{1-k^2}\right).</math>

Also

<math display="block">\lim_{n\to\infty} c_n=0.</math>

Then

<math display="block">E(k) = \frac{\pi}{2a_\infty}\left(1-\sum_{n=0}^{\infty} 2^{n-1} c_n^2\right).</math>

In practice, the arithmetic-geometric mean would simply be computed up to some limit. This formula converges quadratically for all {{math|{{abs|''k''}} ≤ 1}}. To speed up computation further, the relation {{math|1=''c''<sub>''n'' + 1</sub> = {{sfrac|''c<sub>n</sub>''<sup>2</sup>|4''a''<sub>''n'' + 1</sub>}}}} can be used.

Furthermore, if {{math|1=''k''<sup>2</sup> = ''λ''(''i''{{sqrt|''r''}})}} and <math>r \isin \mathbb{Q}^+</math> (where {{mvar|λ}} is the [[Modular lambda function|modular lambda function]]), then {{math|''E''(''k'')}} is expressible in closed form in terms of
<math display="block">K(k)=\frac{\pi}{2\operatorname{agm}\left(1,\sqrt{1-k^2}\right)}</math>
and hence can be computed without the need for the infinite summation term. For example, {{math|1=''r'' = 1}}, {{math|1=''r'' = 3}} and {{math|1=''r'' = 7}} give, respectively,<ref>{{Cite book |last1=Borwein |first1=Jonathan M. |last2=Borwein| first2=Peter B. |title=Pi and the AGM: A Study in Analytic Number Theory and Computational Complexity |publisher=Wiley-Interscience |year=1987 |edition=First |isbn=0-471-83138-7}} p. 26, 161</ref>

<math display="block">E\left(\frac{1}{\sqrt{2}}\right)=\frac{1}{2}K\left(\frac{1}{\sqrt{2}}\right)+\frac{\pi}{4K\left(\frac{1}{\sqrt{2}}\right)},</math>

and

<math display="block">E\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)=\frac{3+\sqrt{3}}{6}K\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)+\frac{\pi\sqrt{3}}{12K\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)},</math>

and

<math display="block">E\left(\frac{3-\sqrt{7}}{4\sqrt{2}}\right)=\frac{7+2\sqrt{7}}{14}K\left(\frac{3-\sqrt{7}}{4\sqrt{2}}\right)+\frac{\pi\sqrt{7}}{28K\left(\frac{3-\sqrt{7}}{4\sqrt{2}}\right)}.</math>

===Derivative and differential equation===
<math display="block">\frac{dE(k)}{dk} = \frac{E(k)-K(k)}{k}</math>
<math display="block">\left(k^2-1\right) \frac{d}{dk} \left( k \;\frac{dE(k)}{dk} \right) = k E(k)</math>

A second solution to this equation is {{math|''E''({{sqrt|1 − ''k''<sup>2</sup>}}) − ''K''({{sqrt|1 − ''k''<sup>2</sup>}})}}.

==Complete elliptic integral of the third kind==
[[Image:Mplwp complete ellipticPi nfixed k.svg|thumb|300px|Plot of the complete elliptic integral of the third kind {{math|Π(''n'',''k'')}} with several fixed values of {{mvar|n}}]]
The '''complete elliptic integral of the third kind''' {{math|Π}} can be defined as

<math display="block">\Pi(n,k) = \int_0^\frac{\pi}{2} \frac{d\theta}{\left(1-n\sin^2\theta\right)\sqrt{1-k^2 \sin^2\theta}}.</math>

Note that sometimes the elliptic integral of the third kind is defined with an inverse sign for the ''characteristic'' {{math|''n''}},
<math display="block">\Pi'(n,k) = \int_0^\frac{\pi}{2} \frac{d\theta}{\left(1+n\sin^2\theta\right)\sqrt{1-k^2 \sin^2\theta}}.</math>

Just like the complete elliptic integrals of the first and second kind, the complete elliptic integral of the third kind can be computed very efficiently using the arithmetic-geometric mean.{{sfn|Carlson|2010|loc=19.8}}

