Page:Annales de mathématiques pures et appliquées, 1826-1827, Tome 17.djvu/110

 Les corrections sont expliquées en page de discussion

(87)$\int _{0}^{\infty }\operatorname {Cos} .bx{\frac {r\operatorname {d} x}{x^{2}-r^{2}}}=-{\frac {\pi }{4}}\operatorname {Sin} .br,\quad \int _{0}^{\infty }\operatorname {Sin} .bx{\frac {x\operatorname {d} x}{x^{2}+r^{2}}}={\frac {\varpi }{2}}\operatorname {Cos} .br.$

(88)$\int _{0}^{\infty }\operatorname {Cos} .bx{\frac {\operatorname {d} x}{1-x^{2}}}=2\int _{0}^{1}\operatorname {Sin} .{\frac {b}{2}}\left(x+{\frac {1}{x}}\right).\operatorname {Sin} .{\frac {b}{2}}\left(x-{\frac {1}{x}}\right).{\frac {\operatorname {d} x}{x^{2}-1}}={\frac {\pi }{2}}\operatorname {Sin} .br.$

(89)$\int _{0}^{\infty }{\frac {\operatorname {Sin} .bx}{x}}\operatorname {d} x={\frac {\varpi }{2}}.\int _{0}^{\infty }{\frac {\operatorname {Sin} .bx}{x}}.{\frac {r^{2}\operatorname {d} x}{x^{2}+r^{2}}}={\frac {\varpi }{2}}\left(1-e^{-br}\right).$

(90)$\int _{-\infty }^{+\infty }\operatorname {l} \left(x^{2}-2rx\operatorname {Cos} .\theta +r^{2}\right){\frac {\operatorname {d} x}{1+x^{2}}}=\varpi \operatorname {l} \left(1+2r\operatorname {Sin} .\theta +r^{2}\right).$

(91)$\int _{-\infty }^{+\infty }\operatorname {Arc} .\operatorname {Tang} .={\frac {r\operatorname {Cos} .\theta -x}{r\operatorname {Sin} .\theta }}.{\frac {\operatorname {d} x}{1+x^{2}}}=\varpi .\operatorname {Arc} .\operatorname {Tang} .={\frac {r\operatorname {Cos} .\theta }{1+r\operatorname {Sin} .\theta }}.$

(92)$\left\{{\begin{aligned}&\int _{0}^{\infty }\operatorname {l} \left(1+{\frac {s^{2}}{x^{2}}}\right).{\frac {r\operatorname {d} x}{x^{2}+r^{2}}}=\varpi \operatorname {l} \left(1+{\frac {s}{r}}\right)\,;\\\\&\int _{0}^{\infty }\operatorname {Arc} .\operatorname {Tang} .\left({\frac {s}{x}}\right).{\frac {x\operatorname {d} x}{x^{2}+r^{2}}}={\frac {\varpi }{2}}\operatorname {l} \left(1+{\frac {s}{x}}\right).\end{aligned}}\right.$

(93)$\left\{{\begin{aligned}&\int _{0}^{\infty }\operatorname {l} \left(1+{\frac {s^{2}}{x^{2}}}\right).{\frac {r\operatorname {d} x}{x^{2}-r^{2}}}=-\varpi \operatorname {Arc} .\operatorname {Tang} .={\frac {s}{r}}\,;\\\\&\int _{0}^{\infty }\operatorname {Arc} .\operatorname {Tang} .\left({\frac {s}{x}}\right).{\frac {x\operatorname {d} x}{x^{2}-r^{2}}}={\frac {\varpi }{4}}\operatorname {l} \left(1+{\frac {s^{2}}{r^{2}}}\right).\end{aligned}}\right.$

(94)$\int _{0}^{\infty }(\operatorname {Arc} .\operatorname {Cot} .x)^{2}\operatorname {d} x=2\int _{0}^{\infty }\operatorname {Arc} .\operatorname {Tang} .{\frac {1}{x}}.{\frac {x\operatorname {d} x}{x^{2}+1}}=\varpi \operatorname {l} (2).$

(95)$\int _{0}^{1}{\frac {\operatorname {Arc} .\operatorname {Tang} .{\frac {1}{x}}-{\frac {1}{x}}\operatorname {Arc} .\operatorname {Tang} .x}{x-{\frac {1}{x}}}}.{\frac {\operatorname {d} x}{x}}={\frac {\varpi }{2}}\operatorname {l} (2).$

(96)$\int _{-\infty }^{+\infty }\left(-x{\sqrt {-1}}\right)^{a-1}\operatorname {l} \left(1+{\frac {s}{x}}{\sqrt {-1}}\right){\frac {r\operatorname {d} x}{x^{2}+r^{2}}}=\varpi r^{a-1}\operatorname {l} \left(1+{\frac {s}{x}}\right).$