Page:Annales de mathématiques pures et appliquées, 1825-1826, Tome 16.djvu/147

$\delta V=(M+Np+Pq+Qr+\ldots )\delta x+N\omega +P{\frac {\operatorname {d} \omega }{\operatorname {d} x}}+Q{\frac {\operatorname {d} ^{2}\omega }{\operatorname {d} x^{2}}}+\ldots$

une fois trouvée, celle de l’intégrale $\int V\operatorname {d} x$ est facile à en déduire. On sait en effet que

$\delta \int V\operatorname {d} x=\int \delta (V\operatorname {d} x)=\int (\delta V.\operatorname {d} x+V\delta \operatorname {d} x)=\int (V\operatorname {d} \delta x+\delta V.\operatorname {d} x),$

substituant pour $\delta V$ la valeur que nous venons d’obtenir, on aura

$\delta \int V\operatorname {d} x=$

$\int \left\{V\operatorname {d} \delta x+\left[{\frac {\operatorname {d} V}{\operatorname {d} x}}\right]\operatorname {d} x\delta x+N\omega \operatorname {d} x+P{\frac {\operatorname {d} \omega }{\operatorname {d} x}}\operatorname {d} x+Q{\frac {\operatorname {d} ^{2}\omega }{\operatorname {d} ^{2}x}}+\ldots \right\}.$

Or, en premier lieu,

$\int \left(V\operatorname {d} \delta x+\left[{\frac {\operatorname {d} V}{\operatorname {d} x}}\right]\operatorname {d} x\delta x\right)=V\delta x\,;$

on trouve ensuite, en intégrant par parties,

${\begin{aligned}\int P{\frac {\operatorname {d} \omega }{\operatorname {d} x}}\operatorname {d} x&=P\omega -\int {\frac {\operatorname {d} P}{\operatorname {d} x}}\omega \operatorname {d} x,\\\\\int Q{\frac {\operatorname {d} ^{2}\omega }{\operatorname {d} x^{2}}}\operatorname {d} x&=Q{\frac {\operatorname {d} \omega }{\operatorname {d} x}}-{\frac {\operatorname {d} Q}{\operatorname {d} x}}\omega +\int {\frac {\operatorname {d} ^{2}Q}{\operatorname {d} x^{2}}}\omega \operatorname {d} x,\\\\\int R{\frac {\operatorname {d} ^{3}\omega }{\operatorname {d} x^{3}}}\operatorname {d} x&=R{\frac {\operatorname {d} ^{2}\omega }{\operatorname {d} x^{2}}}-{\frac {\operatorname {d} R}{\operatorname {d} x}}.{\frac {\operatorname {d} \omega }{\operatorname {d} x}}+{\frac {\operatorname {d} ^{2}R}{\operatorname {d} x^{2}}}\omega -\int {\frac {\operatorname {d} ^{3}R}{\operatorname {d} x^{3}}}\omega \operatorname {d} x,\end{aligned}}$



réunissant alors tous ces termes, on aura finalement

$\delta \int V\operatorname {d} x=V\delta x+\left(P-{\frac {\operatorname {d} Q}{\operatorname {d} x}}+{\frac {\operatorname {d} ^{2}R}{\operatorname {d} x^{2}}}-\ldots \right)\omega +\left(Q-{\frac {\operatorname {d} R}{\operatorname {d} x}}+\ldots \right){\frac {\operatorname {d} \omega }{\operatorname {d} x}}+\ldots$