椭圆积分

在积分学中，椭圆积分最初出现于椭圆的弧长有关的问题中。朱利奥·法尼亚诺（英语：Guilio Fagnano）和欧拉是最早的研究者。现代数学将椭圆积分定义为可以表达为如下形式的任何函数 f {\displaystyle f\,} 的积分 f ( x ) = ∫ c x R [ t , P ( t ) ]   d t {\displaystyle f(x)=\int _{c}^{x}R[t,{\sqrt[{}]{P(t)}}]\ dt\,\!} 其中 R {\displaystyle R\,} 是其两个参数的有理函数， P {\displaystyle P\,} 是一个无重根的 3 {\displaystyle 3\,} 或 4 {\displaystyle 4\,} 阶多项式，而 c {\displaystyle c\,} 是一个常数。 通常，椭圆积分不能用基本函数表达。这个一般规则的例外出现在 P {\displaystyle P\,} 有重根的时候，或者是 R {\displaystyle R\,} , ( x , y ) {\displaystyle \left(x,y\right)\,} 没有 y {\displaystyle y\,} 的奇数幂时。但是，通过适当的简化公式，每个椭圆积分可以变为只涉及有理函数和三个经典形式的积分。（也即，第一，第二，和第三类的椭圆积分）。 除下面给出的形式之外，椭圆积分也可以表达为勒让德形式和Carlson对称形式。通过对施瓦茨-克里斯托费尔映射的研究可以加深对椭圆积分理论的理解。历史上，椭圆函数是作为椭圆积分的逆函数被发现的，特别是这一个： F [ sn ( z ; k ) ; k ] = z {\displaystyle F[{\textrm {sn}}\left(z;k\right);k]=z\,} 其中 sn {\displaystyle {\textrm {sn}}\,} 是雅可比正弦椭圆函数。 椭圆积分通常表述为不同变量的函数。这些变量完全等价（它们给出同样的椭圆积分），但是它们看起来很不相同。很多文献使用单一一种标准命名规则。在定义积分之前，先来检视一下这些变量的命名常规： α {\displaystyle \alpha } 模角; k = sin ⁡ α {\displaystyle k=\sin \alpha } 椭圆模; m = k 2 = sin 2 ⁡ α {\displaystyle m=k^{2}=\sin ^{2}\alpha } 参数; 上述三种常规完全互相确定。规定其中一个和规定另外一个一样。椭圆积分也依赖于另一个变量，可以有如下几种不同的设定方法： ϕ {\displaystyle \phi \,\!} 幅度 x {\displaystyle x\,} 其中 x = sin ⁡ ϕ = sn u {\displaystyle x=\sin \phi ={\textrm {sn}}\;u\,\!} u {\displaystyle u\,} ，其中 x = sn u {\displaystyle x={\textrm {sn}}\;u\,} 而 sn {\displaystyle {\textrm {sn}}\,} 是雅可比椭圆函数之一 规定其中一个决定另外两个。这样，它们可以互换地使用。注意 u {\displaystyle u\,} 也依赖于 m {\displaystyle m\,} 。其它包含 u {\displaystyle u\,} 的关系有 cos ⁡ ϕ = cn u {\displaystyle \cos \phi ={\textrm {cn}}\;u\,\!} 和 1 − m sin 2 ⁡ ϕ = dn u . {\displaystyle {\sqrt {1-m\sin ^{2}\phi }}={\textrm {dn}}\;u.\,\!} 后者有时称为δ幅度并写作 Δ ( ϕ ) = dn u {\displaystyle \Delta (\phi )={\textrm {dn}}\;u\,\!} 。有时文献也称之为补参数，补模或者补模角。这些在四分周期中有进一步的定义。 第一类不完全椭圆积分 F {\displaystyle F\,} 定义为 F ( ϕ ∖ α ) = F ( ϕ | m ) = ∫ 0 ϕ d θ 1 − ( sin ⁡ θ sin ⁡ α ) 2 . {\displaystyle F(\phi \setminus \alpha )=F(\phi |m)=\int _{0}^{\phi }{\frac {{\rm {d}}\theta }{\sqrt {1-(\sin \theta \sin \alpha )^{2}}}}.\,\!} 与此等价，用雅可比的形式，可以设 x = sin ⁡ ϕ   ,   t = sin ⁡ θ {\displaystyle x=\sin \phi ~,~t=\sin \theta \;\!} ；则 F ( ϕ ∖ α ) = F ( x ; k ) = ∫ 0 x d t ( 1 − t 2 ) ( 1 − k 2 t 2 ) {\displaystyle F(\phi \setminus \alpha )=F(x;k)=\int _{0}^{x}{\frac {{\rm {d}}t}{\sqrt {(1-t^{2})(1-k^{2}t^{2})}}}\,\!} 其中，假定任何有竖直条出现的地方，紧跟竖直条的变量是（如上定义的）参数；而且，当反斜杠出现的时候，跟着出现的是模角。 在这个意义下， F ( sin ⁡ ϕ ; sin ⁡ α ) = F ( ϕ | sin 2 ⁡ α ) = F ( ϕ ∖ α )   {\displaystyle F(\sin \phi ;\sin \alpha )=F(\phi |\sin ^{2}\alpha )=F(\phi \setminus \alpha )~\,\!} ，这里的记法来自标准参考书Abramowitz and Stegun。 但是，还有许多不同的用于椭圆积分的记法。取值为椭圆积分的函数没有（像平方根，正弦和误差函数那样的）标准和唯一的名字。甚至关于该领域的文献也常常采用不同的记法。Gradstein, Ryzhik[1] （页面存档备份，存于互联网档案馆）, E q {\displaystyle Eq\,} .(8.111)]采用 F ( ϕ , k ) {\displaystyle F(\phi ,k)\,\!} 。该记法和这里的 F ( ϕ | k 2 )   {\displaystyle F(\phi |k^{2})~\,\!} ；以及下面的 E ( ϕ , k ) = E ( ϕ | k 2 )   {\displaystyle E(\phi ,k)=E(\phi |k^{2})~\,\!} 等价。 和上面的不同对应的是，如果从Mathematica语言翻译代码到Maple语言，必须将EllipticK函数的参数用它的平方根代替。反过来，如果从Maple翻到Mathematica，则参数应该用它的平方代替。Maple中的EllipticK( x {\displaystyle x} )几乎和Mathematica中的EllipticK[ x 2 {\displaystyle x^{2}} ]相等；至少当 0 < x < 1 {\displaystyle 0<x<1\,} 时是相等的。 注意 F ( x ; k ) = u {\displaystyle F(x;k)=u\,\!} 其中 u {\displaystyle u\,} 如上文所定义：由此可见，雅可比椭圆函数是椭圆积分的逆。 ∀ φ 1 , φ 2 ∈ ] − π 2 ; π 2 [ , {\displaystyle \forall \varphi _{1},\varphi _{2}\in \left]-{\frac {\pi }{2}};{\frac {\pi }{2}}\right[,} F ( φ 1 , k ) + F ( φ 2 , k ) = F ( arctan ⁡ ( tan ⁡ φ 1 1 − k 2 sin 2 ⁡ φ 2 ) + arctan ⁡ ( tan ⁡ φ 2 1 − k 2 sin 2 ⁡ φ 1 ) , k ) {\displaystyle F\left(\varphi _{1},k\right)+F\left(\varphi _{2},k\right)=F\left(\arctan \left(\tan \varphi _{1}{\sqrt {1-k^{2}\sin ^{2}\varphi _{2}}}\right)+\arctan \left(\tan \varphi _{2}{\sqrt {1-k^{2}\sin ^{2}\varphi _{1}}}\right),k\right)} arctan ⁡ ( tan ⁡ φ 1 1 − k 2 sin 2 ⁡ φ 2 ) + arctan ⁡ ( tan ⁡ φ 2 1 − k 2 sin 2 ⁡ φ 1 ) ∈ [ − π / 2 ; π / 2 ] ⇒ {\displaystyle \arctan \left(\tan \varphi _{1}{\sqrt {1-k^{2}\sin ^{2}\varphi _{2}}}\right)+\arctan \left(\tan \varphi _{2}{\sqrt {1-k^{2}\sin ^{2}\varphi _{1}}}\right)\in [-\pi /2;\pi /2]\Rightarrow } F ( φ 1 , k ) + F ( φ 2 , k ) = F ( arcsin ⁡ cos ⁡ φ 1 1 − k 2 sin 2 ⁡ φ 1 sin ⁡ φ 2 + cos ⁡ φ 2 1 − k 2 sin 2 ⁡ φ 2 sin ⁡ φ 1 1 − k 2 sin 2 ⁡ φ 1 sin 2 ⁡ φ 2 , k ) {\displaystyle F\left(\varphi _{1},k\right)+F\left(\varphi _{2},k\right)=F\left(\arcsin {\frac {\cos \varphi _{1}{\sqrt {1-k^{2}\sin ^{2}\varphi _{1}}}\sin \varphi _{2}+\cos \varphi _{2}{\sqrt {1-k^{2}\sin ^{2}\varphi _{2}}}\sin \varphi _{1}}{1-k^{2}\sin ^{2}\varphi _{1}\sin ^{2}\varphi _{2}}},k\right)} arctan ⁡ ( tan ⁡ φ 1 1 − k 2 sin 2 ⁡ φ 2 ) + arctan ⁡ ( tan ⁡ φ 2 1 − k 2 sin 2 ⁡ φ 1 ) ∈ [ 0 ; π ] ⇒ {\displaystyle \arctan \left(\tan \varphi _{1}{\sqrt {1-k^{2}\sin ^{2}\varphi _{2}}}\right)+\arctan \left(\tan \varphi _{2}{\sqrt {1-k^{2}\sin ^{2}\varphi _{1}}}\right)\in [0;\pi ]\Rightarrow } F ( φ 1 , k ) + F ( φ 2 , k ) = F ( arccos ⁡ cos ⁡ φ 1 cos ⁡ φ 2 − sin ⁡ φ 1 sin ⁡ φ 2 1 − k 2 sin 2 ⁡ φ 1 1 − k 2 sin 2 ⁡ φ 1 1 − k 2 sin 2 ⁡ φ 2 1 − k 2 sin 2 ⁡ φ 1 sin 2 ⁡ φ 2 , k ) {\displaystyle F\left(\varphi _{1},k\right)+F\left(\varphi _{2},k\right)=F\left(\arccos {\frac {\cos \varphi _{1}\cos \varphi _{2}-\sin \varphi _{1}\sin \varphi _{2}{\sqrt {1-k^{2}\sin ^{2}\varphi _{1}}}{\sqrt {1-k^{2}\sin ^{2}\varphi _{1}}}{\sqrt {1-k^{2}\sin ^{2}\varphi _{2}}}}{1-k^{2}\sin ^{2}\varphi _{1}\sin ^{2}\varphi _{2}}},k\right)} F ( x + n π ; k ) = F ( x ; k ) + 2 n K ( k ) {\displaystyle F(x+n\pi ;k)=F(x;k)+2nK(k)\,\!} F ( x + n π 2 ; k ) = n K ( k ) {\displaystyle F(x+{\frac {n\pi }{2}};k)=nK(k)\,\!} n ∈ Z {\displaystyle n\in \mathbb {Z} \,\!} F ( − x ; k ) = − F ( x ; k ) {\displaystyle F(-x;k)=-F(x;k)\,\!} F ( x ; 0 ) = x {\displaystyle F(x;0)=x\,\!} F ( 0 ; k ) = − F ( x ; k ) {\displaystyle F(0;k)=-F(x;k)\,\!} F ( x ; 1 ) = a r c t a n h sin ⁡ x {\displaystyle F(x;1)={\rm {arctanh}}\sin x\,\!} − π 2 < ℜ ( x ) < π 2 {\displaystyle -{\frac {\pi }{2}}<\Re (x)<{\frac {\pi }{2}}\,\!} d d x F ( x ; k ) = 1 1 − k 2 sin 2 ⁡ x {\displaystyle {\frac {\rm {d}}{{\rm {d}}x}}F(x;k)={\frac {1}{\sqrt {1-k^{2}\sin ^{2}x}}}\,\!} d d k F ( x ; k ) = E ( x ; k ) 2 k ( 1 − k ) − F ( x ; k ) 2 k − sin ⁡ 2 x 4 ( 1 − k ) 1 − k sin 2 ⁡ x {\displaystyle {\frac {\rm {d}}{{\rm {d}}k}}F(x;k)={\frac {E(x;k)}{2k(1-k)}}-{\frac {F(x;k)}{2k}}-{\frac {\sin 2x}{4(1-k){\sqrt {1-k\sin ^{2}x}}}}\,\!} 第二类不完全椭圆积分 E {\displaystyle E\!} 是 E ( ϕ ∖ α ) = E ( ϕ | m ) = ∫ 0 ϕ E ′ ( θ )   d θ = ∫ 0 ϕ 1 − ( sin ⁡ θ sin ⁡ α ) 2   d θ . {\displaystyle E(\phi \setminus \alpha )=E(\phi |m)=\int _{0}^{\phi }\!E'(\theta )\ {\rm {d}}\theta =\int _{0}^{\phi }{\sqrt {1-(\sin \theta \sin \alpha )^{2}}}\ {\rm {d}}\theta .\,\!} 与此等价，采用另外一个记法（作变量替换 t = sin ⁡ θ {\displaystyle t=\sin \theta \,\!} ）， E ( x ; k ) = ∫ 0 x 1 − k 2 t 2 1 − t 2   d t . {\displaystyle E(x;k)=\int _{0}^{x}{\frac {\sqrt {1-k^{2}t^{2}}}{\sqrt {1-t^{2}}}}\ {\rm {d}}t.\,\!} 其它关系包括 E ( ϕ | m ) = ∫ 0 u dn 2 w d w = u − m ∫ 0 u sn 2 w d w = ( 1 − m ) u + m ∫ 0 u cn 2 w d w . {\displaystyle E(\phi |m)=\int _{0}^{u}{\textrm {dn}}^{2}w\;{\rm {d}}w=u-m\int _{0}^{u}{\textrm {sn}}^{2}w\;{\rm {d}}w=(1-m)u+m\int _{0}^{u}{\textrm {cn}}^{2}w\;{\rm {d}}w.\,\!} E ( ϕ | k 2 ) = ( 1 − k 2 ) ∫ 0 ϕ d θ ( 1 − k 2 sin 2 ⁡ θ ) 1 − k 2 sin 2 ⁡ θ + k 2 sin ⁡ θ cos ⁡ θ 1 − k 2 sin 2 ⁡ θ {\displaystyle E(\phi |k^{2})=(1-k^{2})\int _{0}^{\phi }{\frac {{\rm {d}}\theta }{(1-k^{2}\sin ^{2}\theta ){\sqrt {1-k^{2}\sin ^{2}\theta }}}}+{\frac {k^{2}\sin \theta \cos \theta }{\sqrt {1-k^{2}\sin ^{2}\theta }}}\,\!