一种一般化的椭圆积分可以表示为以下函数：
T ( a , b ) ≡ 2 π ∫ 0 π / 2 d θ a 2 cos 2 θ + b 2 sin 2 θ = 2 π ∫ 0 π / 2 d θ cos θ a 2 + b 2 tan 2 θ (1) \begin{aligned} T(a,b)&\equiv {2\over\pi} \int^{\pi/2}_0 {d\theta\over\sqrt{a^2\cos^2\theta+b^2\sin^2\theta}}\\ &= {2\over\pi} \int^{\pi/2}_0 {d\theta\over\cos\theta\sqrt{a^2+b^2\tan^2\theta}} \end{aligned} \tag 1 T(a,b)≡π2∫0π/2a2cos2θ+b2sin2θdθ=π2∫0π/2cosθa2+b2tan2θdθ(1)
使 t ≡ b tan θ , d t = b sec 2 θ d θ = d θ cos θ b 2 + t 2 . t\equiv b\tan\theta,\quad dt=b\sec^2\theta\,d\theta={d\theta \over \cos \theta }\sqrt{b^2+t^2}. t≡btanθ,dt=bsec2θdθ=cosθdθb2+t2. 得到 T ( a , b ) = 2 π ∫ 0 ∞ d t ( a 2 + t 2 ) ( b 2 + t 2 ) = 1 π ∫ − ∞ ∞ d t ( a 2 + t 2 ) ( b 2 + t 2 ) . (2) \begin{aligned} T(a,b)&=&{2\over \pi} \int_0^\infty {dt\over \sqrt{(a^2+t^2)(b^2+t^2)}}\\ &=&{1\over \pi} \int_{-\infty}^\infty {dt\over \sqrt{(a^2+t^2)(b^2+t^2)}}. \end{aligned}\tag 2 T(a,b)==π2∫0∞(a2+t2)(b2+t2)dtπ1∫−∞∞(a2+t2)(b2+t2)dt.(2)
进一步进行变量代换 u ≡ 1 2 ( t − a b / t ) , d u = 1 2 ( 1 + a b / t 2 ) d t u\equiv {\textstyle{1\over 2}}(t-ab/t), \quad du={\textstyle{1\over 2}}(1+ab/t^2)\,dt u≡21(t−ab/t),du=21(1+ab/t2)dt得到
T ( a , b ) = 2 π ∫ − ∞ ∞ d u [ 4 u 2 + ( a + b ) 2 ] ( u 2 + a b ) = 1 π ∫ − ∞ ∞ d u [ u 2 + ( a + b 2 ) 2 ] ( u 2 + a b ) . (3) \begin{aligned} T(a,b)&=& {2\over \pi} \int_{-\infty}^\infty {du\over \sqrt{[4u^2+(a+b)^2](u^2+ab)}}\\ &=& {1\over \pi} \int_{-\infty}^\infty {du\over \sqrt{\left[{u^2+\left({a+b\over 2}\right)^2}\right](u^2+ab)}}. \end{aligned} \tag 3 T(a,b)==π2∫−∞∞[4u2+(a+b)2](u2+ab)duπ1∫−∞∞[u2+(2a+b)2](u2+ab)du.(3)
因此 T ( a , b ) = T ( 1 2 ( a + b ) , a b ) . (4) T(a,b) = T({\textstyle{1\over 2}}(a+b),\sqrt{ab}). \tag 4 T(a,b)=T(21(a+b),ab).(4) 以算术平均和几何平均代替原变量得到相同的结果。
完全椭圆积分可以利用此算术平均和几何平均变换方法进行计算
