Page:Annales de mathématiques pures et appliquées, 1830-1831, Tome 21.djvu/78

$\int ^{m}PQ\operatorname {d} x^{m}=P\int ^{m}Q\operatorname {d} x^{m}-{\frac {m}{1}}{\frac {\operatorname {d} P}{\operatorname {d} x}}\int ^{m+1}Q\operatorname {d} x^{m+1}$

$+{\frac {m}{1}}.{\frac {m+1}{2}}{\frac {\operatorname {d} ^{2}P}{\operatorname {d} x^{2}}}\int ^{m+2}Q\operatorname {d} x^{m+2}+\ldots$(2)

Si l’on pose

$P=(x+a)^{r},\qquad Q=(x+b)^{r}\,;$

on aura

${\frac {\operatorname {d} ^{\mu }(x+b)^{r}}{\operatorname {d} x^{\mu }}}=r^{\mu |-1}.(x+b)^{r-\mu }\,;$

si, dans cette dernière formule, on change le signe de $\mu$., en se rappelant que

$r^{-\mu |-1}={\frac {1}{(r+\mu )^{\mu |-1}}},$

il viendra

$\int ^{\mu }(x+b)^{r}\operatorname {d} x^{\mu }={\frac {(x+b)^{r+\mu }}{(r+\mu )^{\mu |-1}}}\,;$

on a d’ailleurs

${\frac {\operatorname {d} P}{\operatorname {d} x}}=r(x+a)^{r-1},$

${\frac {\operatorname {d} ^{2}P}{\operatorname {d} x^{2}}}=r(r-1)(x+a)^{r-2}=r^{2|-1}(x+a)^{r-2},$

${\frac {\operatorname {d} ^{3}P}{\operatorname {d} x^{3}}}=r(r-1)(r-2)(x+a)^{r-3}=r^{3|-1}(x+a)^{r-3},$

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

Il viendra donc, en remplaçant successivement $\mu$ par $m,m+1,m+2,\ldots$ et substituant dans (2),