} ∀ φ 1 , φ 2 ∈ ] − π 2 ; π 2 [ , {\displaystyle \forall \varphi _{1},\varphi _{2}\in \left]-{\frac {\pi }{2}};{\frac {\pi }{2}}\right[,} E ( φ 1 , k ) + E ( φ 2 , k ) = [ E ( arctan ⁡ ( tan ⁡ φ 1 1 − k 2 sin 2 ⁡ φ 2 ) + arctan ⁡ ( tan ⁡ φ 2 1 − k 2 sin 2 ⁡ φ 1 ) , k ) + k 2 sin ⁡ φ 1 sin ⁡ φ 2 ( cos ⁡ φ 1 1 − k 2 sin 2 ⁡ φ 1 sin ⁡ φ 2 + cos ⁡ φ 2 1 − k 2 sin 2 ⁡ φ 2 sin ⁡ φ 1 ) 1 − k 2 sin ⁡ φ 1 sin ⁡ φ 2 ] {\displaystyle \textstyle E\left(\varphi _{1},k\right)+E\left(\varphi _{2},k\right)=\left[{\begin{aligned}E\left(\arctan \left(\tan \varphi _{1}{\sqrt {1-k^{2}\sin ^{2}\varphi _{2}}}\right)+\arctan \left(\tan \varphi _{2}{\sqrt {1-k^{2}\sin ^{2}\varphi _{1}}}\right),k\right)\\+{\frac {k^{2}\sin \varphi _{1}\sin \varphi _{2}\left(\cos \varphi _{1}{\sqrt {1-k^{2}\sin ^{2}\varphi _{1}}}\sin \varphi _{2}+\cos \varphi _{2}{\sqrt {1-k^{2}\sin ^{2}\varphi _{2}}}\sin \varphi _{1}\right)}{1-k^{2}\sin \varphi _{1}\sin \varphi _{2}}}\end{aligned}}\right]} E ( ϕ + n π ; k ) = E ( ϕ ; k ) + 2 n E ( k ) {\displaystyle E(\phi +n\pi ;k)=E(\phi ;k)+2nE(k)\,\!} E ( − ϕ ; k ) = − E ( ϕ ; k ) {\displaystyle E(-\phi ;k)=-E(\phi ;k)\,\!} d d ϕ E ( ϕ ; k ) = 1 − k 2 sin 2 ⁡ ϕ {\displaystyle {\frac {\rm {d}}{{\rm {d}}\phi }}E(\phi ;k)={\sqrt {1-k^{2}\sin ^{2}\phi }}\,\!} d d k E ( ϕ ; k ) = E ( ϕ ; k ) − F ( ϕ ; k ) 2 k {\displaystyle {\frac {\rm {d}}{{\rm {d}}k}}E(\phi ;k)={\frac {E(\phi ;k)-F(\phi ;k)}{2k}}\,\!} d n d k n E ( ϕ ; k ) = π 2 k n 2 F 1 ( − 1 2 , 1 2 ; 1 − n ; k ) − π cos ⁡ ϕ 2 k 2 n F 2 × 1 × 0 1 × 3 × 2 [ 1 2 ; − 1 2 , 1 2 , 1 ; 1 2 , 1 ; 1 , 3 2 ; 1 − n ; ; − k 2 cos ⁡ ϕ , cos 2 ⁡ ϕ ] + π m 1 − n cos ⁡ ϕ 8 F 3 × 1 × 1 2 × 1 × 1 [ 1 2 , 3 2 , 2 ; 1 2 , 1 ; 2 , 2 − n ; 1 − n ; 3 2 ; 3 2 ; − k 2 cos 2 ⁡ ϕ , k 2 ] {\displaystyle {\frac {{\rm {d}}^{n}}{{\rm {d}}k^{n}}}E(\phi ;k)={\frac {\pi }{2k^{n}}}{}_{2}F_{1}\left(-{\frac {1}{2}},{\frac {1}{2}};1-n;k\right)-{\frac {{\sqrt {\pi }}\cos \phi }{2k^{2n}}}F_{2\times 1\times 0}^{1\times 3\times 2}{\begin{bmatrix}{\frac {1}{2}};-{\frac {1}{2}},{\frac {1}{2}},1;{\frac {1}{2}},1;\\1,{\frac {3}{2}};1-n;;\\-k^{2}\cos \phi ,\cos ^{2}\phi \end{bmatrix}}+{\frac {\pi m^{1-n}\cos \phi }{8}}F_{3\times 1\times 1}^{2\times 1\times 1}{\begin{bmatrix}{\frac {1}{2}},{\frac {3}{2}},2;{\frac {1}{2}},1;\\2,2-n;1-n;{\frac {3}{2}};{\frac {3}{2}};\\-k^{2}\cos ^{2}\phi ,k^{2}\end{bmatrix}}\,\!} 第三类不完全椭圆积分 Π {\displaystyle \Pi \,\!} 是 Π ( n ; ϕ | m ) = ∫ 0 ϕ d θ ( 1 − n sin 2 ⁡ θ ) 1 − ( sin ⁡ θ sin ⁡ o ε ) 2 , {\displaystyle \Pi (n;\phi |m)=\int _{0}^{\phi }{\frac {{\rm {d}}\theta }{(1-n\sin ^{2}\theta ){\sqrt {1-(\sin \theta \sin o\!\varepsilon )^{2}}}}},\,\!} 或者 Π ( n ; ϕ | m ) = ∫ 0 sin ⁡ ϕ d t ( 1 − n t 2 ) ( 1 − k 2 t 2 ) ( 1 − t 2 ) , {\displaystyle \Pi (n;\phi |m)=\int _{0}^{\sin \phi }{\frac {{\rm {d}}t}{(1-nt^{2}){\sqrt {(1-k^{2}t^{2})(1-t^{2})}}}},\,\!} 或者 Π ( n ; ϕ | m ) = ∫ 0 F ( ϕ | m ) d w 1 − n sn 2 ( w | m ) . {\displaystyle \Pi (n;\phi |m)=\int _{0}^{F(\phi |m)}{\frac {{\rm {d}}w}{1-n{\textrm {sn}}^{2}(w|m)}}.\;\,\!} 数字 n {\displaystyle n\,} 称为特征数，可以取任意值，和其它参数独立。但是要注意 Π ( 1 ; π 2 | m ) {\displaystyle \Pi (1;{\frac {\pi }{2}}|m)\,\!} 对于任意 m {\displaystyle m\,\!} 是无穷的。 Π ( n ; ϕ 1 , k ) + Π ( n ; ϕ 2 , k ) = Π [ n ; arccos ⁡ cos ⁡ ϕ 1 cos ⁡ ϕ 2 − sin ⁡ ϕ 1 sin ⁡ ϕ 2 ( 1 − k 2 sin 2 ⁡ ϕ 1 ) ( 1 − k 2 sin 2 ⁡ ϕ 2 ) 1 − k 2 sin 2 ⁡ ϕ 1 sin 2 ⁡ ϕ 2 , k ] − n ( 1 − n ) ( n − k 2 ) arctan ⁡ ( 1 − n ) n ( n − k 2 ) sin ⁡ arccos ⁡ cos ⁡ ϕ 1 cos ⁡ ϕ 2 − sin ⁡ ϕ 1 sin ⁡ ϕ 2 ( 1 − k 2 sin 2 ⁡ ϕ 1 ) ( 1 − k 2 sin 2 ⁡ ϕ 2 ) 1 − k 2 sin 2 ⁡ ϕ 1 sin 2 ⁡ ϕ 2 sin ⁡ ϕ 1 sin ⁡ ϕ 2 n cos ⁡ ϕ 1 cos ⁡ ϕ 2 − n sin ⁡ ϕ 1 sin ⁡ ϕ 2 ( 1 − k 2 sin 2 ⁡ ϕ 1 ) ( 1 − k 2 sin 2 ⁡ ϕ 2 ) 1 − k 2 sin 2 ⁡ ϕ 1 sin 2 ⁡ ϕ 2 1 − k 2 sin 2 ⁡ arccos ⁡ cos ⁡ ϕ 1 cos ⁡ ϕ 2 − sin ⁡ ϕ 1 sin ⁡ ϕ 2 ( 1 − k 2 sin 2 ⁡ ϕ 1 ) ( 1 − k 2 sin 2 ⁡ ϕ 2 ) 1 − k 2 sin 2 ⁡ ϕ 1 sin 2 ⁡ ϕ 2 sin ⁡ ϕ 1 sin ⁡ ϕ 2 + 1 − n sin 2 ⁡ arccos ⁡ cos ⁡ ϕ 1 cos ⁡ ϕ 2 − sin ⁡ ϕ 1 sin ⁡ ϕ 2 ( 1 − k 2 sin 2 ⁡ ϕ 1 ) ( 1 − k 2 sin 2 ⁡ ϕ 2 ) 1 − k 2 sin 2 ⁡ ϕ 1 sin 2 ⁡ ϕ 2 {\displaystyle \Pi (n;\phi _{1},k)+\Pi (n;\phi _{2},k)=\Pi \left[n;\arccos {\frac {\cos \phi _{1}\cos \phi _{2}-\sin \phi _{1}\sin \phi _{2}{\sqrt {(1-k^{2}\sin ^{2}\phi _{1})(1-k^{2}\sin ^{2}\phi _{2})}}}{1-k^{2}\sin ^{2}\phi _{1}\sin ^{2}\phi _{2}}},k\right]-{\sqrt {\frac {n}{(1-n)(n-k^{2})}}}\arctan {\frac {{\sqrt {(1-n)n(n-k^{2})}}\sin \arccos {\frac {\cos \phi _{1}\cos \phi _{2}-\sin \phi _{1}\sin \phi _{2}{\sqrt {(1-k^{2}\sin ^{2}\phi _{1})(1-k^{2}\sin ^{2}\phi _{2})}}}{1-k^{2}\sin ^{2}\phi _{1}\sin ^{2}\phi _{2}}}\sin \phi _{1}\sin \phi _{2}}{{\frac {n\cos \phi _{1}\cos \phi _{2}-n\sin \phi _{1}\sin \phi _{2}{\sqrt {(1-k^{2}\sin ^{2}\phi _{1})(1-k^{2}\sin ^{2}\phi _{2})}}}{1-k^{2}\sin ^{2}\phi _{1}\sin ^{2}\phi _{2}}}{\sqrt {1-k^{2}\sin ^{2}\arccos {\frac {\cos \phi _{1}\cos \phi _{2}-\sin \phi _{1}\sin \phi _{2}{\sqrt {(1-k^{2}\sin ^{2}\phi _{1})(1-k^{2}\sin ^{2}\phi _{2})}}}{1-k^{2}\sin ^{2}\phi _{1}\sin ^{2}\phi _{2}}}}}\sin \phi _{1}\sin \phi _{2}+1-n\sin ^{2}\arccos {\frac {\cos \phi _{1}\cos \phi _{2}-\sin \phi _{1}\sin \phi _{2}{\sqrt {(1-k^{2}\sin ^{2}\phi _{1})(1-k^{2}\sin ^{2}\phi _{2})}}}{1-k^{2}\sin ^{2}\phi _{1}\sin ^{2}\phi _{2}}}}}} ∂ ∂ n Π ( n ; ϕ , k ) = 1 2 ( k 2 − n ) ( n − 1 ) [ E ( ϕ ; k ) + ( k 2 − n ) F ( ϕ ; k ) n + ( n 2 − k 2 ) Π ( n ; ϕ , k ) n − n 1 − k 2 sin ⁡ ϕ sin ⁡ 2 ϕ 2 ( 1 − n sin 2 ⁡ ϕ ) ] {\displaystyle {\frac {\partial }{\partial n}}\Pi (n;\phi ,k)={\frac {1}{2(k^{2}-n)(n-1)}}\left[E(\phi ;k)+{\frac {(k^{2}-n)F(\phi ;k)}{n}}+{\frac {(n^{2}-k^{2})\Pi (n;\phi ,k)}{n}}-{\frac {n{\sqrt {1-k^{2}\sin \phi }}\sin 2\phi }{2(1-n\sin ^{2}\phi )}}\right]} ∂ m ∂ n m Π ( n ; ϕ , k ) = sin ⁡ ϕ n m ∑ q = 0 ∞ q ! ( n sin 2 ⁡ ϕ ) q ( 2 q + 1 ) Γ ( q − m + 1 ) F 1 ( q + 1 2 , 1 2 , 1 2 ; q + 3 2 ; sin 2 ⁡ ϕ , k 2 sin 2 ⁡ ϕ ) {\displaystyle {\frac {{\partial }^{m}}{\partial n^{m}}}\Pi (n;\phi ,k)={\frac {\sin \phi }{n^{m}}}\sum _{q=0}^{\infty }{\frac {q!(n\sin ^{2}\phi )^{q}}{(2q+1)\Gamma (q-m+1)}}F_{1}\left(q+{\frac {1}{2}},{\frac {1}{2}},{\frac {1}{2}};q+{\frac {3}{2}};\sin ^{2}\phi ,k^{2}\sin ^{2}\phi \right)} ∂ ∂ ϕ Π ( n ; ϕ , k ) = 1 ( 1 − k 2 sin 2 ⁡ ϕ ) {\displaystyle {\frac {\partial }{\partial \phi }}\Pi (n;\phi ,k)={\frac {1}{(1-k^{2}\sin ^{2}\phi )}}\!} ∂ ∂ k Π ( n ; ϕ , k ) = k n − k 2 [ E ( ϕ ; k ) k 2 − 1 + Π ( n ; ϕ , k ) − k 2 sin ⁡ 2 ϕ 2 ( k 2 − 1 ) 1 − k 2 sin 2 ⁡ ϕ ] {\displaystyle {\frac {\partial }{\partial k}}\Pi (n;\phi ,k)={\frac {k}{n-k^{2}}}\left[{\frac {E(\phi ;k)}{k^{2}-1}}+\Pi (n;\phi ,k)-{\frac {k^{2}\sin 2\phi }{2(k^{2}-1){\sqrt {1-k^{2}\sin ^{2}\phi }}}}\right]\!} Π ( n ; ϕ , 1 ) = 1 2 n − 2 [ n ln ⁡ 1 + n sin ⁡ ϕ 1 − n sin ⁡ ϕ − 2 ln ⁡ ( sec ⁡ ϕ + tan ⁡ ϕ ) ] {\displaystyle \Pi (n;\phi ,1)={\frac {1}{2n-2}}\left[{\sqrt {n}}\ln {\frac {1+{\sqrt {n}}\sin \phi }{1-{\sqrt {n}}\sin \phi }}-2\ln(\sec \phi +\tan \phi )\right]\!} − π 2 ≤ ℜ ( ϕ ) ≤ π 2 {\displaystyle -{\frac {\pi }{2}}\leq \Re (\phi )\leq {\frac {\pi }{2}}\!} Π ( 0 ; ϕ , k ) = F ( ϕ , k ) {\displaystyle \Pi (0;\phi ,k)=F(\phi ,k)\!} Π ( n ; ϕ , 0 ) = a r c t a n h ( n − 1 tan ⁡ ϕ ) n − 1 {\displaystyle \Pi (n;\phi ,0)={\frac {{\rm {arctanh}}({\sqrt {n-1}}\tan \phi )}{\sqrt {n-1}}}\!} − π 2 ≤ ℜ ( ϕ ) ≤ π 2 {\displaystyle -{\frac {\pi }{2}}\leq \Re (\phi )\leq {\frac {\pi }{2}}\!} Π ( n ; ϕ , n ) = 1 1 − n [ E ( ϕ , n ) − n sin ⁡ 2 ϕ 2 1 − n sin 2 ⁡ ϕ ] {\displaystyle \Pi (n;\phi ,{\sqrt {n}})={\frac {1}{1-n}}\left[E(\phi ,{\sqrt {n}})-{\frac {n\sin 2\phi }{2{\sqrt {1-n\sin ^{2}\phi }}}}\right]\!} Π ( n ; 1 k , k ) = 1 k Π ( n k 2 , 1 k ) {\displaystyle \Pi \left(n;{\frac {1}{k}},k\right)={\frac {1}{k}}\Pi \left({\frac {n}{k^{2}}},{\frac {1}{k}}\right)\!