Page:Mémoires de l’Académie des sciences, Tome 22.djvu/273

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$\left\{{\begin{array}{ll}&f(x)=\left[\mathrm {\cfrac {(A+\alpha )(B-\alpha )}{(A-\alpha )(B+\alpha )}} \right]^{n}{\mathfrak {f}}(x-2na){\text{ entre les limites}}\\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \left\{{\begin{aligned}x=&2na\\x=&(2n+1)a\quad \end{aligned}}\right\},\\&f(x)=-\mathrm {{\cfrac {B-\alpha }{B+\alpha }}\left[{\cfrac {(A+\alpha )(B-\alpha )}{(A-\alpha )(B+\alpha )}}\right]} ^{n}{\mathfrak {f}}(2na+2a-x)\ldots \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \left\{{\begin{aligned}x=&(2n+1)a\\x=&(2n+2)a\quad \end{aligned}}\right\},\\&f(x)=\mathrm {\left[{\cfrac {(A-\alpha )(B+\alpha )}{(A+\alpha )(B-\alpha )}}\right]} ^{n}{\mathfrak {f}}(x+2na)\ldots \ldots \ldots \ldots \ldots .\\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \left\{{\begin{aligned}x=&-(2n-1)a\\x=&-2na\end{aligned}}\right\},\\&f(x)=-\mathrm {{\cfrac {A-\alpha }{A+\alpha }}\left[{\cfrac {(A-\alpha )(B+\alpha )}{(A+\alpha )(B-\alpha )}}\right]} ^{n}{\mathfrak {f}}(-x-2na)\ldots \ldots \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \left\{{\begin{aligned}x=&-2na\\x=&-(2n+1)a\end{aligned}}\right\}.\end{array}}\right.$

On aura d’ailleurs, en vertu du théorème de M. Fourier,

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$\left\{{\begin{aligned}&{\begin{aligned}&{\frac {1}{2\pi }}\int _{-\infty }^{\infty }\int _{k}^{k+a}e^{\lambda (x-\mu )\mathrm {i} }{\mathfrak {f}}(\mu -k)d\mu \,d\lambda \\=&{\frac {1}{2\pi }}\int _{-\infty }^{\infty }\int _{0}^{a}e^{\lambda (x-k-\mu )\mathrm {i} }{\mathfrak {f}}(\mu )d\mu \,d\lambda ={\mathfrak {f}}(x-k)\end{aligned}}\left\{{\begin{aligned}x=&k\\x=&k+a\end{aligned}}\right\}\\&{\text{et}}\\&{\begin{aligned}&{\frac {1}{2\pi }}\int _{-\infty }^{\infty }\int _{k-a}^{k}e^{\lambda (x-\mu )\mathrm {i} }{\mathfrak {f}}(k-\mu )d\mu \,d\lambda \\=&{\frac {1}{2\pi }}\int _{-\infty }^{\infty }\int _{0}^{a}e^{\lambda (x-k+\mu )\mathrm {i} }{\mathfrak {f}}(\mu )d\mu \,d\lambda ={\mathfrak {f}}(k-x)\end{aligned}}\left\{{\begin{aligned}x=&k-a\\x=&k\end{aligned}}\right\},\end{aligned}}\right.$

pour toutes les valeurs de $x,$ comprises entre les limites. indiquées, tandis que les mêmes intégrales doubles s’évanouiront hors de ces limites. On trouvera en conséquence

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$\left\{{\begin{aligned}f(x)=&{\frac {1}{2\pi }}\int _{-\infty }^{\infty }\int _{0}^{a}\left[\mathrm {\frac {(A+\alpha )(B-\alpha )}{(A-\alpha )(B+\alpha )}} e^{-2a\lambda \mathrm {i} }\right]^{n}e^{\lambda (x-\mu )\mathrm {i} }{\mathfrak {f}}(\mu )d\mu \,d\lambda \\&\qquad \qquad \qquad \qquad \qquad \qquad {\text{entre les limites}}\left\{{\begin{aligned}x=&2na\\x=&(2n+1)a\quad \end{aligned}}\right\},\\f(x)=&-{\frac {1}{2\pi }}\int _{-\infty }^{\infty }\int _{0}^{a}\mathrm {\frac {B-\alpha }{B+\alpha }} \left[\mathrm {\frac {(A+\alpha )(B-\alpha )}{(A-\alpha )(B+\alpha )}} e^{-2a\lambda \mathrm {i} }\right]^{n}e^{\lambda (x-2a+\mu )\mathrm {i} }{\mathfrak {f}}(\mu )d\mu \,d\lambda \\&\qquad \qquad \qquad \qquad \qquad \qquad {\text{entre les limites}}\left\{{\begin{aligned}x=&(2n+1)a\\x=&(2n+2)a\quad \end{aligned}}\right\},\\f(x)=&{\frac {1}{2\pi }}\int _{-\infty }^{\infty }\int _{0}^{a}\left[\mathrm {\frac {(A-\alpha )(B+\alpha )}{(A+\alpha )(B-\alpha )}} e^{2a\lambda \mathrm {i} }\right]^{n}e^{\lambda (x-\mu )\mathrm {i} }{\mathfrak {f}}(\mu )d\mu \,d\lambda \\&\qquad \qquad \qquad \qquad \qquad \qquad {\text{entre les limites}}\left\{{\begin{aligned}x=&-(2n-1)a\\x=&-2na\end{aligned}}\right\},\\f(x)=&-{\frac {1}{2\pi }}\int _{-\infty }^{\infty }\int _{0}^{a}\mathrm {\frac {A-\alpha }{A+\alpha }} \left[\mathrm {\frac {(A-\alpha )(B+\alpha )}{(A+\alpha )(B-\alpha )}} e^{2a\lambda \mathrm {i} }\right]^{n}e^{\lambda (x+\mu )\mathrm {i} }{\mathfrak {f}}(\mu )d\mu \,d\lambda \\&\qquad \qquad \qquad \qquad \qquad \qquad {\text{entre les limites}}\left\{{\begin{aligned}x=&-2na\\x=&-(2n+1)a\end{aligned}}\right\}.\end{aligned}}\right.$