Page:Mémoires de l’Académie des sciences, Tome 6.djvu/797

entre les limites $0$ et $a,$

(13)

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

Or, il suffit d’ajouter ces dernières équations pour retrouver la formule (10).

Si l’on remplace $x$ par $a,$ dans les intégrales relatives à $\mu$ que renferment les équations (13), on tirera des formules (3) et (4),

(14)

$\left\{{\begin{aligned}\int _{0}^{a}{\frac {f(\mu )d\mu }{e^{-{\frac {2\pi }{a}}\mu {\sqrt {-1}}}-1}}&=\\{\frac {1}{\sqrt {-1}}}&\int _{0}^{\infty }{\frac {f\left(a+\nu {\sqrt {-1}}\right)-f\left(\nu {\sqrt {-1}}\right)}{e^{{\frac {2\pi }{a}}\nu }-1}}d\nu -{\frac {a}{4}}\left[f(a)+f(0)\right]\\\int _{0}^{a}{\frac {f(\mu )d\mu }{e^{{\frac {2\pi }{a}}\mu {\sqrt {-1}}}-1}}&=\\-{\frac {1}{\sqrt {-1}}}&\int _{0}^{\infty }{\frac {f\left(a-\nu {\sqrt {-1}}\right)-f\left(-\nu {\sqrt {-1}}\right)}{e^{{\frac {2\pi }{a}}\nu }-1}}d\nu -{\frac {a}{4}}\left[f(a)+f(0)\right],\end{aligned}}\right.$

puis, en ajoutant,

(15)

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

On aura donc

(16)

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