Une intégrale elliptique est une intégrale de la forme f ( x ) = ∫ x 0 x R ( t , P ( t ) ) d t {\displaystyle f(x)=\int _{x_{0}}^{x}R\left(t,{\sqrt {P(t)}}\right)\mathrm {d} t} R {\displaystyle R} fonction rationnelle à deux variables, P {\displaystyle P} fonction polynomiale de degré 3 ou 4 avec des racines simples et x 0 {\displaystyle x_{0}} f ( x ) = ∫ x 0 x A ( t ) + B ( t ) P ( t ) C ( t ) + D ( t ) P ( t ) d t {\displaystyle f(x)=\int _{x_{0}}^{x}{A(t)+B(t){\sqrt {P(t)}} \over C(t)+D(t){\sqrt {P(t)}}}\,\mathrm {d} t} A {\displaystyle A} B {\displaystyle B} C {\displaystyle C} D {\displaystyle D}
Adrien-Marie Legendre , qui en a offert la première étude systématique, a montré que des changements de variables adéquats permettent d'exprimer les intégrales elliptiques en fonction de seulement trois formes canoniques[ 1] intégrale elliptique de première , de deuxième et de troisième espèce où 0 < k < 1 {\displaystyle 0<k<1} [ 4]
espèce k 2 ∈ [ 0 ; ∞ [ ∩ φ ∈ ] − ∞ ; ∞ [ {\displaystyle \scriptstyle {\begin{aligned}k^{2}&\in \left[0;\infty \right[\\\cap \\\varphi &\in \left]-\infty ;\infty \right[\end{aligned}}} k 2 ∈ [ 0 ; ∞ [ ∩ φ ∈ [ − π 2 ; π 2 ] {\displaystyle \scriptstyle {\begin{aligned}k^{2}&\in \left[0;\infty \right[\\\cap \\\varphi &\in \left[-{\frac {\pi }{2}};{\frac {\pi }{2}}\right]\end{aligned}}} k 2 ∈ ] − ∞ ; ∞ [ ∩ φ ∈ ] − ∞ ; ∞ [ {\displaystyle \scriptstyle {\begin{aligned}k^{2}&\in \left]-\infty ;\infty \right[\\\cap \\\varphi &\in \left]-\infty ;\infty \right[\end{aligned}}} k 2 ∈ [ 0 ; 1 ] ∩ φ ∈ ] − ∞ ; ∞ [ {\displaystyle \scriptstyle {\begin{aligned}k^{2}&\in \left[0;1\right]\\\cap \\\varphi &\in \left]-\infty ;\infty \right[\end{aligned}}} Forme de Legendre Forme de Jacobi 1re F ( φ , k ) {\displaystyle F\left(\varphi \,,\,k\right)} = F ( sin φ ; k ) {\displaystyle =F\left(\sin \varphi \,;\,k\right)} = F ( φ | k 2 ) {\displaystyle =F\left(\varphi \,|\,k^{2}\right)} = F ( φ ∖ arcsin k ) {\displaystyle =F\left(\varphi \setminus \arcsin k\right)} = ∫ 0 φ d θ 1 − k 2 sin 2 θ {\displaystyle =\int _{0}^{\varphi }\scriptstyle {\frac {\mathrm {d} \theta }{\sqrt {1-k^{2}\sin ^{2}\theta }}}} = t = sin θ ∫ 0 sin φ d t ( 1 − t 2 ) ( 1 − k 2 t 2 ) {\displaystyle {\stackrel {t=\sin \theta }{=}}\int _{0}^{\sin \varphi }\scriptstyle {\frac {\mathrm {d} t}{\sqrt {\left(1-t^{2}\right)\left(1-k^{2}t^{2}\right)}}}} 2e E ( φ , k ) {\displaystyle E\left(\varphi \,,\,k\right)} = E ( sin φ ; k ) {\displaystyle =E\left(\sin \varphi \,;\,k\right)} = E ( φ | k 2 ) {\displaystyle =E\left(\varphi \,|\,k^{2}\right)} = E ( φ ∖ arcsin k ) {\displaystyle =E\left(\varphi \setminus \arcsin k\right)} = ∫ 0 φ 1 − k 2 sin 2 θ d θ {\displaystyle =\int _{0}^{\varphi }\scriptstyle {\sqrt {1-k^{2}\sin ^{2}\theta }}\,\mathrm {d} \theta } = t = sin θ ∫ 0 sin φ 1 − k 2 t 2 1 − t 2 d t {\displaystyle {\stackrel {t=\sin \theta }{=}}\int _{0}^{\sin \varphi }\scriptstyle {\frac {\sqrt {1-k^{2}t^{2}}}{\sqrt {1-t^{2}}}}\,\mathrm {d} t} 3e Π ( n , φ , k ) {\displaystyle \Pi \left(n\,,\,\varphi \,,\,k\right)} = Π ( n , sin φ ; k ) {\displaystyle =\Pi \left(n\,,\,\sin \varphi \,;\,k\right)} = Π ( n , φ | k 2 ) {\displaystyle =\Pi \left(n\,,\,\varphi \,|\,k^{2}\right)} = Π ( n , φ ∖ arcsin k ) {\displaystyle =\Pi \left(n\,,\,\varphi \setminus \arcsin k\right)} = ∫ 0 φ 1 1 − n sin 2 θ d θ 1 − k 2 sin 2 θ {\displaystyle =\int _{0}^{\varphi }\scriptstyle {\frac {1}{1-n\sin ^{2}\theta }}{\frac {\mathrm {d} \theta }{\sqrt {1-k^{2}\sin ^{2}\theta }}}} = t = sin θ ∫ 0 sin φ 1 1 − n t 2 d t ( 1 − t 2 ) ( 1 − k 2 t 2 ) {\displaystyle {\stackrel {t=\sin \theta }{=}}\int _{0}^{\sin \varphi }\scriptstyle {\frac {1}{1-nt^{2}}}{\frac {\mathrm {d} t}{\sqrt {\left(1-t^{2}\right)\left(1-k^{2}t^{2}\right)}}}}
On prendra garde en particulier à ne pas confondre la virgule avec le point-virgule. La notation de l'intégrale elliptique avec un point-virgule ne permet que d'exprimer des intégrales elliptiques où φ ∈ [ − π / 2 ; π / 2 ] {\displaystyle \varphi \in \left[-\pi /2;\pi /2\right]} m {\displaystyle m} k 2 {\displaystyle k^{2}} m < 0 {\displaystyle m<0} on peut toujours se ramener à une forme où k ∈ ] 0 ; 1 [ {\displaystyle k\in ]0;1[} . n ∈ C {\displaystyle n\in \mathbb {C} } [ B 1]
D ( φ , k ) = D ( sin φ ; k ) = D ( φ | k 2 ) = D ( φ ∖ arcsin k ) = ∫ 0 φ sin 2 θ d θ 1 − k 2 sin 2 θ = t = sin θ ∫ = 0 sin φ t 2 d t ( 1 − t 2 ) ( 1 − k 2 t 2 ) {\displaystyle D\left(\varphi \,,\,k\right)=D\left(\sin \varphi \,;\,k\right)=D\left(\varphi \,|\,k^{2}\right)=D\left(\varphi \setminus \arcsin k\right)=\int _{0}^{\varphi }{\frac {\sin ^{2}\theta \,\mathrm {d} \theta }{\sqrt {1-k^{2}\sin ^{2}\theta }}}{\stackrel {t=\sin \theta }{=}}\int _{=0}^{\sin \varphi }{\frac {t^{2}\mathrm {d} t}{\sqrt {\left(1-t^{2}\right)\left(1-k^{2}t^{2}\right)}}}} On appelle :