} Π ( 1 ; ϕ , k ) = 1 − k 2 sin 2 ⁡ ϕ tan ⁡ ϕ − E ( ϕ , k ) 1 − k 2 + F ( ϕ , k ) {\displaystyle \Pi \left(1;\phi ,k\right)={\frac {{\sqrt {1-k^{2}\sin ^{2}\phi }}\tan \phi -E(\phi ,k)}{1-k^{2}}}+F(\phi ,k)\!} 如果幅度为 π 2 {\displaystyle {\frac {\pi }{2}}\,} 或者 x = 1 {\displaystyle x=1\,} ，则称椭圆积分为完全的。 第一类完全椭圆积分 K {\displaystyle K\,} 可以定義为 K ( k ) = ∫ 0 π 2 d θ 1 − k 2 sin 2 ⁡ θ {\displaystyle K(k)=\int _{0}^{\frac {\pi }{2}}{\frac {{\rm {d}}\theta }{\sqrt {1-k^{2}\sin ^{2}\theta }}}} 或者 K ( k ) = ∫ 0 1 d t ( 1 − t 2 ) ( 1 − k 2 t 2 ) . {\displaystyle K(k)=\int _{0}^{1}{\frac {{\rm {d}}t}{\sqrt {(1-t^{2})(1-k^{2}t^{2})}}}.\!} 它是第一类不完全椭圆积分的特例： K ( k ) = F ( 1 ; k ) = F ( π 2 | k 2 ) {\displaystyle K(k)=F(1;\,k)=F\left({\frac {\pi }{2}}\,|\,k^{2}\right)\!} 这个特例可以表达为幂级数 K ( k ) = π 2 ∑ n = 0 ∞ [ ( 2 n ) ! 2 2 n n ! 2 ] 2 k 2 n {\displaystyle K(k)={\frac {\pi }{2}}\sum _{n=0}^{\infty }\left[{\frac {(2n)!}{2^{2n}n!^{2}}}\right]^{2}k^{2n}\!} 它等价于 K ( k ) = π 2 { 1 + ( 1 2 ) 2 k 2 + ( 1 ⋅ 3 2 ⋅ 4 ) 2 k 4 + ⋯ + [ ( 2 n − 1 ) ! ! ( 2 n ) ! ! ] 2 k 2 n + ⋯ } . {\displaystyle 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 {(2n-1)!!}{(2n)!!}}\right]^{2}k^{2n}+\cdots \right\}.\!} 其中 n ! ! {\displaystyle n!!\,} 表示双阶乘。利用高斯的超几何函数，第一类完全椭圆积分可以表达为 K ( k ) = π 2 2 F 1 ( 1 2 , 1 2 ; 1 ; k 2 ) . {\displaystyle K(k)={\frac {\pi }{2}}\,_{2}F_{1}\left({\frac {1}{2}},{\frac {1}{2}};1;k^{2}\right).\,\!} 第一类完全椭圆积分有时称为四分周期。它可以利用算术几何平均值來快速计算。 K ( k ) = π 2 a g m ( 1 , 1 − k 2 ) . {\displaystyle K(k)={\frac {\frac {\pi }{2}}{\mathrm {agm} (1,{\sqrt {1-k^{2}}})}}.} ℜ [ K ( x + y i ) ] = π 2 F 2 × 1 × 1 4 × 0 × 0 [ 3 4 , 3 4 , 5 4 , 5 4 , ; ; ; 1 , 3 2 ; 1 2 ; 3 2 ; − y 2 , x 2 ] + π 8 x F 2 × 1 × 1 4 × 0 × 0 [ 1 4 , 1 4 , 3 4 , 3 4 , ; ; ; 1 , 1 2 ; 1 2 ; 1 2 ; − y 2 , x 2 ] {\displaystyle \Re \left[K(x+y{\rm {i}})\right]={\frac {\pi }{2}}F_{2\times 1\times 1}^{4\times 0\times 0}{\begin{bmatrix}{\frac {3}{4}},{\frac {3}{4}},{\frac {5}{4}},{\frac {5}{4}},;;;\\1,{\frac {3}{2}};{\frac {1}{2}};{\frac {3}{2}};\\-y^{2},x^{2}\end{bmatrix}}+{\frac {\pi }{8}}xF_{2\times 1\times 1}^{4\times 0\times 0}{\begin{bmatrix}{\frac {1}{4}},{\frac {1}{4}},{\frac {3}{4}},{\frac {3}{4}},;;;\\1,{\frac {1}{2}};{\frac {1}{2}};{\frac {1}{2}};\\-y^{2},x^{2}\end{bmatrix}}\,\!} ℑ [ K ( x + y i ) ] = π 8 y F 2 × 1 × 1 4 × 0 × 0 [ 3 4 , 5 4 , 3 4 , 5 4 , ; ; ; 1 , 3 2 ; 3 2 ; 1 2 ; − y 2 , x 2 ] + 9 64 π x y F 2 × 1 × 1 4 × 0 × 0 [ 5 4 , 7 4 , 7 4 , 5 4 , ; ; ; 2 , 3 2 ; 3 2 ; 3 2 ; − y 2 , x 2 ] {\displaystyle \Im \left[K(x+y{\rm {i}})\right]={\frac {\pi }{8}}yF_{2\times 1\times 1}^{4\times 0\times 0}{\begin{bmatrix}{\frac {3}{4}},{\frac {5}{4}},{\frac {3}{4}},{\frac {5}{4}},;;;\\1,{\frac {3}{2}};{\frac {3}{2}};{\frac {1}{2}};\\-y^{2},x^{2}\end{bmatrix}}+{\frac {9}{64}}\pi xyF_{2\times 1\times 1}^{4\times 0\times 0}{\begin{bmatrix}{\frac {5}{4}},{\frac {7}{4}},{\frac {7}{4}},{\frac {5}{4}},;;;\\2,{\frac {3}{2}};{\frac {3}{2}};{\frac {3}{2}};\\-y^{2},x^{2}\end{bmatrix}}\,\!} K ( ± ∞ ) = 0 {\displaystyle K(\pm \infty )=0\,} K ( ± i ∞ ) = 0 {\displaystyle K(\pm {\rm {i}}\infty )=0\,} K ( 0 ) = π 2 {\displaystyle K(0)={\frac {\pi }{2}}\!} K ( 1 ) = ∞ {\displaystyle K(1)=\infty \!} K ( 2 2 ) = 8 π Γ 2 ( − 1 4 ) π {\displaystyle K({\frac {\sqrt {2}}{2}})={\frac {8\pi }{\Gamma ^{2}\left(-{\frac {1}{4}}\right)}}{\sqrt {\pi }}\,} K ( 17 − 12 2 ) = ( 4 + 2 2 ) π Γ 2 ( − 1 4 ) π {\displaystyle K\left({\sqrt {17-12{\sqrt {2}}}}\right)={\frac {(4+2{\sqrt {2}})\pi }{\Gamma ^{2}\left(-{\frac {1}{4}}\right)}}{\sqrt {\pi }}\,} K ( 6 − 2 4 ) = 4 3 ⋅ 3 4 8 π Γ 3 ( 1 3 ) {\displaystyle K\left({\frac {{\sqrt {6}}-{\sqrt {2}}}{4}}\right)={\frac {{\sqrt[{3}]{4}}\cdot {\sqrt[{4}]{3}}}{8\pi }}\Gamma ^{3}\left({\frac {1}{3}}\right)\,} K ( 6 + 2 4 ) = 4 3 ⋅ 27 4 8 π Γ 3 ( 1 3 ) {\displaystyle K\left({\frac {{\sqrt {6}}+{\sqrt {2}}}{4}}\right)={\frac {{\sqrt[{3}]{4}}\cdot {\sqrt[{4}]{27}}}{8\pi }}\Gamma ^{3}\left({\frac {1}{3}}\right)\,} K ( i ) = 2 π 8 π Γ 2 ( 1 4 ) {\displaystyle K(i)={\frac {\sqrt {2\pi }}{8\pi }}\Gamma ^{2}\left({\frac {1}{4}}\right)\,} K ( 2 ) = 4 2 π π Γ 2 ( 1 4 ) + 4 2 π π Γ 2 ( 1 4 ) i {\displaystyle K({\sqrt {2}})={\frac {4{\sqrt {2\pi }}\pi }{\Gamma ^{2}\left({\frac {1}{4}}\right)}}+{\frac {4{\sqrt {2\pi }}\pi }{\Gamma ^{2}\left({\frac {1}{4}}\right)}}{\rm {i}}\,} K ( i k ) = 1 k 2 + 1 K ( k 2 k 2 + 1 ) {\displaystyle K({\rm {i}}k)={\frac {1}{\sqrt {k^{2}+1}}}K\left({\sqrt {\frac {k^{2}}{k^{2}+1}}}\right)\,} 其中 Γ ( 1 4 ) ≈ 3.62561 {\displaystyle \Gamma \left({\frac {1}{4}}\right)\approx 3.62561\,} Γ ( 1 3 ) ≈ 2.67893 {\displaystyle \Gamma \left({\frac {1}{3}}\right)\approx 2.67893\,} 第一类完全椭圆积分满足 E ( k ) K ′ ( k ) + E ′ ( k ) K ( k ) − K ( k ) K ′ ( k ) = π 2 {\displaystyle E(k)K'(k)+E'(k)K(k)-K(k)K'(k)={\frac {\pi }{2}}\,} d d k K n ( k ) = n K n − 1 ( k ) E ( k ) 2 k ( 1 − k ) − n K n ( k ) 2 k {\displaystyle {\frac {\rm {d}}{{\rm {d}}k}}K^{n}(k)={\frac {nK^{n-1}(k)E(k)}{2k(1-k)}}-{\frac {nK^{n}(k)}{2k}}} K ( k 2 ) ≈ π 2 + π 8 k 2 1 − k 2 − π 16 k 4 1 − k 2 {\displaystyle K(k^{2})\approx {\frac {\pi }{2}}+{\frac {\pi }{8}}{\frac {k^{2}}{1-k^{2}}}-{\frac {\pi }{16}}{\frac {k^{4}}{1-k^{2}}}} 這個近似在k<1/2時相對誤差小於3×10−4，若只保留前兩項則誤差在k<1/2時小於0.01 此函數滿足以下微分方程 d d k [ k ( 1 − k 2 ) d K ( k ) d k ] = k K ( k ) {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} k}}\left[k(1-k^{2}){\frac {\mathrm {d} K(k)}{\mathrm {d} k}}\right]=kK(k)} 此微分方程之另一解為 K ( 1 − k 2 ) {\displaystyle K({\sqrt {1-k^{2}}})} ，此解滿足以下關係。 d d k K ( 1 − k 2 ) = E ( k ) k ( 1 − k 2 ) − K ( k ) k {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} k}}K({\sqrt {1-k^{2}}})={\frac {E(k)}{k(1-k^{2})}}-{\frac {K(k)}{k}}} . 第二类完全椭圆积分 E {\displaystyle E\,} 可以定义为 E ( k ) = ∫ 0 π 2 1 − k 2 sin 2 ⁡ θ   d θ {\displaystyle E(k)=\int _{0}^{\frac {\pi }{2}}{\sqrt {1-k^{2}\sin ^{2}\theta }}\ {\rm {d}}\theta \!} 或者 E ( k ) = ∫ 0 1 1 − k 2 t 2 1 − t 2   d t . {\displaystyle E(k)=\int _{0}^{1}{\frac {\sqrt {1-k^{2}t^{2}}}{\sqrt {1-t^{2}}}}\ {\rm {d}}t.\!} 它是第二类不完全椭圆积分的特殊情况： E ( k ) = E ( 1 ; k ) = E ( π 2 | k 2 ) {\displaystyle E(k)=E(1;\,k)=E({\frac {\pi }{2}}\,|\,k^{2})\!} 它可以用幂级数表达 E ( k ) = π 2 ∑ n = 0 ∞ [ ( 2 n ) ! 2 2 n n ! 2 ] 2 k 2 n 1 − 2 n {\displaystyle E(k)={\frac {\pi }{2}}\sum _{n=0}^{\infty }\left[{\frac {(2n)!}{2^{2n}n!^{2}}}\right]^{2}{\frac {k^{2n}}{1-2n}}\!} 也就是 E ( k ) = π 2 { 1 − ( 1 2 ) 2 k 2 1 − ( 1 ⋅ 3 2 ⋅ 4 ) 2 k 4 3 − ⋯ − [ ( 2 n − 1 ) ! ! ( 2 n ) ! ! ] 2 k 2 n 2 n − 1 − ⋯ } . {\displaystyle E(k)={\frac {\pi }{2}}\left\{1-\left({\frac {1}{2}}\right)^{2}{\frac {k^{2}}{1}}-\left({\frac {1\cdot 3}{2\cdot 4}}\right)^{2}{\frac {k^{4}}{3}}-\cdots -\left[{\frac {\left(2n-1\right)!!}{\left(2n\right)!!}}\right]^{2}{\frac {k^{2n}}{2n-1}}-\cdots \right\}.\!} 用高斯超几何函数表示的话，第二类完全椭圆积分可以写作 E ( k ) = π 2 2 F 1 ( − 1 2 , 1 2 ; 1 ; k 2 ) . {\displaystyle E(k)={\frac {\pi }{2}}\,_{2}F_{1}\left(-{\frac {1}{2}},{\frac {1}{2}};1;k^{2}\right).\,\!} 有如下性质 E ( n π 2 ; k ) = n E ( k ) {\displaystyle E({\frac {n\pi }{2}};k)=nE(k)\,\!} n ∈ Z {\displaystyle n\in \mathbb {Z} \,\!} E ( x + y i ) = { π 2 F 2 × 1 × 1 4 × 0 × 0 [ 3 4 , 5 4 , 1 4 , 3 4 , ; − ; − ; 1 , 3 2 ; 1 2 ; 3 2 ; − y 2 , x 2 ] − π 8 x F 2 × 1 × 1 4 × 0 × 0 [ 1 4 , 3 4 , − 1 4 , 1 4 , ; − ; − ; 1 , 1 2 ; 1 2 ; 1 2 ; − y 2 , x 2 ] } + i { − π 8 y F 2 × 1 × 1 4 × 0 × 0 [ 3 4 , 5 4 , 1 4 , 3 4 , ; − ; − ; 1 , 3 2 ; 1 2 ; 3 2 ; − y 2 , x 2 ] − 3 64 π x y F 2 × 1 × 1 4 × 0 × 0 [ 5 4 , 7 4 , 3 4 , 5 4 , ; − ; − ; 2 , 3 2 ; 3 2 ; 3 2 ; − y 2 , x 2 ] } {\displaystyle E(x+y{\rm {i}})=\left\{{\frac {\pi }{2}}F_{2\times 1\times 1}^{4\times 0\times 0}{\begin{bmatrix}{\frac {3}{4}},{\frac {5}{4}},{\frac {1}{4}},{\frac {3}{4}},;-;-;\\1,{\frac {3}{2}};{\frac {1}{2}};{\frac {3}{2}};\\-y^{2},x^{2}\end{bmatrix}}-{\frac {\pi }{8}}xF_{2\times 1\times 1}^{4\times 0\times 0}{\begin{bmatrix}{\frac {1}{4}},{\frac {3}{4}},-{\frac {1}{4}},{\frac {1}{4}},;-;-;\\1,{\frac {1}{2}};{\frac {1}{2}};{\frac {1}{2}};\\-y^{2},x^{2}\end{bmatrix}}\right\}+{\rm {i}}\left\{-{\frac {\pi }{8}}yF_{2\times 1\times 1}^{4\times 0\times 0}{\begin{bmatrix}{\frac {3}{4}},{\frac {5}{4}},{\frac {1}{4}},{\frac {3}{4}},;-;-;\\1,{\frac {3}{2}};{\frac {1}{2}};{\frac {3}{2}};\\-y^{2},x^{2}\end{bmatrix}}-{\frac {3}{64}}\pi xyF_{2\times 1\times 1}^{4\times 0\times 0}{\begin{bmatrix}{\frac {5}{4}},{\frac {7}{4}},{\frac {3}{4}},{\frac {5}{4}},;-;-;\\2,{\frac {3}{2}};{\frac {3}{2}};{\frac {3}{2}};\\-y^{2},x^{2}\end{bmatrix}}\right\}\,\!} E ( 0 ) = π 2 {\displaystyle E(0)={\frac {\pi }{2}}\!} E ( 1 ) = 1 {\displaystyle E(1)=1\!} E ( ∞ ) = i ∞ {\displaystyle E(\infty )={\rm {i}}\infty \,} E ( − ∞ ) = ∞ {\displaystyle E(-\infty )=\infty \,} E ( i ∞ ) = ( 2 2 − 2 2 i ) ∞ {\displaystyle E({\rm {i}}\infty )=({\frac {\sqrt {2}}{2}}-{\frac {\sqrt {2}}{2}}{\rm {i}})\infty \,} E ( i ) = 2 π 2 π Γ 2 ( 3 4 ) + 2 π π 2 4 π Γ 2 ( 3 4 ) = π 2 π Γ 2 ( 1 4 ) + 2 π 8 π Γ 2 ( 1 4 ) {\displaystyle E({\rm {i}})={\frac {\sqrt {2\pi }}{2\pi }}\Gamma ^{2}\left({\frac {3}{4}}\right)+{\frac {{\sqrt {2\pi }}{\pi }^{2}}{4\pi \Gamma ^{2}\left({\frac {3}{4}}\right)}}={\frac {\pi {\sqrt {2\pi }}}{\Gamma ^{2}\left({\frac {1}{4}}\right)}}+{\frac {\sqrt {2\pi }}{8\pi }}\Gamma ^{2}\left({\frac {1}{4}}\right)\,} E ( − i ∞ ) = ( 2 2 + 2 2 i ) ∞ {\displaystyle E(-{\rm {i}}\infty )=({\frac {\sqrt {2}}{2}}+{\frac {\sqrt {2}}{2}}{\rm {i}})\infty \,} E ( 2 2 ) = π 3 2 Γ ( 1 4 ) − 2 + 1 8 π Γ ( 1 4 ) 2 {\displaystyle E\left({\tfrac {\sqrt {2}}{2}}\right)=\pi ^{\frac {3}{2}}\Gamma \left({\tfrac {1}{4}}\right)^{-2}+{\tfrac {1}{8{\sqrt {\pi }}}}\Gamma \left({\tfrac {1}{4}}\right)^{2}} E ( 6 − 2 4 ) = 2 3 ⋅   3 4 3 Γ 3 ( 1 3 ) π 2 + 4 3 ( 3 3 4 + 27 4 ) 48 π Γ 3 ( 1 3 ) {\displaystyle E\left({\frac {{\sqrt {6}}-{\sqrt {2}}}{4}}\right)={\frac {{\sqrt[{3}]{2}}\cdot \ {\sqrt[{4}]{3}}}{3\Gamma ^{3}\left({\frac {1}{3}}\right)}}{\pi }^{2}+{\frac {{\sqrt[{3}]{4}}\left(3{\sqrt[{4}]{3}}+{\sqrt[{4}]{27}}\right)}{48{\pi }}}\Gamma ^{3}\left({\frac {1}{3}}\right)\!} E ( 6 + 2 4 ) = 2 3 ⋅   27 4 3 Γ 3 ( 1 3 ) π 2 + 4 3 ( 27 4 − 3 4 ) 16 π Γ 3 ( 1 3 ) {\displaystyle E\left({\frac {{\sqrt {6}}+{\sqrt {2}}}{4}}\right)={\frac {{\sqrt[{3}]{2}}\cdot \ {\sqrt[{4}]{27}}}{3\Gamma ^{3}\left({\frac {1}{3}}\right)}}{\pi }^{2}+{\frac {{\sqrt[{3}]{4}}\left({\sqrt[{4}]{27}}-{\sqrt[{4}]{3}}\right)}{16{\pi }}}\Gamma ^{3}\left({\frac {1}{3}}\right)\!} E ( 2 − 1 ) = π 8 [ Γ ( 1 8 ) Γ ( 5 8 ) + Γ ( 5 8 ) Γ ( 9 8 ) ] {\displaystyle E({\sqrt {2}}-1)={\frac {\sqrt {\pi }}{8}}\left[{\frac {\Gamma ({\frac {1}{8}})}{\Gamma ({\frac {5}{8}})}}+{\frac {\Gamma ({\frac {5}{8}})}{\Gamma ({\frac {9}{8}})}}\right]\!} E ( 2 ) = 1 2 π Γ 2 ( 3 4 ) + 1 2 π Γ 2 ( 3 4 ) i {\displaystyle E({\sqrt {2}})={\sqrt {\frac {1}{2\pi }}}\Gamma ^{2}\left({\frac {3}{4}}\right)+{\sqrt {\frac {1}{2\pi }}}\Gamma ^{2}\left({\frac {3}{4}}\right){\rm {i}}} 其中 Γ ( 1 8 ) ≈ 7.53394 {\displaystyle \Gamma \left({\frac {1}{8}}\right)\approx 7.53394\,} Γ ( 5 8 ) ≈ 1.43452 {\displaystyle \Gamma \left({\frac {5}{8}}\right)\approx 1.43452\,} Γ ( 9 8 ) ≈ 0.94174 {\displaystyle \Gamma \left({\frac {9}{8}}\right)\approx 0.94174\,} Γ ( 3 4 ) ≈ 1.22541 {\displaystyle \Gamma \left({\frac {3}{4}}\right)\approx 1.22541\,} d d k E ( k ) = E ( k ) − K ( k ) k {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} k}}E(k)={\frac {E(k)-K(k)}{k}}} ∫ E ( k ) d k = 2 3 [ k K ( k ) − K ( k ) + k E ( k ) + E ( k ) ] {\displaystyle \int E(k){\rm {d}}k={\frac {2}{3}}\left[kK(k)-K(k)+kE(k)+E(k)\right]} ( k 2 − 1 ) d d k [ k d E ( k ) d k ] = k E ( k ) {\displaystyle (k^{2}-1){\frac {\mathrm {d} }{\mathrm {d} k}}\left[k\;{\frac {\mathrm {d} E(k)}{\mathrm {d} k}}\right]=kE(k)} 此微分方程之另解為 E ( 1 − k 2 ) − K ( 1 − k 2 ) {\displaystyle E({\sqrt {1-k^{2}}})-K({\sqrt {1-k^{2}}})} 。 