Page:Laplace - Œuvres complètes, Gauthier-Villars, 1878, tome 8.djvu/457

On trouvera pareillement

${\begin{aligned}-\int {\frac {ardt\sin(\varphi '-\varphi )}{r'^{2}\left(1+s'^{2}\right)^{\frac {3}{2}}}}=&{\frac {1}{i^{2}(n'-n)}}\cos(n't-nt+\mathrm {B} )\\&-\alpha e\left[{\frac {1}{2i^{2}(n'-2n)}}\cos(n't-2nt+\mathrm {B} -\theta )\right.\\&\qquad \qquad \qquad \left.-{\frac {3}{2i^{2}n'}}\cos(n't+\mathrm {B} +\theta )\right]\\&-{\frac {2\alpha e'}{i^{2}(2n'-n)}}\cos(2n't-nt+\mathrm {B} +\theta '),\end{aligned}}$

${\begin{aligned}{\frac {a^{2}\cos(\varphi '-\varphi )}{r'^{2}\left(1+s'^{2}\right)^{\frac {3}{2}}}}=&{\frac {1}{i^{2}}}\cos(n't-nt+\mathrm {B} )\\&-{\frac {\alpha e}{i^{2}}}\left[\cos(n't-2nt+\mathrm {B} -\theta )-\cos(n't+\mathrm {B} +\theta )\right]\\&-{\frac {2\alpha e'}{i^{2}}}\cos(2n't-nt+\mathrm {B} +\theta ').\end{aligned}}$

On aura enfin

$a^{2}[s'-s\cos(\varphi '-\varphi )\left[{\frac {1}{r'^{2}\left(1+s'^{2}\right)^{\frac {3}{2}}}}-{\frac {r'}{\sideset {^{1}}{^{3}}v}}\right]$

$={\frac {\alpha \gamma 'i}{2}}\left\{{\begin{aligned}\left[{\frac {2}{i^{3}}}-2(b)\right]\sin(n't&+\varpi ')\\-(b_{1})&\left[\ \ \sin(2n't-\ \ nt+\ \ \mathrm {B} +\varpi ')\right.\\&\left.+\sin(\ \ n\ t-\qquad \quad \ \mathrm {B} +\varpi ')\right]\\-(b_{2})&\left[\ \ \sin(3n't-2nt+2\mathrm {B} +\varpi ')\right.\\&\left.-\sin(\ \ n't-2nt+2\mathrm {B} -\varpi ')\right]\\-(b_{3})&\left[\ \ \sin(4n't-3nt+3\mathrm {B} +\varpi ')\right.\\&\left.-\sin(2n't-3nt+3\mathrm {B} -\varpi ')\right]\\&-\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \end{aligned}}\right\}$

$=+{\frac {\alpha \gamma i}{4}}\left\{{\begin{aligned}2(b)\sin(nt+\varpi )&\\+\left[2(b_{1})+(b_{2})-{\frac {2}{i^{3}}}\right]&\left[\ \ \sin(\ \ n't\qquad \ \ \ +\ \,\mathrm {B} +\varpi )\right.\\&\left.-\sin(\ \ n't-2nt+\ \ \mathrm {B} -\varpi )\right]\\+\ \,\left[(b_{1})+(b_{2})\right]\quad \qquad &\left[\ \ \sin(2n't-\ \ nt+2\mathrm {B} +\varpi )\right.\\&\left.-\sin(2n't-3nt+2\mathrm {B} +\varpi ')\right]\\+\ldots \ldots \ldots \ldots \ldots \ldots &\ldots \ldots \ldots \ldots \ldots \end{aligned}}\right\}$