k ∈ ] 0 ; 1 [ {\displaystyle k\in ]0;1[} module elliptique ou excentricité m = k 2 {\displaystyle m=k^{2}} paramètre k ′ = 1 − k 2 {\displaystyle k'={\sqrt {1-k^{2}}}} comodule arcsin k {\displaystyle \arcsin k} angle modulaire φ {\displaystyle \varphi } amplitude n {\displaystyle n} caractéristique L'intégrale est dite :
incomplète si φ {\displaystyle \varphi } complète si φ = π / 2 {\displaystyle \varphi =\pi /2} Les intégrales elliptiques complètes de 1re , 2e et 3e espèce sont respectivement[ 5]
K ( k ) = F ( π 2 , k ) E ( k ) = E ( π 2 , k ) Π ( n , k ) = Π ( n , π 2 , k ) {\displaystyle {\begin{aligned}K\left(k\right)&=F\left({\tfrac {\pi }{2}},k\right)\\E\left(k\right)&=E\left({\tfrac {\pi }{2}},k\right)\\\Pi \left(n,k\right)&=\Pi \left(n,{\tfrac {\pi }{2}},k\right)\end{aligned}}} K m ( m ) = F ( π 2 | m ) E m ( m ) = E ( π 2 | m ) Π ( n | m ) = Π ( n , π 2 | m ) {\displaystyle {\begin{aligned}K_{m}\left(m\right)&=F\left({\tfrac {\pi }{2}}|m\right)\\E_{m}\left(m\right)&=E\left({\tfrac {\pi }{2}}|m\right)\\\Pi \left(n\,|\,m\right)&=\Pi \left(n,{\tfrac {\pi }{2}}|m\right)\end{aligned}}}
On définit aussi[ 6]
K ′ ( k ) = K ( k ′ ) E ′ ( k ) = E ( k ′ ) Π ′ ( k ) = Π ( k ′ ) {\displaystyle {\begin{aligned}K'\left(k\right)&=K\left(k'\right)\\E'\left(k\right)&=E\left(k'\right)\\\Pi '\left(k\right)&=\Pi \left(k'\right)\end{aligned}}}
On définit[ A 2]
Δ = Δ ( θ ) = Δ ( θ , k ) = 1 − k 2 sin 2 θ {\displaystyle \Delta =\Delta (\theta )=\Delta (\theta ,k)={\sqrt {1-k^{2}\sin ^{2}\theta }}} Le "nom elliptique"[ traduction souhaitée 1]
q ( k ) = exp [ − π K ′ ( k ) K ( k ) ] {\displaystyle q\left(k\right)=\exp \left[-\pi {\frac {K'\left(k\right)}{K\left(k\right)}}\right]}
Intégrales complètes
K ( k ) {\displaystyle K(k)} E ( k ) {\displaystyle E(k)} Π ( n , k ) {\displaystyle \Pi (n,k)} pour différentes valeurs de
n {\displaystyle n} K m ( m ) {\displaystyle K_{m}(m)} et
E m ( m ) {\displaystyle E_{m}(m)} La pente n'est pas nulle en 0.
pointillés ligne continue tirets f ( θ , k ) = 1 1 − k 2 sin 2 θ {\displaystyle \color {red}f\left(\theta \,,\,k\right)={\frac {1}{\sqrt {1-k^{2}\sin ^{2}\theta }}}} F ( φ , k ) = ∫ 0 φ f ( θ , k ) d θ {\displaystyle \color {orange}F\left(\varphi \,,\,k\right)=\int _{0}^{\varphi }f\left(\theta \,,\,k\right)\mathrm {d} \theta } K ( k ) = F ( π 2 , k ) d θ {\displaystyle \color {brown}K\left(k\right)=F\left({\tfrac {\pi }{2}},k\right)\mathrm {d} \theta } e ( θ , k ) = 1 − k 2 sin 2 θ {\displaystyle \color {blue}e\left(\theta \,,\,k\right)={\sqrt {1-k^{2}\sin ^{2}\theta }}} E ( φ , k ) = ∫ 0 φ e ( θ , k ) d θ {\displaystyle \color {cyan}E\left(\varphi \,,\,k\right)=\int _{0}^{\varphi }e\left(\theta \,,\,k\right)\mathrm {d} \theta } E ( k ) = E ( π 2 , k ) d θ {\displaystyle \color {purple}E\left(k\right)=E\left({\tfrac {\pi }{2}},k\right)\mathrm {d} \theta } π ( n , θ , k ) = 1 1 − n sin 2 θ 1 1 − k 2 sin 2 θ {\displaystyle \color {green}\pi \left(n\,,\,\theta \,,\,k\right)={\frac {1}{1-n\sin ^{2}\theta }}{\frac {1}{\sqrt {1-k^{2}\sin ^{2}\theta }}}} Π ( n , φ , k ) = ∫ 0 φ π ( n , θ , k ) d θ {\displaystyle \definecolor {lightGreen}{rgb}{0,1,0.25098039215686274}\color {lightGreen}\Pi \left(n\,,\,\varphi \,,\,k\right)=\int _{0}^{\varphi }\pi \left(n\,,\,\theta \,,\,k\right)\mathrm {d} \theta } Π ( n , k ) = ∫ 0 φ π ( n , π 2 , k ) d θ {\displaystyle \color {black}\Pi \left(n\,,\,k\right)=\int _{0}^{\varphi }\pi \left(n\,,\,{\tfrac {\pi }{2}}\,,\,k\right)\mathrm {d} \theta }
F ( φ | m ) {\displaystyle F(\varphi |m)} pour diverses valeurs de
m {\displaystyle m} E ( φ | m ) {\displaystyle E(\varphi |m)} pour diverses valeurs de
m {\displaystyle m}
L'intégrale est appelée elliptique car des intégrales de cette forme apparaissent lors du calcul du périmètre des ellipses et de la surface des ellipsoïdes . Il existe également des applications de grande envergure en physique. Par exemple :
Legendre appelait ces intégrales des fonctions elliptiques. Après les travaux de Niels Abel et de Carl Gustav Jakob Jacobi , en 1827, le nom de fonction elliptique est maintenant réservé aux applications réciproques de ces intégrales ou découlant de ces applications réciproques : les fonctions elliptiques de Jacobi , les fonctions elliptiques de Weierstrass et les fonctions elliptiques d'Abel .
