Page:Mémoires de l’Académie des sciences, Tome 6.djvu/795

Cela posé, revenons à l’équation (2). Cette équation, pouvant s’écrire comme il suit :

(7)

${\begin{aligned}f(x)={\frac {1}{a}}\int _{0}^{a}f(\mu )d\mu &+{\frac {1}{a}}\int _{0}^{a}e^{{\frac {2\pi }{a}}(x-\mu ){\sqrt {-1}}}f(\mu )d\mu +{\text{etc}}.\\&+{\frac {1}{a}}\int _{0}^{a}e^{{\frac {2\pi }{a}}(\mu -x){\sqrt {-1}}}f(\mu )d\mu +{\text{etc}}.,\end{aligned}}$

on en déduira, à l’aide des équations (5) et (6),

(8)

$f(x)={\frac {1}{a}}\int _{0}^{a}f(\mu )d\mu$

${\begin{aligned}&+{\frac {1}{a{\sqrt {-1}}}}\int _{0}^{\infty }\left\{e^{-{\frac {2\pi }{a}}(x-a){\sqrt {-1}}}f\left(a+\nu {\sqrt {-1}}\right)-e^{-{\frac {2\pi }{a}}x{\sqrt {-1}}}f\left(\nu {\sqrt {-1}}\right)\right\}e^{-{\frac {2\pi }{a}}\nu }d\nu +{\text{etc}}.\\&-{\frac {1}{a{\sqrt {-1}}}}\int _{0}^{\infty }\left\{e^{{\frac {2\pi }{a}}(x-a){\sqrt {-1}}}f\left(a-\nu {\sqrt {-1}}\right)-e^{{\frac {2\pi }{a}}x{\sqrt {-1}}}f\left(-\nu {\sqrt {-1}}\right)\right\}e^{-{\frac {2\pi }{a}}\nu }d\nu +{\text{etc}}.\\&={\frac {1}{a}}\int _{0}^{a}f(\mu )d\mu \\&+{\frac {1}{a{\sqrt {-1}}}}\int _{0}^{\infty }\left\{e^{-{\frac {2\pi x}{a}}{\sqrt {-1}}}e^{-{\frac {2\pi }{a}}\nu }+{\text{etc}}.\right\}\left[f\left(a+\nu {\sqrt {-1}}\right)-f\left(\nu {\sqrt {-1}}\right)\right]d\nu \\&-{\frac {1}{a{\sqrt {-1}}}}\int _{0}^{\infty }\left\{e^{{\frac {2\pi x}{a}}{\sqrt {-1}}}e^{-{\frac {2\pi }{a}}\nu }+{\text{etc}}.\right\}\left[f\left(a-\nu {\sqrt {-1}}\right)-f\left(-\nu {\sqrt {-1}}\right)\right]d\nu ,\end{aligned}}$

et par suite

(9)

$f(x)={\frac {1}{a}}\int _{0}^{a}f(\mu )d\mu$

$+{\frac {1}{a{\sqrt {-1}}}}\int _{0}^{\infty }\left\{{\frac {f\left(a+\nu {\sqrt {-1}}\right)-f\left(\nu {\sqrt {-1}}\right)}{e^{-{\frac {2\pi x}{a}}{\sqrt {-1}}}e^{{\frac {2\pi }{a}}\nu }-1}}-{\frac {f\left(a-\nu {\sqrt {-1}}\right)-f\left(-\nu {\sqrt {-1}}\right)}{e^{-{\frac {2\pi x}{a}}{\sqrt {-1}}}e^{{\frac {2\pi }{a}}\nu }-1}}\right\}d\nu .$