Page:Annales de mathématiques pures et appliquées, 1821-1822, Tome 12.djvu/155

149

DÉFINIES.

$-{\frac {\alpha }{\operatorname {Log} .(1-\alpha )}}=\int _{0}^{1}\operatorname {d} n-\alpha \int _{0}^{1}{\frac {n}{1}}\operatorname {d} n+\alpha ^{2}\int _{0}^{1}{\frac {n}{1}}.{\frac {n-1}{2}}\operatorname {d} n$

$-\alpha ^{3}\int _{0}^{1}{\frac {n}{1}}.{\frac {n-1}{2}}.{\frac {n-2}{3}}\operatorname {d} n+\ldots \,;$

ainsi, en posant, pour plus de simplicité,

$-{\frac {\alpha }{\operatorname {Log} .(1-\alpha )}}=1+B_{1}\alpha +B_{2}\alpha ^{2}+B_{3}\alpha ^{3}+B_{4}\alpha ^{4}+\ldots ,$

on aura

${\begin{aligned}&B_{1}=-\int _{0}^{1}{\frac {n}{1}}\operatorname {d} n=-{\frac {1}{2}}\\\\&B_{2}=+\int _{0}^{1}{\frac {n}{1}}.{\frac {n-1}{2}}\operatorname {d} n=-{\frac {1}{12}}\\\\&B_{3}=-\int _{0}^{1}{\frac {n}{1}}.{\frac {n-1}{2}}.{\frac {n-2}{3}}\operatorname {d} n=-{\frac {1}{24}}\\\\&B_{4}=\ \ \int _{0}^{1}{\frac {n}{1}}.{\frac {n-1}{2}}.{\frac {n-2}{3}}.{\frac {n-3}{4}}\operatorname {d} n=-{\frac {19}{720}}\\\\&B_{5}=\ \ \int _{0}^{1}{\frac {n}{1}}.{\frac {n-1}{2}}.{\frac {n-2}{3}}.{\frac {n-3}{4}}.{\frac {n-4}{5}}\operatorname {d} n=-{\frac {3}{160}}\\\\\end{aligned}}$

. . . . . . . . . . . . . . . . . . . . . . . . .

et il est clair que les mêmes coefficiens donnent

${\frac {\alpha }{\operatorname {Log} .(1+x)}}=1-B_{1}\alpha +B_{2}\alpha ^{2}-B_{3}\alpha ^{3}+B_{4}\alpha ^{4}-\ldots \qquad$(2)

Cela posé, si l’on écrit $\operatorname {Log} .\left[1-(1-x)\right]$ au lieu de $\operatorname {Log} .x,$ on obtiendra, à l’aide de la formule (1),