Adrien-Marie Legendre a montré que des changements de variables permettent de ramener les intégrales de la forme[ A 3] f ( x ) = ∫ x 0 x A ( t ) + B ( t ) P ( t ) C ( t ) + D ( t ) P ( t ) d t {\displaystyle f(x)=\int _{x_{0}}^{x}{A(t)+B(t){\sqrt {P(t)}} \over C(t)+D(t){\sqrt {P(t)}}}\,\mathrm {d} t}
En effet, on peut décomposer l'intégrande ainsi :
f ( x ) = ∫ x 0 x A ( t ) C ( t ) − B ( t ) D ( t ) P ( t ) C 2 ( t ) − D 2 ( t ) P ( t ) d t + ∫ x 0 x [ B ( t ) C ( t ) − A ( t ) D ( t ) ] P ( t ) [ C 2 ( t ) − D 2 ( t ) P ( t ) ] P ( t ) d t = ∫ x 0 x E ( t ) d t + ∫ x 0 x F ( t ) I ( t ) d t + ∫ x 0 x G ( t ) P ( t ) d t + ∫ x 0 x H ( t ) I ( t ) P ( t ) d t = ∑ i = 0 deg ( E ) ρ i ∫ x 0 x t i d t + ∑ i = 1 deg ( I ) κ i ∫ x 0 x 1 ( t − t i ) p i d t + ∑ i = 0 deg ( G ) λ i ∫ x 0 x t i P ( t ) d t + ∑ i = 1 deg ( I ) μ i ∫ x 0 x 1 ( t − t i ) p i P ( t ) d t = ∑ i = 0 deg ( E ) [ ρ i t i + 1 i + 1 ] x 0 x + ∑ i = 1 deg ( I ) [ ( p i = 1 ) κ i ln ( t − t i ) + ( p i ≠ 1 ) − κ i ( p i − 1 ) ( t − t i ) p i − 1 ] x 0 x + ∑ i = 0 deg ( G ) λ i ∫ x 0 x t i P ( t ) d t + ∑ i = 1 deg ( I ) μ i ∫ x 0 x 1 ( t − t i ) p i P ( t ) d t {\displaystyle {\begin{aligned}f(x)&=\int _{x_{0}}^{x}{\frac {A(t)C(t)-B(t)D(t)P(t)}{C^{2}(t)-D^{2}(t)P(t)}}\,\mathrm {d} t+\int _{x_{0}}^{x}{\frac {\left[B(t)C(t)-A(t)D(t)\right]P(t)}{\left[C^{2}(t)-D^{2}(t)P(t)\right]{\sqrt {P(t)}}}}\,\mathrm {d} t\\&=\int _{x_{0}}^{x}E(t)\,\mathrm {d} t+\int _{x_{0}}^{x}{\frac {F(t)}{I(t)}}\,\mathrm {d} t+\int _{x_{0}}^{x}{\frac {G(t)}{\sqrt {P(t)}}}\,\mathrm {d} t+\int _{x_{0}}^{x}{\frac {H(t)}{I(t){\sqrt {P(t)}}}}\,\mathrm {d} t\\&=\sum _{i=0}^{\deg(E)}\rho _{i}\int _{x_{0}}^{x}t^{i}\,\mathrm {d} t+\sum _{i=1}^{\deg(I)}\kappa _{i}\int _{x_{0}}^{x}{\frac {1}{\left(t-t_{i}\right)^{p_{i}}}}\,\mathrm {d} t+\sum _{i=0}^{\deg(G)}\lambda _{i}\int _{x_{0}}^{x}{\frac {t^{i}}{\sqrt {P(t)}}}\,\mathrm {d} t+\sum _{i=1}^{\deg(I)}\mu _{i}\int _{x_{0}}^{x}{\frac {1}{(t-t_{i})^{p_{i}}{\sqrt {P(t)}}}}\,\mathrm {d} t\\&=\sum _{i=0}^{\deg(E)}\left[\rho _{i}{\frac {t^{i+1}}{i+1}}\right]_{x_{0}}^{x}+\sum _{i=1}^{\deg(I)}\left[{\begin{matrix}\left(p_{i}=1\right)\kappa _{i}\ln \left(t-t_{i}\right)\\+\left(p_{i}\neq 1\right){\frac {-\kappa _{i}}{\left(p_{i}-1\right)\left(t-t_{i}\right)^{p_{i}-1}}}\end{matrix}}\right]_{x_{0}}^{x}+\sum _{i=0}^{\deg(G)}\lambda _{i}\int _{x_{0}}^{x}{\frac {t^{i}}{\sqrt {P(t)}}}\,\mathrm {d} t+\sum _{i=1}^{\deg(I)}\mu _{i}\int _{x_{0}}^{x}{\frac {1}{(t-t_{i})^{p_{i}}{\sqrt {P(t)}}}}\,\mathrm {d} t\end{aligned}}} où E {\displaystyle E} F {\displaystyle F} G {\displaystyle G} H {\displaystyle H} I {\displaystyle I} deg ( F ) ⩽ deg ( I ) − 1 {\displaystyle \deg(F)\leqslant \deg(I)-1} deg ( H ) ⩽ deg ( I ) − 1 {\displaystyle \deg(H)\leqslant \deg(I)-1} P ( t ) = α + β t + γ t 2 + δ t 3 + ϵ t 4 {\displaystyle P(t)=\alpha +\beta \,t+\gamma \,t^{2}+\delta \,t^{3}+\epsilon \,t^{4}}
Chacune des intégrales du troisième terme peut se ramener à une expression de la forme :
∫ x 0 x a t 2 + b t + c P ( t ) d t {\displaystyle \int _{x_{0}}^{x}{\frac {at^{2}+bt+c}{\sqrt {P(t)}}}\mathrm {d} t} Démonstration
On pose :
Γ i ( x ) = ∫ x 0 x t i P ( t ) d t {\displaystyle \Gamma ^{i}(x)=\int _{x_{0}}^{x}{\frac {t^{i}}{\sqrt {P(t)}}}\mathrm {d} t} Legendre fait remarquer que, puisque :
x i − 3 P ( x ) − x 0 i − 3 P ( x 0 ) = ∫ x 0 x d d t [ t i − 3 P ( t ) ] d t = ∫ x 0 x [ ( i − 3 ) P ( t ) + 1 2 t P ′ ( t ) ] t i − 4 P ( t ) d t = ∫ x 0 x [ ( i − 3 ) ( α + β t + γ t 2 + δ t 3 + ϵ t 4 ) + 1 2 ( β t + 2 γ t 2 + 3 δ t 3 + 4 ϵ t 4 ) ] t i − 4 P ( t ) d t = ∫ x 0 x [ ( i − 3 ) α + ( i − 5 2 ) β t + ( i − 2 ) γ t 2 + ( i − 3 2 ) δ t 3 + ( i − 1 ) ϵ t 4 ] t i − 4 P ( t ) d t = ( i − 3 ) α Γ i − 4 ( x ) + ( i − 5 2 ) β Γ i − 3 ( x ) + ( i − 2 ) γ Γ i − 2 ( x ) + ( i − 3 2 ) δ Γ i − 1 ( x ) + ( i − 1 ) ϵ Γ i ( x ) {\displaystyle {\begin{aligned}x^{i-3}{\sqrt {P(x)}}-x_{0}^{i-3}{\sqrt {P(x_{0})}}&=\int _{x_{0}}^{x}{\frac {\mathrm {d} }{\mathrm {d} t}}\left[t^{i-3}{\sqrt {P(t)}}\right]\mathrm {d} t=\int _{x_{0}}^{x}{\frac {\left[(i-3)P(t)+{\tfrac {1}{2}}tP'(t)\right]t^{i-4}}{\sqrt {P(t)}}}\mathrm {d} t\\&=\int _{x_{0}}^{x}{\frac {\left[{\begin{aligned}(i-3)\left(\alpha +\beta \,t+\;\,\gamma \,t^{2}+\;\;\delta \,t^{3}+\;\;\epsilon \,t^{4}\right)\\+{\tfrac {1}{2}}\;\left(\qquad \beta t+2\gamma \,t^{2}+3\delta \,t^{3}+4\epsilon \,t^{4}\right)\end{aligned}}\right]t^{i-4}}{\sqrt {P(t)}}}\mathrm {d} t\\&=\int _{x_{0}}^{x}{\frac {\left[\left(i-3\right)\alpha +\left(i-{\tfrac {5}{2}}\right)\beta \,t+\left(i-2\right)\gamma \,t^{2}+\left(i-{\tfrac {3}{2}}\right)\delta \,t^{3}+\left(i-1\right)\epsilon \,t^{4}\right]t^{i-4}}{\sqrt {P(t)}}}\mathrm {d} t\\&=\left(i-3\right)\alpha \,\Gamma ^{i-4}(x)+\left(i-{\tfrac {5}{2}}\right)\beta \,\Gamma ^{i-3}(x)+\left(i-2\right)\gamma \,\Gamma ^{i-2}(x)+\left(i-{\tfrac {3}{2}}\right)\delta \,\Gamma ^{i-1}(x)+\left(i-1\right)\epsilon \,\Gamma ^{i}(x)\end{aligned}}} ⇒ Γ i ( x ) = x i − 3 P ( x ) − x 0 i − 3 P ( x 0 ) − ( i − 3 2 ) δ Γ i − 1 ( x ) − ( i − 2 ) γ Γ i − 2 ( x ) − ( i − 5 2 ) β Γ i − 3 ( x ) − ( i − 3 ) α Γ i − 4 ( x ) ( i − 1 ) ϵ {\displaystyle \Rightarrow \Gamma ^{i}(x)={\frac {x^{i-3}{\sqrt {P(x)}}-x_{0}^{i-3}{\sqrt {P(x_{0})}}-\left(i-{\tfrac {3}{2}}\right)\delta \,\Gamma ^{i-1}(x)-\left(i-2\right)\gamma \,\Gamma ^{i-2}(x)-\left(i-{\tfrac {5}{2}}\right)\beta \,\Gamma ^{i-3}(x)-\left(i-3\right)\alpha \,\Gamma ^{i-4}(x)}{(i-1)\epsilon }}} Γ i {\displaystyle \Gamma ^{i}} Γ i − 1 {\displaystyle \Gamma ^{i-1}} Γ i − 1 {\displaystyle \Gamma ^{i-1}} Γ i − 2 {\displaystyle \Gamma ^{i-2}} Γ 3 {\displaystyle \Gamma ^{3}} Γ 2 {\displaystyle \Gamma ^{2}} Γ 1 {\displaystyle \Gamma ^{1}} Γ 0 {\displaystyle \Gamma ^{0}}
Chacune des intégrales du quatrième terme peut se ramener à une expression de la forme :
∫ x 0 x a t 2 + b t + c + d t − r P ( t ) d t {\displaystyle \int _{x_{0}}^{x}{\frac {at^{2}+bt+c+{\frac {d}{t-r}}}{\sqrt {P(t)}}}\mathrm {d} t} Démonstration
Si on pose ω = t − r {\displaystyle \omega =t-r}
P ( t ) = α + β ( ω + r ) + γ ( ω + r ) 2 + δ ( ω + r ) 3 + ϵ ( ω + r ) 4 = α ′ + β ′ ω + γ ′ ω 2 + δ ′ ω 3 + ϵ ′ ω 4 = P ω ( ω ) {\displaystyle P(t)=\alpha +\beta \left(\omega +r\right)+\gamma \left(\omega +r\right)^{2}+\delta \left(\omega +r\right)^{3}+\epsilon \left(\omega +r\right)^{4}=\alpha '+\beta '\omega +\gamma '\omega ^{2}+\delta '\omega ^{3}+\epsilon '\omega ^{4}=P_{\omega }\left(\omega \right)} On pose :
Υ r i ( x ) = ∫ x 0 − r x − r 1 ω i P ω ( ω ) d ω {\displaystyle \Upsilon _{r}^{i}(x)=\int _{x_{0}-r}^{x-r}{\frac {1}{\omega ^{i}{\sqrt {P_{\omega }(\omega )}}}}\,\mathrm {d} \omega } et on a :
− P ω ( x − r ) ( x − r ) i − 1 + P ω ( x 0 − r ) ( x 0 − r ) i − 1 = ∫ x 0 − r x − r d d ω [ − ω 1 − i P ω ( ω ) ] d ω = ∫ x 0 − r x − r ( i − 1 ) P ω ( ω ) − 1 2 ω P ω ′ ( ω ) ω i P ω ( ω ) d ω = ∫ x 0 − r x − r [ ( i − 1 ) ( α ′ + β ′ ω + γ ′ ω 2 + δ ′ ω 3 + ϵ ′ ω 4 ) − 1 2 ( β ′ ω + 2 γ ′ ω 2 + 3 δ ′ ω 3 + 4 ϵ ′ ω 4 ) ] ω i P ω ( ω ) d ω = ∫ x 0 − r x − r [ ( i − 1 ) α ′ + ( i − 3 2 ) β ′ ω + ( i − 2 ) γ ′ ω 2 + ( i − 5 2 ) δ ′ ω 3 + ( q i − 3 ) ϵ ′ ω 4 ] ω i P ω ( ω ) d ω = ( i − 1 ) α ′ Υ r i ( x ) + ( i − 3 2 ) β ′ Υ r i − 1 ( x ) + ( i − 2 ) γ ′ Υ r i − 2 ( x ) + ( i − 5 2 ) δ ′ Υ r i − 3 ( x ) + ( i − 3 ) ϵ ′ Υ r i − 4 ( x ) {\displaystyle {\begin{aligned}-{\frac {\sqrt {P_{\omega }(x-r)}}{\left(x-r\right)^{i-1}}}+{\frac {\sqrt {P_{\omega }(x_{0}-r)}}{\left(x_{0}-r\right)^{i-1}}}&=\int _{x_{0}-r}^{x-r}{\frac {\mathrm {d} }{\mathrm {d} \omega }}\left[-\omega ^{1-i}{\sqrt {P_{\omega }\left(\omega \right)}}\right]\mathrm {d} \omega =\int _{x_{0}-r}^{x-r}{\frac {(i-1)P_{\omega }(\omega )-{\tfrac {1}{2}}\omega P_{\omega }'(\omega )}{\omega ^{i}{\sqrt {P_{\omega }(\omega )}}}}\mathrm {d} \omega \\&=\int _{x_{0}-r}^{x-r}{\frac {\left[{\begin{aligned}(i-1)\left(\alpha '+\beta '\omega +\;\;\,\gamma '\omega ^{2}+\;\;\delta '\omega ^{3}+\;\;\,\epsilon '\omega ^{4}\right)\\-{\tfrac {1}{2}}\left(\qquad \;\beta '\omega +2\,\gamma '\omega ^{2}+3\,\delta '\omega ^{3}+4\,\epsilon '\omega ^{4}\right)\\\end{aligned}}\right]}{\omega ^{i}{\sqrt {P_{\omega }(\omega )}}}}\mathrm {d} \omega \\&=\int _{x_{0}-r}^{x-r}{\frac {\left[\left(i-1\right)\alpha '+\left(i-{\tfrac {3}{2}}\right)\beta '\omega +\left(i-2\right)\gamma '\omega ^{2}+\left(i-{\tfrac {5}{2}}\right)\delta '\omega ^{3}+\left(q_{i}-3\right)\epsilon '\omega ^{4}\right]}{\omega ^{i}{\sqrt {P_{\omega }(\omega )}}}}\mathrm {d} \omega \\&=\left(i-1\right)\alpha '\,\Upsilon _{r}^{i}(x)+\left(i-{\tfrac {3}{2}}\right)\beta '\,\Upsilon _{r}^{i-1}(x)+\left(i-2\right)\gamma '\,\Upsilon _{r}^{i-2}(x)+\left(i-{\tfrac {5}{2}}\right)\delta '\,\Upsilon _{r}^{i-3}(x)+\left(i-3\right)\epsilon '\,\Upsilon _{r}^{i-4}(x)\end{aligned}}} ⇒ Υ r i ( x ) = − P ω ( x − r ) ( x − r ) i − 1 + P ω ( x 0 − r ) ( x 0 − r ) i − 1 − ( i − 3 2 ) β ′ Υ r i − 1 ( x ) + ( i − 2 ) γ ′ Υ r i − 2 ( x ) + ( i − 5 2 ) δ ′ Υ r i − 3 ( x ) + ( i − 3 ) ϵ ′ Υ r i − 4 ( x ) ( i − 1 ) α ′ {\displaystyle \Rightarrow \Upsilon _{r}^{i}(x)={\frac {-{\frac {\sqrt {P_{\omega }(x-r)}}{\left(x-r\right)^{i-1}}}+{\frac {\sqrt {P_{\omega }(x_{0}-r)}}{\left(x_{0}-r\right)^{i-1}}}-\left(i-{\tfrac {3}{2}}\right)\beta '\,\Upsilon _{r}^{i-1}(x)+\left(i-2\right)\gamma '\,\Upsilon _{r}^{i-2}(x)+\left(i-{\tfrac {5}{2}}\right)\delta '\,\Upsilon _{r}^{i-3}(x)+\left(i-3\right)\epsilon '\,\Upsilon _{r}^{i-4}(x)}{(i-1)\alpha '}}} Ici encore, les Υ r i {\displaystyle \Upsilon _{r}^{i}} Υ r 1 {\displaystyle \Upsilon _{r}^{1}} Υ r 0 {\displaystyle \Upsilon _{r}^{0}} Υ r − 1 {\displaystyle \Upsilon _{r}^{-1}} Υ r − 2 {\displaystyle \Upsilon _{r}^{-2}}
Si α {\displaystyle \alpha } β {\displaystyle \beta } γ {\displaystyle \gamma } δ {\displaystyle \delta } ϵ {\displaystyle \epsilon } r {\displaystyle r}
z = 1 t − r ⇔ t = 1 z + r {\displaystyle z={\frac {1}{t-r}}\Leftrightarrow t={\frac {1}{z}}+r} On peut ensuite faire disparaître les puissances impaires sous le radical. Si on écrit :
P ( t ) = α + β t + γ t 2 + δ t 3 + ϵ t 4 = ( α 1 + 2 β 1 t + γ 1 t 2 ) ( α 2 + 2 β 2 t + γ 2 t 2 ) = ϵ ( t − t a ) ( t − t b ) ( t − t c ) ( t − t d ) {\displaystyle P(t)=\alpha +\beta \,t+\gamma \,t^{2}+\delta \,t^{3}+\epsilon \,t^{4}=\left(\alpha _{1}+2\beta _{1}t+\gamma _{1}t^{2}\right)\left(\alpha _{2}+2\beta _{2}t+\gamma _{2}t^{2}\right)=\epsilon \,\left(t-t_{a}\right)\left(t-t_{b}\right)\left(t-t_{c}\right)\left(t-t_{d}\right)} Une première méthode est de poser : y = α 2 + 2 β 2 t + γ 2 t 2 α 1 + 2 β 1 t + γ 1 t 2 ⇒ P ( t ) = ( α 1 + 2 β 1 t + γ 1 t 2 ) y {\displaystyle y={\sqrt {\frac {\alpha _{2}+2\beta _{2}t+\gamma _{2}t^{2}}{\alpha _{1}+2\beta _{1}t+\gamma _{1}t^{2}}}}\Rightarrow {\sqrt {P(t)}}=\left(\alpha _{1}+2\beta _{1}t+\gamma _{1}t^{2}\right)y} Ainsi, on a : t = β 1 y 2 − β 2 ± ( β 1 y 2 − β 2 ) 2 − ( α 1 y 2 − α 2 ) ( γ 1 y 2 − γ 2 ) γ 2 − γ 1 y 2 {\displaystyle t={\frac {\beta _{1}\,y^{2}-\beta _{2}\pm {\sqrt {\left(\beta _{1}\,y^{2}-\beta _{2}\right)^{2}-\left(\alpha _{1}\,y^{2}-\alpha _{2}\right)\left(\gamma _{1}\,y^{2}-\gamma _{2}\right)}}}{\gamma _{2}-\gamma _{1}\,y^{2}}}} d t = y [ 2 γ 1 ( β 1 y 2 − β 2 ) − 2 β 1 ( γ 1 y 2 − γ 2 ) ( γ 1 y 2 − γ 2 ) 2 ± ( γ 1 y 2 − γ 2 ) [ 2 ( α 1 γ 1 − β 1 2 ) y 2 + 2 β 1 β 2 + α 1 γ 2 − α 2 γ 1 ] + 2 γ 1 [ ( β 1 y 2 − β 2 ) 2 − ( α 1 y 2 − α 2 ) ( γ 1 y 2 − γ 2 ) ] ( γ 1 y 2 − γ 2 ) 2 ( β 1 y 2 − β 2 ) 2 − ( α 1 y 2 − α 2 ) ( γ 1 y 2 − γ 2 ) ] d y {\displaystyle \mathrm {d} t=y\left[{\tfrac {2\,\gamma _{1}\,\left(\beta _{1}\,y^{2}-\beta _{2}\right)-2\,\beta _{1}\,\left(\gamma _{1}\,y^{2}-\gamma _{2}\right)}{\left(\gamma _{1}\,y^{2}-\gamma _{2}\right)^{2}}}\pm {\tfrac {\left(\gamma _{1}\,y^{2}-\gamma _{2}\right)\left[2\left(\alpha _{1}\gamma _{1}-\beta _{1}^{2}\right)y^{2}+2\,\beta _{1}\beta _{2}\,+\alpha _{1}\gamma _{2}-\alpha _{2}\gamma _{1}\right]+2\gamma _{1}\left[\left(\beta _{1}\,y^{2}-\beta _{2}\right)^{2}-\left(\alpha _{1}\,y^{2}-\alpha _{2}\right)\left(\gamma _{1}\,y^{2}-\gamma _{2}\right)\right]}{\left(\gamma _{1}\,y^{2}-\gamma _{2}\right)^{2}{\sqrt {\left(\beta _{1}\,y^{2}-\beta _{2}\right)^{2}-\left(\alpha _{1}\,y^{2}-\alpha _{2}\right)\left(\gamma _{1}\,y^{2}-\gamma _{2}\right)}}}}\right]\mathrm {d} y} Une deuxième méthode qui permet d'avoir P ( t ) = ( α 3 + γ 3 y 2 ) ( α 4 + γ 4 y 2 ) {\displaystyle P(t)=\left(\alpha _{3}+\gamma _{3}y^{2}\right)\left(\alpha _{4}+\gamma _{4}y^{2}\right)} t = p + q y 1 + y {\displaystyle t={\frac {p+q\,y}{1+y}}} Si α {\displaystyle \alpha } β {\displaystyle \beta } γ {\displaystyle \gamma } δ {\displaystyle \delta } ϵ {\displaystyle \epsilon } p {\displaystyle p} q {\displaystyle q} P ( y ) {\displaystyle P(y)} y {\displaystyle y} Démonstration
Deux cas se présentent :
P ( t ) {\displaystyle P(t)} En réinjectant l'expression de t {\displaystyle t} P ( t ) {\displaystyle P(t)} y {\displaystyle y} { α 1 + β 1 ( p + q ) + γ 1 p q = 0 α 2 + β 2 ( p + q ) + γ 2 p q = 0 {\displaystyle {\begin{cases}\alpha _{1}\,+\beta _{1}\,\left(p+q\right)+\gamma _{1}\,p\,q=0\\\alpha _{2}\,+\beta _{2}\,\left(p+q\right)+\gamma _{2}\,p\,q=0\end{cases}}} Si on suppose que α 1 + 2 β 1 t + γ 1 t 2 {\displaystyle \alpha _{1}+2\,\beta _{1}\,t+\gamma _{1}\,t^{2}} β 1 2 − α 1 γ 1 < 0 {\displaystyle \beta _{1}^{2}-\alpha _{1}\,\gamma _{1}<0} ( p − q ) 2 = ( p + q ) 2 − 4 p q = ( α 1 + γ 1 p q β 1 ) 2 − 4 p q = ( α 1 + γ 1 p q β 1 − 2 β 1 γ 1 ) 2 + 4 α 1 γ 1 − 4 β 1 2 γ 1 2 > 0 ⇒ p , q ∈ R {\displaystyle \left(p-q\right)^{2}=\left(p+q\right)^{2}-4p\,q=\left({\frac {\alpha _{1}+\gamma _{1}\,p\,q}{\beta _{1}}}\right)^{2}-4\,p\,q=\left({\frac {\alpha _{1}+\gamma _{1}\,p\,q}{\beta _{1}}}-{\frac {2\beta _{1}}{\gamma _{1}}}\right)^{2}+{\frac {4\,\alpha _{1}\gamma _{1}-4\,\beta _{1}^{2}}{\gamma _{1}^{2}}}>0\Rightarrow p,q\in \mathbb {R} } P ( t ) {\displaystyle P(t)} La paire d'équation se réécrit : { t a t b − 1 2 ( t a + t b ) ( p + q ) + p q = 0 t c t d − 1 2 ( t c + t d ) ( p + q ) + p q = 0 ⇒ { p + q = 2 ( t a t b − t c t d ) t a + t b − t c − t d p q = − t c t d + ( t c + t d ) ( t a t b − t c t d ) t a + t b − t c − t d = t a t b t c + t a t b t d − t a t c t d − t b t c t d t a + t b − t c − t d {\displaystyle {\begin{cases}t_{a}t_{b}-{\frac {1}{2}}\left(t_{a}+t_{b}\right)\left(p+q\right)+p\,q=0\\t_{c}t_{d}-{\frac {1}{2}}\left(t_{c}+t_{d}\right)\left(p+q\right)+p\,q=0\end{cases}}\Rightarrow {\begin{cases}p+q={\frac {2\left(t_{a}t_{b}-t_{c}t_{d}\right)}{t_{a}+t_{b}-t_{c}-t_{d}}}\\p\,q=-t_{c}t_{d}+{\frac {\left(t_{c}+t_{d}\right)\left(t_{a}t_{b}-t_{c}t_{d}\right)}{t_{a}+t_{b}-t_{c}-t_{d}}}={\frac {t_{a}t_{b}t_{c}+t_{a}t_{b}t_{d}-t_{a}t_{c}t_{d}-t_{b}t_{c}t_{d}}{t_{a}+t_{b}-t_{c}-t_{d}}}\end{cases}}} ⇒ ( p − q 2 ) 2 = ( p + q 2 ) 2 − p q = ( t a t b − t c t d ) 2 + ( − t a t b t c − t a t b t d + t a t c t d + t b t c t d ) ( t a + t b − t c − t d ) ( t a + t b − t c − t d ) 2 = 2 t a t b t c t d + t a 2 t b 2 + t c 2 t d 2 + t a 2 ( t c t d − t b t c − t b t d ) + t b 2 ( t c t d − t a t c − t a t d ) + t c 2 ( t a t b − t a t d − t b t d ) + t d 2 ( t a t b − t a t c − t b t c ) ( t a + t b − t c − t d ) 2 = ( t a 2 + t c t d − t a t c − t a t d ) ( t b 2 + t c t d − t b t c − t b t d ) ( t a + t b − t c − t d ) 2 = ( t a − t c ) ( t a − t d ) ( t b − t c ) ( t b − t d ) ( t a + t b − t c − t d ) 2 {\displaystyle {\begin{aligned}\Rightarrow \left({\frac {p-q}{2}}\right)^{2}&=\left({\frac {p+q}{2}}\right)^{2}-pq={\tfrac {\left(t_{a}t_{b}-t_{c}t_{d}\right)^{2}+\left(-t_{a}t_{b}t_{c}-t_{a}t_{b}t_{d}+t_{a}t_{c}t_{d}+t_{b}t_{c}t_{d}\right)\left(t_{a}+t_{b}-t_{c}-t_{d}\right)}{\left(t_{a}+t_{b}-t_{c}-t_{d}\right)^{2}}}\\&={\tfrac {2t_{a}t_{b}t_{c}t_{d}+t_{a}^{2}t_{b}^{2}+t_{c}^{2}t_{d}^{2}+t_{a}^{2}\left(t_{c}t_{d}-t_{b}t_{c}-t_{b}t_{d}\right)+t_{b}^{2}\left(t_{c}t_{d}-t_{a}t_{c}-t_{a}t_{d}\right)+t_{c}^{2}\left(t_{a}t_{b}-t_{a}t_{d}-t_{b}t_{d}\right)+t_{d}^{2}\left(t_{a}t_{b}-t_{a}t_{c}-t_{b}t_{c}\right)}{\left(t_{a}+t_{b}-t_{c}-t_{d}\right)^{2}}}\\&={\tfrac {\left(t_{a}^{2}+t_{c}t_{d}-t_{a}t_{c}-t_{a}t_{d}\right)\left(t_{b}^{2}+t_{c}t_{d}-t_{b}t_{c}-t_{b}t_{d}\right)}{\left(t_{a}+t_{b}-t_{c}-t_{d}\right)^{2}}}={\tfrac {\left(t_{a}-t_{c}\right)\left(t_{a}-t_{d}\right)\left(t_{b}-t_{c}\right)\left(t_{b}-t_{d}\right)}{\left(t_{a}+t_{b}-t_{c}-t_{d}\right)^{2}}}\end{aligned}}} ⇒ p , q = t a t b − t c t d ± ( t a − t c ) ( t a − t d ) ( t b − t c ) ( t b − t d ) t a + t b − t c − t d {\displaystyle \Rightarrow p,q={\frac {t_{a}t_{b}-t_{c}t_{d}\pm {\sqrt {\left(t_{a}-t_{c}\right)\left(t_{a}-t_{d}\right)\left(t_{b}-t_{c}\right)\left(t_{b}-t_{d}\right)}}}{t_{a}+t_{b}-t_{c}-t_{d}}}} Si t a ⩾ t c ⩾ t b ⩾ t d {\displaystyle t_{a}\geqslant t_{c}\geqslant t_{b}\geqslant t_{d}} p , q ∉ R {\displaystyle p,q\notin \mathbb {R} } t a ⩾ t b ⩾ t c ⩾ t d {\displaystyle t_{a}\geqslant t_{b}\geqslant t_{c}\geqslant t_{d}} t a ⩾ t c ⩾ t d ⩾ t b {\displaystyle t_{a}\geqslant t_{c}\geqslant t_{d}\geqslant t_{b}} p , q ∈ R {\displaystyle p,q\in \mathbb {R} } P ( t ) {\displaystyle P(t)} p {\displaystyle p} q {\displaystyle q} p {\displaystyle p} q {\displaystyle q}
En posant t 1 = t 2 {\displaystyle t_{1}=t^{2}} d t 1 = 2 t d t {\displaystyle \mathrm {d} t_{1}=2\,t\,\mathrm {d} t} ∫ x 0 x b t P ( t ) d t {\displaystyle \int _{x_{0}}^{x}{\frac {b\,t}{\sqrt {P(t)}}}\mathrm {d} t}
De même, ∫ x 0 x d t − r P ( t ) d t {\displaystyle \int _{x_{0}}^{x}{\frac {\frac {d}{t-r}}{\sqrt {P(t)}}}\mathrm {d} t} ∫ x 0 x d ′ d t ( 1 − n t 2 ) P ( t ) {\displaystyle \int _{x_{0}}^{x}{\frac {d'\mathrm {d} t}{\left(1-n\,t^{2}\right){\sqrt {P(t)}}}}}
Démonstration
On pose :
{ d ′ = − d r n = 1 r 2 {\displaystyle {\begin{cases}d'&=-{\frac {d}{r}}\\n&={\frac {1}{r^{2}}}\end{cases}}} On a :
∫ x 0 x d t − r P ( t ) d t = ∫ x 0 x d ( t + r ) t 2 − r 2 P ( t ) d t = t 1 = t 2 − d r ∫ x 0 x
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Deutsch
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![{\displaystyle \scriptstyle {\begin{aligned}k^{2}&\in \left[0;1\right]\\\cap \\\varphi &\in \left]-\infty ;\infty \right[\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4877a6064c9804ace5d026f4715467d6ee76d5da)
![{\displaystyle \varphi \in \left[-\pi /2;\pi /2\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb4a93743841cd7f9f51aa244a3c54f68a0d36ba)
![{\displaystyle k\in ]0;1[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aee3c3fcdebeaa5c3c862d3d254bc04993d74ed5)
![{\displaystyle q\left(k\right)=\exp \left[-\pi {\frac {K'\left(k\right)}{K\left(k\right)}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/84623646a930baeae6b1e2578e193690b1175772)
![{\displaystyle {\begin{aligned}f(x)&=\int _{x_{0}}^{x}{\frac {A(t)C(t)-B(t)D(t)P(t)}{C^{2}(t)-D^{2}(t)P(t)}}\,\mathrm {d} t+\int _{x_{0}}^{x}{\frac {\left[B(t)C(t)-A(t)D(t)\right]P(t)}{\left[C^{2}(t)-D^{2}(t)P(t)\right]{\sqrt {P(t)}}}}\,\mathrm {d} t\\&=\int _{x_{0}}^{x}E(t)\,\mathrm {d} t+\int _{x_{0}}^{x}{\frac {F(t)}{I(t)}}\,\mathrm {d} t+\int _{x_{0}}^{x}{\frac {G(t)}{\sqrt {P(t)}}}\,\mathrm {d} t+\int _{x_{0}}^{x}{\frac {H(t)}{I(t){\sqrt {P(t)}}}}\,\mathrm {d} t\\&=\sum _{i=0}^{\deg(E)}\rho _{i}\int _{x_{0}}^{x}t^{i}\,\mathrm {d} t+\sum _{i=1}^{\deg(I)}\kappa _{i}\int _{x_{0}}^{x}{\frac {1}{\left(t-t_{i}\right)^{p_{i}}}}\,\mathrm {d} t+\sum _{i=0}^{\deg(G)}\lambda _{i}\int _{x_{0}}^{x}{\frac {t^{i}}{\sqrt {P(t)}}}\,\mathrm {d} t+\sum _{i=1}^{\deg(I)}\mu _{i}\int _{x_{0}}^{x}{\frac {1}{(t-t_{i})^{p_{i}}{\sqrt {P(t)}}}}\,\mathrm {d} t\\&=\sum _{i=0}^{\deg(E)}\left[\rho _{i}{\frac {t^{i+1}}{i+1}}\right]_{x_{0}}^{x}+\sum _{i=1}^{\deg(I)}\left[{\begin{matrix}\left(p_{i}=1\right)\kappa _{i}\ln \left(t-t_{i}\right)\\+\left(p_{i}\neq 1\right){\frac {-\kappa _{i}}{\left(p_{i}-1\right)\left(t-t_{i}\right)^{p_{i}-1}}}\end{matrix}}\right]_{x_{0}}^{x}+\sum _{i=0}^{\deg(G)}\lambda _{i}\int _{x_{0}}^{x}{\frac {t^{i}}{\sqrt {P(t)}}}\,\mathrm {d} t+\sum _{i=1}^{\deg(I)}\mu _{i}\int _{x_{0}}^{x}{\frac {1}{(t-t_{i})^{p_{i}}{\sqrt {P(t)}}}}\,\mathrm {d} t\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/709a3ab6367845c01fba4b13bcfab241f3ed9627)