T ( a , b ) ≡ 2 π ∫ 0 π / 2 d θ a 2 cos 2 θ + b 2 sin 2 θ = 2 π ∫ 0 π / 2 d θ a cos 2 θ + ( b a ) 2 sin 2 θ = 2 a π ∫ 0 π / 2 d θ 1 − ( 1 − b 2 a 2 ) 2 sin 2 θ . (5) \begin{aligned} T(a,b)&\equiv& {2\over \pi} \int^{\pi/2}_0 {d\theta\over\sqrt{a^2\cos^2\theta+b^2\sin^2\theta}}\\ &=& {2\over\pi}\int^{\pi/2}_0 {d\theta \over a\sqrt{\cos ^2\theta +\left({b\over a}\right)^2 \sin^2 \theta}}\\ &=&{2\over a\pi}\int^{\pi/2}_0 {d\theta \over \sqrt{1-\left({1-{b^2\over a^2}}\right)^2\sin^2 \theta}}. \end{aligned} \tag 5 T(a,b)≡==π2∫0π/2a2cos2θ+b2sin2θdθπ2∫0π/2acos2θ+(ab)2sin2θdθaπ2∫0π/21−(1−a2b2)2sin2θdθ.(5)
因此
T ( a , b ) = 2 a π K ( 1 − b 2 a 2 ) (6) T(a,b)= {2\over a\pi}K\left({1-{b^2\over a^2}}\right) \tag 6 T(a,b)=aπ2K(1−a2b2)(6)
K ( k ) K(k) K(k) 是第一类完全椭圆积分，令 a ≡ a 0 ≡ 1 , b ≡ b 0 ≡ k ′ , k = 1 − k ′ 2 , a\equiv a_0\equiv 1,\quad b\equiv b_0\equiv k', \quad k=\sqrt{1-k'^2}, a≡a0≡1,b≡b0≡k′,k=1−k′2, 等式 (6) 为
2 π K ( 1 − k ′ 2 ) = 2 π K ( k ) , {2\over \pi}K(\sqrt{1-k'^2}\,) = {2\over \pi}K(k), π2K(1−k′2)=π2K(k),
定义
K ′ ( k ) ≡ K ( 1 − k 2 ) = K ( k ′ ) . (7) K'(k)\equiv K(\sqrt{1-k^2}\,)=K(k'). \tag 7 K′(k)≡K(1−k2)=K(k′).(7)
应用变量算术平均和几何平均进行变量代换
1 a K ( 1 − b 2 a 2 ) = 2 a + b K ( 1 − 4 a b ( a + b ) 2 ) = 2 a + b K ( a 2 + b 2 − 2 a b ( a + b ) 2 ) = 2 a + b K ( a − b a + b ) (8) \begin{aligned} {1\over a}K\left({\sqrt{1-{b^2\over a^2}}\,}\right)&=& {2\over a+b}K\left({\sqrt{1-{4ab\over (a+b)^2}}\,}\right)\\ &=& {2\over a+b} K\left({\sqrt{a^2+b^2-2ab\over(a+b)^2}\,}\right)\\ &=& {2\over a+b}K\left({a-b\over a+b}\right) \end{aligned} \tag 8 a1K(1−a2b2)===a+b2K(1−(a+b)24ab)a+b2K((a+b)2a2+b2−2ab)a+b2K(a+ba−b)(8)
得到
K ( 1 − b 2 a 2 ) = 2 1 + b a K ( 1 − b a 1 + b a ) . (9) K\left({\sqrt{1-{b^2\over a^2}}\,}\right)= {2\over 1+{b\over a}}K\left({1-{b\over a}\over 1+{b\over a}}\right). \tag 9 K(1−a2b2)=1+ab2K(1+ab1−ab).(9) 令
k ′ ≡ b a , k ≡ 1 − k ′ 2 , k'\equiv {b\over a}, \quad k\equiv \sqrt{1-k'^2}, k′≡ab,k≡1−k′2, 得到
K ( k ) = 2 1 + k ′ K ( 1 − k ′ 1 + k ′ ) . (10) K(k)={2\over 1+k'} K\left({1-k'\over 1+k'}\right). \tag {10} K(k)=1+k′2K(1+k′1−k′).(10)
令
l ≡ ( 1 − k ′ ) / ( 1 + k ′ ) , l\equiv (1-k')/(1+k'), l≡(1−k′)/(1+k′), 得 k ′ = 1 − l 1 + l , k = 1 − k ′ 2 = 2 l 1 + l , k'={1-l\over 1+l}, \quad k=\sqrt{1-k'^2}= {2\sqrt{l}\over 1+l}, k′=1+l1−l,k=1−k′2=1+l2l, 因此 K ( k ) = 1 k + 1 K ( 2 k 1 + k ) . (11) K(k)={1\over k+1} K\left({2\sqrt{k}\over 1+k}\right). \tag {11} K(k)=k+11K(1+k2k).(11) 类似地可以得到
