Page:Mémoires de l’Académie des sciences, Tome 22.djvu/274

Par suite, si l’on appelle $\eta$ un nombre qui diffère infiniment peu de l’unité, on trouvera pour la valeur générale de $f(x)$

(76)

${\begin{aligned}f(x)&={\frac {1}{2\pi }}\int _{-\infty }^{\infty }\int _{0}^{a}{\frac {e^{\lambda (x-\mu )\mathrm {i} }}{1-\eta \mathrm {\cfrac {(A+\alpha )(B-\alpha )}{(A-\alpha )(B+\alpha )}} e^{-2a\lambda \mathrm {i} }}}{\mathfrak {f}}(\mu )d\mu \,d\lambda \\&-{\frac {1}{2\pi }}\int _{-\infty }^{\infty }\int _{0}^{a}{\frac {e^{\lambda (x-2a+\mu )\mathrm {i} }}{1-\eta \mathrm {\cfrac {(A+\alpha )(B-\alpha )}{(A-\alpha )(B+\alpha )}} e^{-2a\lambda \mathrm {i} }}}\mathrm {\frac {B-\alpha }{B+\alpha }} {\mathfrak {f}}(\mu )d\mu \,d\lambda \\&+{\frac {1}{2\pi }}\int _{-\infty }^{\infty }\int _{0}^{a}{\frac {e^{\lambda (x-\mu +2a)\mathrm {i} }}{1-\eta \mathrm {\cfrac {(A-\alpha )(B+\alpha )}{(A+\alpha )(B-\alpha )}} e^{2a\lambda \mathrm {i} }}}\mathrm {\frac {(A-\alpha )(B+\alpha )}{(A+\alpha )(B-\alpha )}} {\mathfrak {f}}(\mu )d\mu \,d\lambda \\&-{\frac {1}{2\pi }}\int _{-\infty }^{\infty }\int _{0}^{a}{\frac {e^{\lambda (x+\mu )\mathrm {i} }}{1-\eta \mathrm {\cfrac {(A-\alpha )(B+\alpha )}{(A+\alpha )(B-\alpha )}} e^{2a\lambda \mathrm {i} }}}\mathrm {\frac {A-\alpha }{A+\alpha }} {\mathfrak {f}}(\mu )d\mu \,d\lambda ,\end{aligned}}$

ou, ce qui revient au même,

(77)

$f(x)=$

${\frac {\mathrm {A} -\alpha }{2\pi }}\int _{-\infty }^{\infty }\int _{0}^{a}\left[(\mathrm {B} +\alpha )e^{\lambda (a-\mu )\mathrm {i} }-(\mathrm {B} -\alpha )e^{\lambda (\mu -a)\mathrm {i} }\right]{\mathfrak {F}}(\alpha )e^{\lambda x\mathrm {i} }{\mathfrak {f}}(\mu )d\mu \,d\lambda ,$

la valeur de ${\mathfrak {F}}(\alpha )$ étant donnée par l’équation

${\begin{aligned}{\mathfrak {F}}(\alpha )&={\frac {1}{(\mathrm {A} -\alpha )(\mathrm {B} +\alpha )e^{a\lambda \mathrm {i} }-\eta (\mathrm {A} +\alpha )(\mathrm {B} -\alpha )e^{-a\lambda \mathrm {i} }}}\\&+{\frac {1}{(\mathrm {A} +\alpha )(\mathrm {B} -\alpha )e^{-a\lambda \mathrm {i} }-\eta (\mathrm {A} -\alpha )(\mathrm {B} +\alpha )e^{a\lambda \mathrm {i} }}}\\&=4(1-\eta ){\frac {(\mathrm {A} +\alpha )(\mathrm {B} -\alpha )e^{-a\lambda \mathrm {i} }+(\mathrm {A} -\alpha )(\mathrm {B} +\alpha )e^{a\lambda \mathrm {i} }}{(1-\eta )^{2}\left[(\mathrm {A} +\alpha )(\mathrm {B} -\alpha )e^{-a\lambda \mathrm {i} }+(\mathrm {A} -\alpha )(\mathrm {B} +\alpha )e^{a\lambda \mathrm {i} }\right]^{2}-{\text{etc.}}}}.\end{aligned}}$

c’est-à-dire, par l’équation

(78)

${\mathfrak {F}}(\alpha )=$

$4(1-\eta ){\frac {(\mathrm {A} +\alpha )(\mathrm {B} -\alpha )e^{-a\lambda \mathrm {i} }+(\mathrm {A} -\alpha )(\mathrm {B} +\alpha )e^{a\lambda \mathrm {i} }}{\left({\begin{aligned}&(1-\eta )^{2}\left[(\mathrm {A} +\alpha )(\mathrm {B} -\alpha )e^{-a\lambda \mathrm {i} }+(\mathrm {A} -\alpha )(\mathrm {B} +\alpha )e^{a\lambda \mathrm {i} }\right]^{2}\\-&(1+\eta )^{2}\left[(\mathrm {A} +\alpha )(\mathrm {B} -\alpha )e^{-a\lambda \mathrm {i} }-(\mathrm {A} -\alpha )(\mathrm {B} +\alpha )e^{a\lambda \mathrm {i} }\right]^{2}\end{aligned}}\right)}}.$

Si, dans l’équation (77), on fait passer le facteur $\mathrm {A} -\alpha$ sous le signe $\iint ,$ on pourra remplacer ensuite $\alpha ,$ qui indique