5. Recouvrement : mise en place
\[ S(R)=\int\phi_A\phi_B\,d^3r. \]En utilisant les expressions précédentes,
\[ S(R) = N^2\frac{R^5}{32} \int_1^\infty e^{-R\mu/2} \int_{-1}^{1} \int_0^{2\pi} (\mu^2-1)(1-\nu^2) (\mu^2-\nu^2) \cos^2\varphi \,d\varphi\,d\nu\,d\mu. \]Or
\[ \int_0^{2\pi}\cos^2\varphi\,d\varphi=\pi, \]et \(N^2=1/(32\pi)\). Ainsi,
\[ S(R) = \frac{R^5}{1024} \int_1^\infty e^{-R\mu/2} \left[ \int_{-1}^{1} (\mu^2-1)(1-\nu^2)(\mu^2-\nu^2) \,d\nu \right]d\mu. \]Intégration sur \(\nu\)
On développe
\[ (1-\nu^2)(\mu^2-\nu^2) = \mu^2-(1+\mu^2)\nu^2+\nu^4. \]En utilisant
\[ \int_{-1}^{1}d\nu=2, \qquad \int_{-1}^{1}\nu^2d\nu=\frac23, \qquad \int_{-1}^{1}\nu^4d\nu=\frac25, \]on trouve
\[ \int_{-1}^{1} (\mu^2-1)(1-\nu^2)(\mu^2-\nu^2)d\nu = \frac4{15} (\mu^2-1)(5\mu^2-1). \]Donc
\[ S(R) = \frac{R^5}{3840} \int_1^\infty \left( 5\mu^4-6\mu^2+1 \right) e^{-R\mu/2}d\mu. \]