Page:Annales de mathématiques pures et appliquées, 1824-1825, Tome 15.djvu/375

$\int _{u'}^{u''}\left\{{\frac {e^{v''{\sqrt {-1}}}}{e^{v''}}}\operatorname {f} \left({\frac {ue^{v''{\sqrt {-1}}}}{e^{v''}}}\right)-{\frac {e^{v'{\sqrt {-1}}}}{e^{v'}}}\operatorname {f} \left({\frac {ue^{v'{\sqrt {-1}}}}{e^{v'}}}\right)\right\}\operatorname {d} u$

$=\left(1+{\sqrt {-1}}\right)\int _{v'}^{v''}\left\{{\frac {u''e^{v{\sqrt {-1}}}}{e^{v}}}\operatorname {f} \left({\frac {u''e^{v{\sqrt {-1}}}}{e^{v}}}\right)-{\frac {u'e^{v{\sqrt {-1}}}}{e^{v}}}\operatorname {f} \left({\frac {u'e^{v{\sqrt {-1}}}}{e^{v}}}\right)\right\}\operatorname {d} v.$

Si, dans cette dernière formule, on fait $v'=0,sv''=\infty ,u'=0,u''=1,$ on aura

$\int _{0}^{1}-\operatorname {f} (u)\operatorname {d} u=\left(-1+{\sqrt {-1}}\right)\int _{0}^{\infty }\left\{{\frac {e^{v{\sqrt {-1}}}}{e^{v}}}\operatorname {f} \left({\frac {e^{v{\sqrt {-1}}}}{e^{v}}}\right)\right\}\operatorname {d} v\,;$

d’où

$\int _{0}^{\infty }{\frac {e^{v{\sqrt {-1}}}}{e^{v}}}\operatorname {f} \left({\frac {e^{v{\sqrt {-1}}}}{e^{v}}}\right)\operatorname {d} v={\frac {1}{2}}\int _{0}^{1}\operatorname {f} (u)\operatorname {d} u+{\frac {\sqrt {-1}}{2}}\int _{0}^{1}\operatorname {f} (u)\operatorname {d} u.$

C’est aux notations près la formule (Q) de M. Vernier, (pag. 185).