Page:Annales de mathématiques pures et appliquées, 1830-1831, Tome 21.djvu/86

 Les corrections sont expliquées en page de discussion

$t={\sqrt {x}},\qquad 2t\operatorname {d} t=\operatorname {d} x,\qquad \operatorname {d} t={\frac {\operatorname {d} x}{2t}}={\frac {\operatorname {d} x}{2{\sqrt {x}}}},$

et par suite

$e^{-t^{2}}\operatorname {d} t=e^{-x}{\frac {\operatorname {d} x}{2{\sqrt {x}}}}={\frac {1}{2}}x^{-{\frac {1}{2}}}.e^{-x}\operatorname {d} x\,;$

posant alors

$P=x^{-{\frac {1}{2}}},\qquad Q=e^{-x},$

on trouvera

${\begin{alignedat}{1}{\frac {\operatorname {d} P}{\operatorname {d} x}}=&-{\frac {1}{2}}x^{-{\frac {3}{2}}},\\\\{\frac {\operatorname {d} ^{2}P}{\operatorname {d} x^{2}}}=&+{\frac {1.3}{2.2}}x^{-{\frac {5}{2}}},\\\\{\frac {\operatorname {d} ^{3}P}{\operatorname {d} x^{3}}}=&-{\frac {1.3.5}{2.2.2}}x^{-{\frac {7}{2}}},\\\ldots \ldots \ldots \ldots \ldots \ldots \\{\frac {\operatorname {d} ^{\mu }P}{\operatorname {d} x^{\mu }}}=&(-1)^{\mu }{\frac {1^{\mu |2a}}{2^{\mu }}}x^{-{\frac {2\mu +1}{2}}}\,;\end{alignedat}}$

on aura d’ailleurs

$\int ^{\mu }Q\operatorname {d} x^{\mu }=(-1)^{\mu }e^{-x}\,;$

substituant donc dans la formule (2), on aura

${\frac {1}{2}}\int ^{\mu }{\frac {\operatorname {d} x^{m}}{e^{x}{\sqrt {x}}}}={\frac {(-1)^{m}}{e^{x}}}\left\{x^{-{\frac {1}{2}}}-{\frac {m}{1}}.{\frac {1}{2}}x^{-{\frac {3}{2}}}+{\frac {m}{1}}{\frac {m+1}{2}}.{\frac {1.3}{2.2}}x^{-{\frac {5}{2}}}-\ldots \right\},$

d’où, en posant $m=1$ et en remettant pour $x$ sa valeur $t^{2}$,