第三类完全椭圆积分 Π {\displaystyle \Pi \,} 可以定义为 Π ( n , k ) = ∫ 0 π 2   d θ ( 1 − n sin 2 ⁡ θ ) 1 − k 2 sin 2 ⁡ θ {\displaystyle \Pi (n,k)=\int _{0}^{\frac {\pi }{2}}{\frac {\ {\rm {d}}\theta }{(1-n\sin ^{2}\theta ){\sqrt {1-k^{2}\sin ^{2}\theta }}}}} 注意有时第三类椭圆积分被定义为带相反符号的 n {\displaystyle n\,} ，也即 Π ′ ( n , k ) = ∫ 0 π 2   d θ ( 1 + n sin 2 ⁡ θ ) 1 − k ′ 2 sin 2 ⁡ θ . {\displaystyle \Pi '(n,k)=\int _{0}^{\frac {\pi }{2}}{\frac {\ {\rm {d}}\theta }{(1+n\sin ^{2}\theta ){\sqrt {1-k'^{2}\sin ^{2}\theta }}}}.} 用阿佩尔函数可表示为 Π ( m , n ) = π 2 F 1 ( 1 2 ; 1 , 1 2 ; 1 ; m , n ) {\displaystyle \Pi (m,n)={\frac {\pi }{2}}F_{1}\left({\frac {1}{2}};1,{\frac {1}{2}};1;m,n\right)\,} 第三类完全椭圆积分和第一类椭圆积分之间的关系 Π [ ( 1 + x ) ( 1 − 3 x ) ( 1 − x ) ( 1 + 3 x ) , ( 1 + x ) 3 ( 1 − 3 x ) ( 1 − x ) 3 ( 1 + 3 x ) ] − 1 + 3 x 6 x K [ ( 1 + x ) 3 ( 1 − 3 x ) ( 1 − x ) 3 ( 1 + 3 x ) ] = {\displaystyle \Pi \left[{\frac {(1+x)(1-3x)}{(1-x)(1+3x)}},{\frac {(1+x)^{3}(1-3x)}{(1-x)^{3}(1+3x)}}\right]-{\frac {1+3x}{6x}}K\left[{\frac {(1+x)^{3}(1-3x)}{(1-x)^{3}(1+3x)}}\right]=\,} { 0 for  0 < x < 1 − π ( x − 1 ) ( x − 1 ) ( 1 + 3 x ) 12 x for  x < 0 , x > 1 {\displaystyle {\begin{cases}0&{\mbox{for }}0<x<1\!\,\\-{\frac {\pi (x-1){\sqrt {(x-1)(1+3x)}}}{12x}}&{\mbox{for }}x<0,x>1\!\,\\\end{cases}}} 如 K ( 2 2 ) = π 4 π Γ 2 ( 1 4 ) = 3 − 6 3 − 9 2 Π ( 1 − 2 3 − 3 2 , 1 2 ) {\displaystyle K\left({\frac {\sqrt {2}}{2}}\right)={\frac {\sqrt {\pi }}{4\pi }}\Gamma ^{2}\left({\frac {1}{4}}\right)={\frac {3-{\sqrt {6{\sqrt {3}}-9}}}{2}}\Pi \left({\frac {1-{\sqrt {2{\sqrt {3}}-3}}}{2}},{\frac {1}{2}}\right)\,} = 3 + 6 3 − 9 2 Π ( 1 + 2 3 − 3 2 , 1 2 ) − π 2 + 3 + 7 + 38 9 3 {\displaystyle ={\frac {3+{\sqrt {6{\sqrt {3}}-9}}}{2}}\Pi \left({\frac {1+{\sqrt {2{\sqrt {3}}-3}}}{2}},{\frac {1}{2}}\right)-\pi {\sqrt {2+{\sqrt {3}}+{\sqrt {7+{\frac {38}{9}}{\sqrt {3}}}}}}\,} ∂ ∂ n Π ( n , k ) = 1 2 ( k 2 − n ) ( n − 1 ) [ E ( k ) + ( k 2 − n ) K ( k ) n + ( n 2 − k 2 ) Π ( n , k ) n ] {\displaystyle {\frac {\partial }{\partial n}}\Pi (n,k)={\frac {1}{2(k^{2}-n)(n-1)}}\left[E(k)+{\frac {(k^{2}-n)K(k)}{n}}+{\frac {(n^{2}-k^{2})\Pi (n,k)}{n}}\right]} ∂ ∂ k Π ( n , k ) = k n − k 2 [ E ( k ) k 2 − 1 + Π ( n , k ) ] {\displaystyle {\frac {\partial }{\partial k}}\Pi (n,k)={\frac {k}{n-k^{2}}}\left[{\frac {E(k)}{k^{2}-1}}+\Pi (n,k)\right]} Π ( 0 , 0 ) = π 2 {\displaystyle \Pi (0,0)={\frac {\pi }{2}}\,} Π ( n , 0 ) = π 2 1 − n {\displaystyle \Pi (n,0)={\frac {\pi }{2{\sqrt {1-n}}}}\,} Π ( n , 1 ) = − ∞ sgn ⁡ n − 1 {\displaystyle \Pi (n,1)=-{\frac {\infty }{\operatorname {sgn} {n-1}}}\,} Π ( n , n ) = E ( n ) 1 − n {\displaystyle \Pi (n,{\sqrt {n}})={\frac {E(n)}{1-n}}\,} Π ( 0 , n ) = K ( n ) {\displaystyle \Pi (0,{\sqrt {n}})=K(n)\,} Π ( ± ∞ , n ) = 0 {\displaystyle \Pi (\pm \infty ,{\sqrt {n}})=0\,} Π ( n , ± ∞ ) = 0 {\displaystyle \Pi (n,\pm \infty )=0\,} 勒讓得關係指出了第一类和第二类完全椭圆积分之间的联系: K ( k ) E ( 1 − k 2 ) + E ( k ) K ( 1 − k 2 ) − K ( k ) K ( 1 − k 2 ) = π 2 . {\displaystyle K(k)E\left({\sqrt {1-k^{2}}}\right)+E(k)K\left({\sqrt {1-k^{2}}}\right)-K(k)K\left({\sqrt {1-k^{2}}}\right)={\frac {\pi }{2}}.} 椭圆曲线 施瓦茨-克里斯托费尔映射 雅可比橢圓函數 魏爾斯特拉斯橢圓函數 Θ函數 算术-几何平均数 Milton Abramowitz and Irene A. Stegun, Handbook of Mathematical Functions, (1964) Dover Publications, New York. ISBN 0-486-61272-4. (See chapter 17). Harris Hancock Lectures on the theory of Elliptic functions (New York, J. Wiley & sons, 1910) Alfred George Greenhill The applications of elliptic functions (New York, Macmillan, 1892) Louis V. King On The Direct Numerical Calculation Of Elliptic Functions And Integrals (Cambridge University Press, 1924)