===Partial derivatives===
<math display="block">\begin{align}
\frac{\partial\Pi(n,k)}{\partial n} &= \frac{1}{2\left(k^2-n\right)(n-1)}\left(E(k)+\frac{1}{n}\left(k^2-n\right)K(k) + \frac{1}{n} \left(n^2-k^2\right)\Pi(n,k)\right) \\[8pt]

\frac{\partial\Pi(n,k)}{\partial k} &= \frac{k}{n-k^2}\left(\frac{E(k)}{k^2-1}+\Pi(n,k)\right)
\end{align}</math>

==Jacobi zeta function==
In 1829, Jacobi defined the '''Jacobi zeta function''':
<math display="block">Z(\varphi,k)=E(\varphi,k)-\frac{E(k)}{K(k)}F(\varphi,k).</math>
It is periodic in <math>\varphi</math> with minimal period <math>\pi</math>. It is related to the [[Jacobi elliptic functions|Jacobi zn function]] by <math>Z(\varphi,k)=\operatorname{zn}(F(\varphi,k),k)</math>. In the literature (e.g. Whittaker and Watson (1927)), sometimes <math>Z</math> means Wikipedia's <math>\operatorname{zn}</math>. Some authors (e.g. King (1924)) use <math>Z</math> for both Wikipedia's <math>Z</math> and <math>\operatorname{zn}</math>.

==Legendre's relation==

The [[Legendre's relation]] or ''Legendre Identity'' shows the relation of the integrals K and E of an elliptic modulus and its anti-related counterpart<ref>{{cite web|access-date=2022-11-29|language=de|title=Legendre-Relation|url=https://www.spektrum.de/lexikon/mathematik/legendre-relation/5873}}<!-- auto-translated by Module:CS1 translator --></ref><ref>{{cite web|access-date=2022-11-29|title=Legendre Relation|url=https://archive.lib.msu.edu/crcmath/math/math/l/l179.htm}}<!-- auto-translated by Module:CS1 translator --></ref> in an integral equation of second degree:

For two modules that are Pythagorean counterparts to each other, this relation is valid:

<math display="block">K(\varepsilon) E\left(\sqrt{1-\varepsilon^2}\right) + E(\varepsilon) K\left(\sqrt{1-\varepsilon^2}\right) - K(\varepsilon) K\left(\sqrt{1-\varepsilon^2}\right) = \frac {\pi}{2}</math>

For example:
: <math>K({\color{blueviolet}\tfrac{3}{5}})E({\color{blue}\tfrac{4}{5}}) + E({\color{blueviolet}\tfrac{3}{5}})K({\color{blue}\tfrac{4}{5}}) - K({\color{blueviolet}\tfrac{3}{5}})K({\color{blue}\tfrac{4}{5}}) = \tfrac{1}{2}\pi</math>

And for two modules that are tangential counterparts to each other, the following relationship is valid:

: <math>(1 + \varepsilon)K(\varepsilon)E(\tfrac{1 - \varepsilon}{1 + \varepsilon}) + \tfrac{2}{1 + \varepsilon}E(\varepsilon)K (\tfrac{1 - \varepsilon}{1 + \varepsilon}) - 2K(\varepsilon)K(\tfrac{1 - \varepsilon}{1 + \varepsilon}) = \tfrac{1}{2}\pi </math>

For example:
: <math>\tfrac{4}{3}K({\color{blue}\tfrac{1}{3}})E({\color{green}\tfrac{1}{2}}) + \tfrac{3}{2}E({\color{blue}\tfrac{1}{3}})K({\color{green}\tfrac{1}{2}}) - 2K({\color{blue}\tfrac{1}{3}})K({\color{green}\tfrac{1}{2}}) = \tfrac{1}{2}\pi</math>

The Legendre's relation for tangential modular counterparts results directly from the Legendre's identity for Pythagorean modular counterparts by using the [[Landen's transformation|Landen modular transformation]] on the Pythagorean counter modulus.

