Page:Mémoires de l’Académie des sciences, Tome 3.djvu/408

il vient

${\begin{aligned}{\frac {d^{2}i\varphi }{dt^{2}}}=&at\iint e^{t{\sqrt {b}}cos.u'}{\frac {d^{2}\operatorname {F} }{dx^{2}}}cos.^{2}u\,d\omega \\&\qquad \qquad \qquad +2t{\sqrt {ab}}\iint e^{t{\sqrt {b}}cos.u'}{\frac {d\operatorname {F} }{dx}}cos.u\,cos.u'd\omega \\&\ +2{\sqrt {a}}\iint e^{t{\sqrt {b}}cos.u'}{\frac {d\operatorname {F} }{dx}}cos.u\,d\omega \\&\qquad \qquad \qquad +bt\iint e^{t{\sqrt {b}}cos.u'}\operatorname {F} cos.^{2}u'd\omega \\&\ +2{\sqrt {b}}\iint e^{t{\sqrt {b}}cos.u'}\operatorname {F} cos.u'\,d\omega .\end{aligned}}$

L’intégration par parties donne

$+2\int e^{t{\sqrt {b}}cos.u'}\operatorname {F} cos.u'sin.u'\,du'=e^{t{\sqrt {b}}cos.u'}\operatorname {F} sin.^{2}u'$

$+t{\sqrt {b}}\int e^{t{\sqrt {b}}cos.u'}\operatorname {F} sin.^{3}u'du'$

$-t{\sqrt {a}}\int e^{t{\sqrt {b}}cos.u'}{\frac {d\operatorname {F} }{dx}}sin.^{2}u'cos.u'sin.v'du',$

le premier terme de cette expression s’évanouit aux deux limites $u'=0$ et $u'=\pi \,;$ en prenant donc $d\omega =sin.u'du'dv',$ et observant que $sin.u'sin.v'=cos.u,$ il en résultera

$2{\sqrt {b}}\iint e^{t{\sqrt {b}}cos.u'}\operatorname {F} cos.u'\,d\omega =bt\iint e^{t{\sqrt {b}}cos.u'}\operatorname {F} sin.^{2}u'd\omega$

$-t{\sqrt {ab}}\iint e^{t{\sqrt {b}}cos.u'}{\frac {d\operatorname {F} }{dx}}cos.u\,cos.u'd\omega \,;$

d’où l’on conclut ensuite

${\frac {d^{2}\varphi }{dt^{2}}}=at\iint e^{t{\sqrt {b}}cos.u'}{\frac {d^{2}\operatorname {F} }{dx^{2}}}cos.^{2}u\,d\omega$

$+t{\sqrt {ab}}\iint e^{t{\sqrt {b}}cos.u'}{\frac {d\operatorname {F} }{dx}}cos.u\,cos.u'd\omega$