Page:Laplace - Œuvres complètes, Gauthier-Villars, 1878, tome 7.djvu/675

nous aurons d’abord

$0={\frac {i+1}{a}}-s-{\frac {nc^{a}}{c^{a}-1}},$

ensuite

${\begin{aligned}f\ =&{\frac {i+1}{2a^{2}}}+{\frac {n}{2}}{\frac {c^{a}}{c^{a}-1}}-{\frac {n}{2}}\left({\frac {c^{a}}{c^{a}-1}}\right)^{2},\\f'\,=&{\sqrt {-1}}\left[{\frac {i+1}{3a^{3}}}-{\frac {n}{6}}{\frac {c^{a}}{c^{a}-1}}+{\frac {n}{2}}\left({\frac {c^{a}}{c^{a}-1}}\right)^{2}-{\frac {n}{3}}\left({\frac {c^{a}}{c^{a}-1}}\right)^{3}\right]\\f''=&-{\frac {i+1}{4a^{4}}}-{\frac {n}{24}}{\frac {c^{a}}{c^{a}-1}}\\&\qquad +{\frac {7n}{24}}\left({\frac {c^{a}}{c^{a}-1}}\right)^{2}-{\frac {n}{2}}\left({\frac {c^{a}}{c^{a}-1}}\right)^{2}+{\frac {n}{24}}\left({\frac {c^{a}}{c^{a}-1}}\right)^{4},\\\ldots &\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \end{aligned}}$

On a ensuite, par le retour des séries,

$a\varpi ={\frac {t}{\sqrt {f}}}\left(1-{\frac {f't}{2f{\sqrt {f}}}}+{\frac {5f'^{2}-4ff''}{8f^{3}}}t^{2}+\ldots \right)\,;$

on a donc, en prenant les intégrales depuis $\varpi$ et $t$ égaux à $-\infty$ jusqu’à $t$ et $\varpi$ égaux à $+\infty ,$

${\begin{aligned}\int &{\frac {d\varpi c^{as\left(1-\varpi {\sqrt {-1}}\right)}\left(c^{a\left(1-\varpi {\sqrt {-1}}\right)}-1\right)^{n}}{\left(1-\varpi {\sqrt {-1}}\right)^{n}}}\\&={\frac {c^{as}(c^{a}-1)^{n}}{a}}\int {\frac {c^{-t^{2})}dt}{\sqrt {f}}}\left(1-{\frac {f't}{f{\sqrt {f}}}}+3{\frac {5f'^{2}-4ff''}{8f^{3}}}t^{2}+\ldots \right)\\&={\frac {\sqrt {\pi }}{\sqrt {f}}}\left(1+{\frac {15f'^{2}-12ff''}{16f^{3}}}+\ldots \right){\frac {c^{as}(c^{a}-1)^{n}}{a}}.\end{aligned}}$

Si l’on suppose $s=1,\ n=0$ et si l’on change $a$ en $a',$ on aura

$a'=i+1,\quad f={\frac {1}{2(i+1)}},\quad f'={\frac {\sqrt {-1}}{3(i+1)^{2}}},\quad f''={\frac {1}{4(i+1)^{3}}},\quad \ldots \,;$

on aura donc

$\int {\frac {d\varpi c^{a'\left(1-\varpi {\sqrt {-1}}\right)}}{\left(1-\varpi {\sqrt {-1}}\right)^{i+1}}}={\frac {c^{i+1}}{i+1}}\left(1-{\frac {1}{12i}}+\ldots \right){\sqrt {2(i+1)\pi }}.$