![{\displaystyle \varphi ={\frac {\sqrt {g}}{\pi ^{2}}}\iiiint f(\alpha ,\beta )e^{-zu}.cos.(u\rho .cos.\omega ).sin.\left(t{\sqrt {gu}}\right).{\sqrt {u}}du\,d\omega \,d\alpha \,d\beta .\quad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/9d60b4dafa069591c9125b5b3f7846909a4d50cb)
(a)
(32) On peut effectivement vérifier que cette expression satisfait encore à l’équation (1) ; car si l’on fait
![{\displaystyle \int e^{-zu}cos.(\rho .cos.\omega )d\omega =\mathrm {R} ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/313a5ee761c37e42f1200aea23ae3656e918cdd7)
il est évident que cette vérification reviendra à faire voir que
![{\displaystyle {\frac {d^{2}\mathrm {R} }{dz^{2}}}+{\frac {d^{2}\mathrm {R} }{dy^{2}}}+{\frac {d^{2}\mathrm {R} }{dx^{2}}}=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b7f287eeacd2484c24e273084875931144137403)
Or, on a
![{\displaystyle {\begin{aligned}{\frac {d^{2}\mathrm {R} }{dz^{2}}}=&u^{2}\int e^{-zu}cos.(u\rho .cos.\omega )d\omega ,\\{\frac {d^{2}\mathrm {R} }{dy^{2}}}=&-u^{2}{\frac {d\rho ^{2}}{dy^{2}}}.\int e^{-zu}cos.(u\rho .cos.\omega )cos.^{2}\omega \,d\omega \\&-u\ \ {\frac {d^{2}\rho }{dy^{2}}}.\int e^{-zu}sin.(u\rho .cos.\omega )cos.\omega \,d\omega ,\\{\frac {d^{2}\mathrm {R} }{dx^{2}}}=&-u^{2}{\frac {d\rho ^{2}}{dx^{2}}}.\int e^{-zu}cos.(u\rho .cos.\omega )cos.^{2}\omega \,d\omega \\&-u\ \ {\frac {d^{2}\rho }{dx^{2}}}.\int e^{-zu}sin.(u\rho .cos.\omega )cos.\omega \,d\omega \,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7c95f422f7695a61cb3200aeb4fe942d2007a1a9)
de plus la valeur de
est
![{\displaystyle \rho ={\sqrt {(x-\alpha )^{2}+(y-\beta )^{2}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1cf68e70765a8f7807f9728320855ee58c037e7f)
d’où l’on tire
![{\displaystyle {\frac {d\rho ^{2}}{dy^{2}}}+{\frac {d\rho ^{2}}{dx^{2}}}=1,\qquad {\frac {d^{2}\rho }{dy^{2}}}+{\frac {d^{2}\rho }{dx^{2}}}={\frac {1}{\rho }}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d25bac171440c47fa069b5258605646b8906e492)
on aura donc, en faisant la somme des trois équations précédentes, et réduisant