Je reprends pour cela l’équation
et j’observe qu’elle se réduit à celle-ci, en faisant
![{\displaystyle {\begin{aligned}0=\int _{0}^{\pi }\int _{0}^{\pi }{\frac {1}{\sin \varphi }}&\left[2^{2}\sin ^{5}p\sin q\cos q(\cos q\sin \varphi -\sin q\cos \varphi )^{2}{\frac {d^{2}y}{1.2.dx^{2}}}\right.\\&+2^{3}\sin ^{7}p\sin ^{2}q\cos q(\cos q\sin \varphi -\sin q\cos \varphi )^{3}{\frac {d^{3}y}{1.2.3.dx^{3}}}\\&+\left.\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots {\begin{aligned}\\\\\end{aligned}}\right]dpdq.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e78fdb415c2c4aca66f47444116bd30dd1b53265)
Si l’on suppose, dans cette équation,
![{\displaystyle y=gx^{\mu }\qquad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/c6b29a5e27f8d76743b243126ace13b5f957aaa1)
et
![{\displaystyle \qquad \sin ^{2}\varphi =-x^{2},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8bb2aea5674817e057c88152cee0d7951f416fb8)
il est visible qu’elle donnera l’équation
or, dans cette supposition, on a
![{\displaystyle inn\varphi =x{\sqrt {-1}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/78cba4217cfe75b2c68a109c301395e60598ebd4)
partant [1]
![{\displaystyle {\begin{aligned}(\cos q\sin \varphi -\sin q\cos \varphi )^{i}=&x^{i}\left({\sqrt {-1}}\right)^{i}\left(\cos q+{\sqrt {-1}}\sin q\right)^{i}\\=&x^{i}\left({\sqrt {-1}}\right)^{i}\left(\cos iq+{\sqrt {-1}}\sin iq\right).\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cafc48cc260bb93cb4a79f858f3e8860570db44a)
Le terme
![{\displaystyle \int _{0}^{\pi }\int _{0}^{\pi }{\frac {1}{\sin \varphi }}{\frac {d^{2n}y}{1.2.3\ldots 2ndx^{2n}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6048aafccb00f7913bc4132967ce0dfae82997f4)
![{\displaystyle \times 2^{2n}\sin ^{4n+1}p\sin ^{2n-1}q\cos q(\cos q\sin \varphi -\sin q\cos \varphi )^{2n}dpdq}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bc59a4fd4d261b5bdee289762d406a57dc0d371c)
deviendra donc
![{\displaystyle {\frac {\mu (\mu -1)(\mu -2)\ldots (\mu -2n+1)}{1.2.3\ldots 2n}}\int _{0}^{\pi }\int _{0}^{\pi }{\frac {1}{\sqrt {-1}}}x^{2n-1}(-1)^{n}\sin ^{4n+1}p.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bd0fe4f542ba3e2dbc2d2d1486480034da695cc2)
![{\displaystyle \times 2^{2n}\sin ^{2n-1}q\cos q\left(\cos 2nq+{\sqrt {-1}}\sin 2nq\right)dpdq\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d8fd867f61ba99a27f91d36dc9ed6d1045f0e31c)
or on a
![{\displaystyle 2^{2n-2}\sin ^{2n-1}q=\pm \left[\sin(2n-1)q-(2n-1)\sin(2n-3)q+\ldots \right],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1cc81bec4c28300ce7ab1e6c49e05fe9b9038239)
- ↑ On devrait avoir
![{\displaystyle (\cos q\sin \varphi -\sin q\cos \varphi )^{i}=x^{i}\left({\sqrt {-1}}\right)^{i}\left(\cos q+{\sqrt {-1}}{\sqrt {1+{\frac {1}{x^{2}}}}}\sin q\right)^{i}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8ba72a77997687109290edbb252460484e79c6d3)
mais, comme on ne cherche que le terme du degré le plus élevé en
on peut prendre, avec Laplace,
![{\displaystyle x^{i}\left({\sqrt {-1}}\right)^{i}\left(\cos q+{\sqrt {-1}}\sin q\right)^{i}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cd4fc726c296ef7b59833482d203f0e6233c1669)
(Note de l’Éditeur.)