=== Special identity for the lemniscatic case ===
For the lemniscatic case, the elliptic modulus or specific eccentricity ε is equal to half the square root of two. Legendre's identity for the lemniscatic case can be proved as follows:

According to the [[Chain rule]] these derivatives hold:

: <math>\frac{\mathrm{d}}{\mathrm{d}y} \,K\bigl(\frac{1}{2}\sqrt{2}\bigr) - F\biggl[\arccos (xy);\frac{1}{2}\sqrt{2}\biggr] = \frac{\sqrt{2}\,x}{\sqrt{1 - x^4 y^4}}</math>
: <math>\frac{\mathrm{d}}{\mathrm{d}y} \,2E\bigl(\frac{1}{2}\sqrt{2}\bigr) - K\bigl(\frac {1}{2}\sqrt{2}\bigr) - 2E\biggl[\arccos(xy);\frac{1}{2}\sqrt{2}\biggr] + F\biggl[\arccos(xy );\frac{1}{2}\sqrt{2}\biggr] = \frac{\sqrt{2}\,x^3 y^2}{\sqrt{1 - x^4 y^4}} </math>

By using the [[Fundamental theorem of calculus]] these formulas can be generated:

: <math>K\bigl(\frac{1}{2}\sqrt{2}\bigr) - F\biggl[\arccos (x);\frac{1}{2}\sqrt{2}\biggr] = \int_{0}^{1} \frac{\sqrt{2}\,x}{\sqrt{1 - x^4 y^4}} \,\mathrm{d}y </math>
: <math>2E\bigl(\frac{1}{2}\sqrt{2}\bigr) - K\bigl(\frac {1}{2}\sqrt{2}\bigr) - 2E\biggl[\arccos(x);\frac{1}{2}\sqrt{2}\biggr] + F\biggl[\arccos(x);\frac{1}{2}\sqrt{2}\biggr] = \int_{0}^{1} \frac{\sqrt{2}\,x^3 y^2}{\sqrt{1 - x^4 y^4}} \,\mathrm{d}y </math>

The [[Linear combination]] of the two now mentioned integrals leads to the following formula:

: <math>
\frac{\sqrt{2}}{\sqrt{1 - x^4}} \biggl\{2E\bigl(\frac{1}{2}\sqrt{2}\bigr) - K\bigl(\frac{1}{2}\sqrt{2}\bigr) - 2E\biggl[\arccos(x);\frac{1}{2}\sqrt{2}\biggr] + F\biggl[\arccos( x);\frac{1}{2}\sqrt{2}\biggr]\biggr\} \,+
</math>
: <math>
+ \,\frac{\sqrt{2} \,x^2}{\sqrt{1 - x^4}} \biggl\{K\bigl(\frac{1}{2}\sqrt{2}\bigr) - F\biggl[\arccos(x);\frac{1}{2}\sqrt{2}\biggr]\biggr\} = \int_{0}^{1} \frac{2\,x ^3 (y^2 + 1)}{\sqrt{(1 - x^4)(1 - x^4\,y^4)}} \,\mathrm{d}y
</math>

By forming the original antiderivative related to x from the function now shown using the [[Product rule]] this formula results:

: <math> \biggl\{K\bigl(\frac{1}{2}\sqrt{2}\bigr) - F\biggl[\arccos(x);\frac{1}{2}\sqrt{ 2}\biggr]\biggr\}\biggl\{2E\bigl(\frac{1}{2}\sqrt{2}\bigr) - K\bigl(\frac{1}{2}\sqrt{2 }\bigr) - 2E\biggl[\arccos(x);\frac{1}{2}\sqrt{2}\biggr] + F\biggl[\arccos(x);\frac{1}{2} \sqrt{2}\biggr]\biggr\} =
</math>
: <math> = \int_{0}^{1} \frac{1}{y^2}(y^2 + 1)\biggl[\text{artanh}(y^2) - \text{artanh} \bigl(\frac{\sqrt{1 - x^4}\,y^2}{\sqrt{1 - x^4 y^4}}\bigr)\biggr] \mathrm{d}y </math>

