Page:Mémoires de l’Académie des sciences, Tome 8.djvu/373

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${\frac {\varpi (t+x)}{x}}={\frac {\alpha }{2}}{\frac {(t+x)^{2}-1}{x}}$

sera inférieur à l’unité. D’ailleurs, si l’on pose

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$u={\frac {\alpha }{2}}{\frac {(t+x)^{2}-1}{x}},\quad x=re^{s{\sqrt {-1}}},\quad {\frac {dx}{ds}}=x{\sqrt {-1}},$

on aura

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$w=\operatorname {l} (u)=\operatorname {l} \left({\frac {\alpha }{2}}\right)+\operatorname {l} \left[(t+x)^{2}-1\right]-\operatorname {l} (x),$

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$\left\{{\begin{aligned}w'\ &={\frac {dw}{ds}}=\left[{\frac {2(t+x)}{(t+x)^{2}-1}}-{\frac {1}{x}}\right]x{\sqrt {-1}}={\sqrt {-1}}\left[{\frac {2tx+2x^{2}}{(t+x)^{2}-1}}-1\right],\\w''&={\frac {d^{2}w}{ds^{2}}}=p''+q''{\sqrt {-1}}=-\left\{{\frac {2(t+2x)x}{(t+x)^{2}-1}}-\left({\frac {2tx+2x^{2}}{(t+x)^{2}-1}}\right)^{2}\right\}.\end{aligned}}\right.$

Par suite l'équation $w'=0,$ donnera"

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$\left\{{\begin{aligned}&{\frac {2tx+2x^{2}}{(t+x)^{2}-1}}-1=0,\\&x^{2}=t^{2}-1=-\sin .^{2}\theta ,\\&x\;=\pm \sin .\theta .{\sqrt {-1}}\,;\end{aligned}}\right.$

et l’on en conclura, en plaçant le signe $+$ devant $\sin .\theta .{\sqrt {-1}},$

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$\left\{{\begin{aligned}\mathrm {W} ''&=\mathrm {P''+Q''} {\sqrt {-1}}=-\left[{\frac {2(t+2x)x}{(t+x)^{2}-1}}-1\right]={\frac {3x^{2}+1-t^{2}}{1-t^{2}-x^{2}-2tx}}\\&={\frac {1}{\sqrt {-1}}}{\frac {\sin .\theta }{\cos .\theta +{\sqrt {-1}}\sin .\theta }}={\frac {\sin .\theta \left(\cos .\theta -{\sqrt {-1}}\sin .\theta \right)}{\sqrt {-1}}}\,;\end{aligned}}\right.$

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$\mathrm {P} ''=-\sin .^{2}\theta ,\ \ \mathrm {B} =\sin .\theta ,\ \ \mathrm {Q} ''=-\sin .\theta \cos .\theta ,\ \ \mathrm {\frac {Q''}{P''}} =\cot .\theta =\operatorname {tang} .\left({\frac {\pi }{2}}-\theta \right)\,;$

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$\left\{{\begin{aligned}\mathrm {U} &={\frac {\alpha }{2}}{\frac {(t+x)^{2}-1}{x}}={\frac {\alpha }{2}}{\frac {\cos .2\theta -1+\sin .2\theta .{\sqrt {-1}}}{\sin .\theta {\sqrt {-1}}}}\\&=\alpha \left[\cos .\theta +\sin .\theta .{\sqrt {-1}}\right]\,;\end{aligned}}\right.$