puissances supérieures ; dans ce cas, on aura
![{\displaystyle \int \Gamma (u)ds=\int r\Gamma (u)d\varphi \,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3d7f3317fe0a8453849849064e0a053707d074a0)
or
![{\displaystyle \Gamma (u)=\Gamma \left[{\frac {1-\alpha e\cos(\varphi +\varepsilon )}{a}}\right]=\Gamma \left({\frac {1}{a}}\right)-{\frac {\alpha e}{a}}\cos(\varphi +\varepsilon )\Gamma '\left({\frac {1}{a}}\right),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/42dafa55ef17299a3ebea4435d4f2378c58c3fde)
en exprimant
par
donc
![{\displaystyle \int \Gamma (u)ds=a\Gamma \left({\frac {1}{a}}\right)\varphi +\sin(\varphi +\varepsilon )\left[\alpha ea\Gamma \left({\frac {1}{a}}\right)-\alpha e\Gamma '\left({\frac {1}{a}}\right)\right]\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/de988b4291f6230b7e4d79bcef3c789e408f1849)
donc
![{\displaystyle {\begin{aligned}\mathrm {A} =&{\frac {2(\mathrm {S+P} )}{h^{2}}}a\Gamma \left({\frac {1}{a}}\right),\\\mathrm {B} =&{\frac {2(\mathrm {S+P} )}{h^{2}}}\alpha e\left[a\Gamma \left({\frac {1}{a}}\right)-\Gamma '\left({\frac {1}{a}}\right)\right]\,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5483134f0784d3a411d927b05a9ee91b0723c941)
mais on a (art. IX)
![{\displaystyle {\frac {\mathrm {S+P} }{h^{2}}}={\frac {1}{a}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/563e493ac5d349073948db8f8d8a9933f382cb69)
partant,
![{\displaystyle \mathrm {A} =2\Gamma \left({\frac {1}{a}}\right),\qquad {\frac {2\alpha e}{a}}\left[a\Gamma \left({\frac {1}{a}}\right)-\Gamma '\left({\frac {1}{a}}\right)\right]\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5ecce4e2273ffd7edc5c184b15a77472fc3bacf2)
on aura donc les deux équations
![{\displaystyle {\begin{aligned}d{\frac {\cfrac {1}{a}}{dx}}=&{\frac {2\Gamma \left({\cfrac {1}{a}}\right)}{a^{\frac {3}{2}}}},\\d{\frac {\cfrac {\alpha e}{a}}{dx}}=&{\frac {\alpha e}{a^{\frac {5}{2}}}}\left[a\Gamma \left({\frac {1}{a}}\right)-\Gamma '\left({\frac {1}{a}}\right)\right]\,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b13c7f637dc7a409b4c52fb1d1cb3423438e2b75)
partant,
![{\displaystyle {\begin{aligned}{\frac {da}{dx{\sqrt {a}}}}=&-2\Gamma \left({\frac {1}{a}}\right),\\{\frac {de}{dx}}=&-{\frac {e}{a^{\frac {3}{2}}}}\left[a\Gamma \left({\frac {1}{a}}\right)-\Gamma '\left({\frac {1}{a}}\right)\right].\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fcef424681ffe7af44665d89274e638ecc3553e3)
Puisque nous supposons
très petit, il faut que la valeur de
soit telle, qu’elle reste toujours fort petite, sans quoi la solution précé-