Page:Annales de mathématiques pures et appliquées, 1824-1825, Tome 15.djvu/185

${\frac {1}{2\varpi {\sqrt {-1}}}}\int _{0}^{\varpi }\ldots$

${\frac {\operatorname {F} \left(\alpha +pe^{x{\sqrt {-1}}}\right).\left(\operatorname {Cos} .rx+{\sqrt {-1}}\operatorname {Sin} .rx\right)-\operatorname {F} \left(\alpha +pe^{-x{\sqrt {-1}}}\right).\left(\operatorname {Cos} .rx-{\sqrt {-1}}\operatorname {Sin} .rx\right)}{\operatorname {Sin} .x}}\operatorname {d} x\,;$

ou encore

${\frac {1}{2\varpi }}\int _{0}^{\varpi }{\frac {e^{rx{\sqrt {-1}}}.\operatorname {F} \left(\alpha +pe^{x{\sqrt {-1}}}\right)-e^{-rx{\sqrt {-1}}}.\operatorname {F} \left(\alpha +pe^{-x{\sqrt {-1}}}\right)}{{\sqrt {-1}}.\operatorname {Sin} .x}}\operatorname {d} x\,;$

de sorte qu’on pourra écrire, si $r$ est un nombre pair,

${\frac {1}{\varpi }}\int _{0}^{\varpi }{\frac {e^{rx{\sqrt {-1}}}.\operatorname {F} \left(\alpha +pe^{x{\sqrt {-1}}}\right)-e^{-rx{\sqrt {-1}}}.\operatorname {F} \left(\alpha +pe^{-x{\sqrt {-1}}}\right)}{{\sqrt {-1}}.\operatorname {Sin} .x}}\operatorname {d} x$

$=\operatorname {F} (\alpha +p)-\operatorname {F} (\alpha -p)\,;$(E)

et, si $r$ est un nombre impair,

${\frac {1}{\varpi }}\int _{0}^{\varpi }{\frac {e^{rx{\sqrt {-1}}}.\operatorname {F} \left(\alpha +pe^{x{\sqrt {-1}}}\right)-e^{-rx{\sqrt {-1}}}.\operatorname {F} \left(\alpha +pe^{-x{\sqrt {-1}}}\right)}{{\sqrt {-1}}.\operatorname {Sin} .x}}\operatorname {d} x$

$=\operatorname {F} (\alpha +p)+\operatorname {F} (\alpha -p)$(F)

On remarquera que les seconds membres de ces équations sont indépendans de $r.$

Si dans la formule (B), on fait tour-à-tour $r=0$ et $r=1,$ elle donnera

${\frac {1}{\varpi }}\int _{0}^{\varpi }{\frac {\operatorname {F} \left(\alpha +pe^{x{\sqrt {-1}}}\right)-\operatorname {F} \left(\alpha +pe^{-x{\sqrt {-1}}}\right)}{{\sqrt {-1}}.\operatorname {Sin} .x}}\operatorname {d} x$

$=\operatorname {F} (\alpha +p)-\operatorname {F} (\alpha -p)$(G)

${\frac {1}{\varpi }}\int _{0}^{\varpi }{\frac {\left\{\operatorname {F} \left(\alpha +pe^{x{\sqrt {-1}}}\right)-\operatorname {F} \left(\alpha +pe^{-x{\sqrt {-1}}}\right)\right\}\operatorname {Cos} .x}{{\sqrt {-1}}.\operatorname {Sin} .x}}\operatorname {d} x$

$=\operatorname {F} (\alpha +p)+\operatorname {F} (\alpha -p)-2\operatorname {F} (\alpha )$(H)

mais on ne pourrait obtenir des résultats analogues de la formule (A) qu’en y supposant $r$ infini.