E ( k ) = 1 + k 2 E ( 2 k 1 + k ) + k ′ 2 2 K ( k ) (12) E(k)={1+k\over 2}E\left({2\sqrt{k}\over 1+k}\right)+{k'^2\over 2}K(k) \tag {12} E(k)=21+kE(1+k2k)+2k′2K(k)(12) E ( k ) = ( 1 + k ′ ) E ( 1 − k ′ 1 + k ′ ) − k ′ K ( k ) . (13) E(k)=(1+k')E\left({1-k'\over 1+k'}\right)-k'K(k). \tag {13} E(k)=(1+k′)E(1+k′1−k′)−k′K(k).(13) 这里 E ( k ) E(k) E(k) 为第二类完全椭圆积分。
K ′ ( k ) = K ( k ′ ) = 2 1 + k K ( 1 − k 1 + k ) = 2 1 + k K ′ ( 1 − ( 1 − k 1 + k ) 2 ) = 2 1 + k K ′ ( 2 k 1 + k ) (14) \begin{aligned} K'(k)&=&K(k') = {2\over 1+k} K\left({1-k\over 1+k}\right)\\ &=& {2\over 1+k}K'\left({\sqrt{1-\left({1-k\over 1+k}\right)^2}\,}\right)\\ &=& {2\over 1+k}K'\left({2\sqrt{k}\over 1+k}\right) \end{aligned} \tag {14} K′(k)===K(k′)=1+k2K(1+k1−k)1+k2K′⎝⎛1−(1+k1−k)2⎠⎞1+k2K′(1+k2k)(14)
K ′ ( k ) = 1 1 + k ′ K ( 2 k ′ 1 + k ′ ) = 1 1 + k ′ K ′ ( 1 − k ′ 1 + k ′ ) , (15) K'(k)={1\over 1+k'} K\left({2\sqrt{k'}\over 1+k'}\right)= {1\over 1+k'} K'\left({1-k'\over 1+k'}\right), \tag {15} K′(k)=1+k′1K(1+k′2k′)=1+k′1K′(1+k′1−k′),(15)
E ′ ( k ) = ( 1 + k ) E ′ ( 2 k 1 + k ) − k K ′ ( k ) (16) E'(k)=(1+k)E'\left({2\sqrt{k}\over 1+k}\right)-kK'(k) \tag {16} E′(k)=(1+k)E′(1+k2k)−kK′(k)(16)
E ′ ( k ) = ( 1 + k ′ 2 ) E ′ ( 1 − k ′ 1 + k ′ ) + k 2 2 K ′ ( k ) . (17) E'(k)=\left({1+k'\over 2}\right)E'\left({1-k'\over 1+k'}\right)+{k^2\over 2}K'(k).\tag {17} E′(k)=(21+k′)E′(1+k′1−k′)+2k2K′(k).(17)
K ′ ( k ) K ( k ) = 2 K ′ ( 2 k 1 + k ) K ( 2 k 1 + k ) = 1 2 K ′ ( 1 − k ′ 1 + k ′ ) K ( 1 − k ′ 1 + k ′ ) (18) {K'(k)\over K(k)}=2 {K'\left({2\sqrt{k}\,\over 1+k}\right)\over K\left({2\sqrt{k}\,\over 1+k}\right)} = {1\over 2} {K'\left({1-k'\over 1+k'}\right)\over K\left({1-k'\over 1+k'}\right)} \tag {18} K(k)K′(k)=2K(1+k2k)K′(1+k2k)=21K(1+k′1−k′)K′(1+k′1−k′)(18)
Ref.
https://archive.lib.msu.edu/crcmath/math/math/e/e099.htm
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