If the value <math>x = 1</math> is inserted in this integral identity, then the following identity emerges:

: <math> K\bigl(\frac{1}{2}\sqrt{2}\bigr)\biggl[2\,E\bigl(\frac{1}{2}\sqrt{2}\bigr) - K\bigl (\frac{1}{2}\sqrt{2}\bigr)\biggr] = \int_{0}^{1} \frac{1}{y^2}(y^2 + 1) \,\text{artanh}(y^2) \,\mathrm{d}y = </math>

: <math> = \biggl[2\arctan(y) - \frac{1}{y}(1 - y^2)\,\text{artanh}(y^2)\biggr]_{y = 0}^{y = 1} = 2\arctan(1) = \frac{\pi}{2} </math>

This is how this lemniscatic excerpt from Legendre's identity appears:

: <math>2E\bigl(\frac{1}{2}\sqrt{2}\bigr)K\bigl(\frac{1}{2}\sqrt{2}\bigr) - K\bigl(\frac{1}{2}\sqrt{2}\bigr)^2 = \frac{\pi}{2}</math>

=== Generalization for the overall case ===
Now the modular general case<ref>{{cite web|access-date=2023-02-10|language=en|title=integration - Proving Legendres Relation for elliptic curves|url=https://math.stackexchange.com/questions/701515/proving-legendres-relation-for-elliptic-curves}}<!-- auto-translated by Module:CS1 translator --></ref><ref>{{citation|access-date=2023-02-10|author=Internet Archive|date=1991|isbn=0-387-97509-8|publisher=New York : Springer-Verlag|title=Paul Halmos celebrating 50 years of mathematics|url=https://archive.org/details/paulhalmoscelebr0000unse}}<!-- auto-translated by Module:CS1 translator --></ref>  is worked out. For this purpose, the derivatives of the complete elliptic integrals are derived after the modulus <math> \varepsilon </math> and then they are combined. And then the Legendre's identity balance is determined.

Because the derivative of the ''circle function'' is the negative product of the ''identical mapping function'' and the reciprocal of the circle function:

: <math>\frac{\mathrm{d}}{\mathrm{d}\varepsilon}\sqrt{1 - \varepsilon^2} = -\,\frac{\varepsilon}{\sqrt{1 - \varepsilon^2}}</math>

These are the derivatives of K and E shown in this article in the sections above:

: <math>\frac{\mathrm{d}}{\mathrm{d}\varepsilon} K(\varepsilon) = \frac{1}{\varepsilon(1-\varepsilon^2)} \bigl[E( \varepsilon) - (1-\varepsilon^2)K(\varepsilon)\bigr]</math>
: <math>\frac{\mathrm{d}}{\mathrm{d}\varepsilon} E(\varepsilon) = - \,\frac{1}{\varepsilon}\bigl[K(\varepsilon) - E (\varepsilon)\bigr]</math>

In combination with the derivative of the circle function these derivatives are valid then:

: <math>\frac{\mathrm{d}}{\mathrm{d}\varepsilon}K(\sqrt{1 - \varepsilon^2}) = \frac{1}{\varepsilon(1-\varepsilon^ 2)} \bigl[\varepsilon^2 K(\sqrt{1 - \varepsilon^2}) - E(\sqrt{1 - \varepsilon^2})\bigr]</math>
: <math>\frac{\mathrm{d}}{\mathrm{d}\varepsilon }E(\sqrt{1 - \varepsilon ^2}) = \frac{\varepsilon }{1 - \varepsilon ^2} \bigl[K(\sqrt{1 - \varepsilon^2}) - E(\sqrt{1 - \varepsilon^2})\bigr]</math>

Legendre's identity includes products of any two complete elliptic integrals. For the derivation of the function side from the equation scale of Legendre's identity, the [[Product rule]] is now applied in the following:

: <math>\frac{\mathrm{d}}{\mathrm{d}\varepsilon}K(\varepsilon)E(\sqrt{1 - \varepsilon^2}) = \frac{1}{\varepsilon( 1-\varepsilon^2)} \bigl[E(\varepsilon)E(\sqrt{1 - \varepsilon^2}) - K(\varepsilon)E(\sqrt{1 - \varepsilon^2}) + \varepsilon^2 K(\varepsilon)K(\sqrt{1 - \varepsilon^2})\bigr]</math>
: <math>\frac{\mathrm{d}}{\mathrm{d}\varepsilon}E(\varepsilon)K(\sqrt{1 - \varepsilon^2}) = \frac{1}{\varepsilon( 1-\varepsilon^2)} \bigl[- E(\varepsilon)E(\sqrt{1 - \varepsilon^2}) + E(\varepsilon)K(\sqrt{1 - \varepsilon^2}) - (1 - \varepsilon^2) K(\varepsilon)K(\sqrt{1 - \varepsilon^2})\bigr]</math>
: <math>\frac{\mathrm{d}}{\mathrm{d}\varepsilon}K(\varepsilon)K(\sqrt{1 - \varepsilon^2}) = \frac{1}{\varepsilon( 1-\varepsilon^2)} \bigl[E(\varepsilon)K(\sqrt{1 - \varepsilon^2}) - K(\varepsilon)E(\sqrt{1 - \varepsilon^2}) - ( 1 - 2\varepsilon^2) K(\varepsilon)K(\sqrt{1 - \varepsilon^2})\bigr]</math>

Of these three equations, adding the top two equations and subtracting the bottom equation gives this result:

: <math>\frac{\mathrm{d}}{\mathrm{d}\varepsilon} \bigl[K(\varepsilon)E(\sqrt{1 - \varepsilon^2}) + E(\varepsilon)K (\sqrt{1 - \varepsilon^2}) - K(\varepsilon)K(\sqrt{1 - \varepsilon^2})\bigr] = 0</math>

In relation to the <math> \varepsilon </math> the equation balance constantly gives the value zero.

The previously determined result shall be combined with the Legendre equation to the modulus <math>\varepsilon = 1/\sqrt{2}</math> that is worked out in the section before:

: <math>2E\bigl(\frac{1}{2}\sqrt{2}\bigr)K\bigl(\frac{1}{2}\sqrt{2}\bigr) - K\bigl(\frac{1}{2}\sqrt{2}\bigr)^2 = \frac{\pi}{2}</math>

The combination of the last two formulas gives the following result:

: <math>K(\varepsilon)E(\sqrt{1 - \varepsilon^2}) + E(\varepsilon)K(\sqrt{1 - \varepsilon^2}) - K(\varepsilon)K(\sqrt{1 - \varepsilon^2}) = \tfrac{1}{2}\pi</math>

Because if the derivative of a continuous function constantly takes the value zero, then the concerned function is a constant function. This means that this function results in the same function value for each abscissa value <math> \varepsilon </math> and the associated function graph is therefore a horizontal straight line.

==See also==
{{div col}}
* [[Elliptic curve]]
* [[Schwarz–Christoffel mapping]]
* [[Carlson symmetric form]]
* Jacobi's elliptic functions
* [[Weierstrass's elliptic functions]]
* [[Theta function|Jacobi theta function]]
* [[Ramanujan theta function]]
* [[Arithmetic–geometric mean]]
* [[Pendulum (mathematics)#Arbitrary-amplitude period|Pendulum period]]
* [[Earth:Meridian arc#Relation to elliptic integrals|Meridian arc]]
{{div col end}}