![{\displaystyle {\begin{aligned}x^{i-3}{\sqrt {P(x)}}-x_{0}^{i-3}{\sqrt {P(x_{0})}}&=\int _{x_{0}}^{x}{\frac {\mathrm {d} }{\mathrm {d} t}}\left[t^{i-3}{\sqrt {P(t)}}\right]\mathrm {d} t=\int _{x_{0}}^{x}{\frac {\left[(i-3)P(t)+{\tfrac {1}{2}}tP'(t)\right]t^{i-4}}{\sqrt {P(t)}}}\mathrm {d} t\\&=\int _{x_{0}}^{x}{\frac {\left[{\begin{aligned}(i-3)\left(\alpha +\beta \,t+\;\,\gamma \,t^{2}+\;\;\delta \,t^{3}+\;\;\epsilon \,t^{4}\right)\\+{\tfrac {1}{2}}\;\left(\qquad \beta t+2\gamma \,t^{2}+3\delta \,t^{3}+4\epsilon \,t^{4}\right)\end{aligned}}\right]t^{i-4}}{\sqrt {P(t)}}}\mathrm {d} t\\&=\int _{x_{0}}^{x}{\frac {\left[\left(i-3\right)\alpha +\left(i-{\tfrac {5}{2}}\right)\beta \,t+\left(i-2\right)\gamma \,t^{2}+\left(i-{\tfrac {3}{2}}\right)\delta \,t^{3}+\left(i-1\right)\epsilon \,t^{4}\right]t^{i-4}}{\sqrt {P(t)}}}\mathrm {d} t\\&=\left(i-3\right)\alpha \,\Gamma ^{i-4}(x)+\left(i-{\tfrac {5}{2}}\right)\beta \,\Gamma ^{i-3}(x)+\left(i-2\right)\gamma \,\Gamma ^{i-2}(x)+\left(i-{\tfrac {3}{2}}\right)\delta \,\Gamma ^{i-1}(x)+\left(i-1\right)\epsilon \,\Gamma ^{i}(x)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7f6a16c1f3bee06d941af9f2b775eea59e5e6f9b)
![{\displaystyle {\begin{aligned}-{\frac {\sqrt {P_{\omega }(x-r)}}{\left(x-r\right)^{i-1}}}+{\frac {\sqrt {P_{\omega }(x_{0}-r)}}{\left(x_{0}-r\right)^{i-1}}}&=\int _{x_{0}-r}^{x-r}{\frac {\mathrm {d} }{\mathrm {d} \omega }}\left[-\omega ^{1-i}{\sqrt {P_{\omega }\left(\omega \right)}}\right]\mathrm {d} \omega =\int _{x_{0}-r}^{x-r}{\frac {(i-1)P_{\omega }(\omega )-{\tfrac {1}{2}}\omega P_{\omega }'(\omega )}{\omega ^{i}{\sqrt {P_{\omega }(\omega )}}}}\mathrm {d} \omega \\&=\int _{x_{0}-r}^{x-r}{\frac {\left[{\begin{aligned}(i-1)\left(\alpha '+\beta '\omega +\;\;\,\gamma '\omega ^{2}+\;\;\delta '\omega ^{3}+\;\;\,\epsilon '\omega ^{4}\right)\\-{\tfrac {1}{2}}\left(\qquad \;\beta '\omega +2\,\gamma '\omega ^{2}+3\,\delta '\omega ^{3}+4\,\epsilon '\omega ^{4}\right)\\\end{aligned}}\right]}{\omega ^{i}{\sqrt {P_{\omega }(\omega )}}}}\mathrm {d} \omega \\&=\int _{x_{0}-r}^{x-r}{\frac {\left[\left(i-1\right)\alpha '+\left(i-{\tfrac {3}{2}}\right)\beta '\omega +\left(i-2\right)\gamma '\omega ^{2}+\left(i-{\tfrac {5}{2}}\right)\delta '\omega ^{3}+\left(q_{i}-3\right)\epsilon '\omega ^{4}\right]}{\omega ^{i}{\sqrt {P_{\omega }(\omega )}}}}\mathrm {d} \omega \\&=\left(i-1\right)\alpha '\,\Upsilon _{r}^{i}(x)+\left(i-{\tfrac {3}{2}}\right)\beta '\,\Upsilon _{r}^{i-1}(x)+\left(i-2\right)\gamma '\,\Upsilon _{r}^{i-2}(x)+\left(i-{\tfrac {5}{2}}\right)\delta '\,\Upsilon _{r}^{i-3}(x)+\left(i-3\right)\epsilon '\,\Upsilon _{r}^{i-4}(x)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d336186f5517656be09fd3dbcedc6e3155831f36)
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![{\displaystyle {\begin{aligned}\int _{x_{0}}^{x}{\frac {bt}{\sqrt {a't^{4}+2b't^{2}+c'}}}\mathrm {d} t&=\int _{x_{0}}^{x}{\frac {bt}{\sqrt {a'\left(t^{2}+{\frac {b'}{a'}}+{\frac {\sqrt {b'^{2}-a'c'}}{a'}}\right)\left(t^{2}+{\frac {b'}{a'}}-{\frac {\sqrt {b'^{2}-a'c'}}{a'}}\right)}}}\mathrm {d} t\\&{\stackrel {t_{1}=t^{2}+{\frac {b'}{a'}}}{=}}\int _{x_{0}^{2}+{\frac {b'}{a'}}}^{x^{2}+{\frac {b'}{a'}}}{\frac {b}{2{\sqrt {a'\left(t_{1}^{2}-{\frac {b'^{2}-a'c'}{a'^{2}}}\right)}}}}\mathrm {d} t_{1}\\&={\frac {b}{2}}{\sqrt {\frac {-a'}{b'^{2}-a'c'}}}\int _{t_{1}=x_{0}^{2}+{\frac {b'}{a'}}}^{x^{2}+{\frac {b'}{a'}}}{\frac {1}{\sqrt {1-{\frac {a'^{2}}{b'^{2}-a'c'}}t_{1}^{2}}}}\mathrm {d} t_{1}{\stackrel {t_{2}={\frac {a'}{\sqrt {b'^{2}-a'c'}}}t_{1}}{=}}{\frac {b}{2{\sqrt {-a'}}}}\int _{\frac {a'x_{0}^{2}+b'}{\sqrt {b'^{2}-a'c'}}}^{\frac {a'x^{2}+b'}{\sqrt {b'^{2}-a'c'}}}{\frac {1}{\sqrt {1-t_{2}^{2}}}}\mathrm {d} t_{2}\\&={\frac {b}{2{\sqrt {-a'}}}}\left[\arcsin {\frac {a'x^{2}+b'}{\sqrt {b'^{2}-a'c'}}}-\arcsin {\frac {a'x_{0}^{2}+b'}{\sqrt {b'^{2}-a'c'}}}\right]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8b254398ecc3f9d9c340b428798d9a4187e9bbf9)
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