==References==
===Notes===
{{reflist|group=note}}

===References===
{{reflist}}

===Sources===
{{refbegin}}
*{{AS ref|17|587}}
*{{cite book |last1=Byrd|first1=P. F.|last2=Friedman|first2=M.D.|date=1971|title=Handbook of Elliptic Integrals for Engineers and Scientists|publisher=Springer-Verlag|location=New York|edition=2nd|isbn=0-387-05318-2}}
*{{cite journal | first=B. C. | last=Carlson | year=1995 | journal=Numerical Algorithms | title=Numerical Computation of Real or Complex Elliptic Integrals | volume=10 | issue=1 | pages=13–26 | arxiv=math/9409227 | bibcode=1995NuAlg..10...13C | doi=10.1007/BF02198293 | s2cid=11580137 }}
*{{dlmf|first=B. C.|last=Carlson|id=19}}
*{{cite book | url=http://apps.nrbook.com/bateman/Vol2.pdf | title=Higher transcendental functions. Vol II | last2=Magnus | first2=Wilhelm | last3=Oberhettinger | first3=Fritz | last4=Tricomi | first4=Francesco G. | publisher=McGraw-Hill Book Company, Inc., New York-Toronto-London | year=1953 | mr=0058756 | last1=Erdélyi | first1=Arthur | }}
*{{cite book |author-first1=Izrail Solomonovich |author-last1=Gradshteyn |author-first2=Iosif Moiseevich |author-last2=Ryzhik |author-link2=Iosif Moiseevich Ryzhik |author-first3=Yuri Veniaminovich |author-last3=Geronimus |author-link3=Yuri Veniaminovich Geronimus |author-first4=Michail Yulyevich |author-last4=Tseytlin |author-link4=Michail Yulyevich Tseytlin |author-first5=Alan |author-last5=Jeffrey |editor-first1=Daniel |editor-last1=Zwillinger |editor-first2=Victor Hugo |editor-last2=Moll |translator=Scripta Technica, Inc. |title=Table of Integrals, Series, and Products |publisher=Academic Press, Inc. |date=2015 |orig-year=October 2014 |edition=8 |language=en |isbn=978-0-12-384933-5 |lccn=2014010276 <!-- |url=https://books.google.com/books?id=NjnLAwAAQBAJ |access-date=2016-02-21 --> |title-link=Gradshteyn and Ryzhik |chapter=8.1. }}
* {{cite book | first=Alfred George | last=Greenhill |  url=https://archive.org/details/applicationselli00greerich | title=The applications of elliptic functions | location=New York |publisher=Macmillan | year=1892 }}
*{{cite book | last=Hancock | first=Harris |  year=1910 | title=Lectures on the Theory of Elliptic Functions | url=https://archive.org/details/lecturestheorell00hancrich | publisher=J. Wiley & sons | location=New York }}
*{{cite book | first=Louis V. | last=King | url=https://archive.org/details/onthenumerical032686mbp | title=On The Direct Numerical Calculation Of Elliptic Functions And Integrals | publisher=Cambridge University Press | year=1924 }}
* {{Citation | last1=Press | first1=W. H. | last2=Teukolsky | first2=S. A. | last3=Vetterling | first3=W. T. | last4=Flannery | first4=B. P. | year=2007 | title=Numerical Recipes: The Art of Scientific Computing | edition=3rd | publisher=Cambridge University Press | publication-place=New York | isbn=978-0-521-88068-8 | chapter=Section 6.12. Elliptic Integrals and Jacobian Elliptic Functions | chapter-url=http://apps.nrbook.com/empanel/index.html#pg=309 }}
{{refend}}

==External links==
*{{springer|title=Elliptic integral|id=p/e035490}}
*[http://mathworld.wolfram.com/EllipticIntegral.html Eric W. Weisstein, "Elliptic Integral" (Mathworld)]
*[https://github.com/moiseevigor/elliptic Matlab code for elliptic integrals evaluation] by elliptic project
*[http://www.exstrom.com/math/elliptic/ellipint.html Rational Approximations for Complete Elliptic Integrals] (Exstrom Laboratories)
*[https://scholar.rose-hulman.edu/cgi/viewcontent.cgi?article=1148&context=rhumj A Brief History of Elliptic Integral Addition Theorems]

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