Page:Annales de mathématiques pures et appliquées, 1826-1827, Tome 17.djvu/115

 Les corrections sont expliquées en page de discussion

(115)$\int _{0}^{\infty }{\frac {e^{ax}-e^{-ax}}{e^{x}-e^{-x}}}\operatorname {d} x={\frac {\varpi }{2}}\operatorname {Tang} .{\frac {a\varpi }{2}}.$

(116)$\int _{0}^{\infty }{\frac {e^{\lambda x}-e^{-\lambda x}}{e^{x}+2\operatorname {Cos} .\theta +e^{-x}}}\operatorname {Cos} .(\mu x).\operatorname {d} x$

$={\frac {\varpi }{\operatorname {Sin} .\theta }}.{\frac {\left[e^{\nu (\pi +\theta )}+e^{-\nu (\varpi +\theta )}\right]\operatorname {Cos} .\lambda (\varpi -\theta )-\left[e^{\nu (\pi -\theta )}+e^{-\nu (\varpi -\theta )}\right]\operatorname {Cos} .\lambda (\varpi +\theta )}{e^{2\nu \varpi }-2\operatorname {Cos} .(2\lambda \varpi )+e^{-2\nu \varpi }}}.$

(117)$\int _{0}^{\infty }{\frac {e^{\lambda x}-e^{-\lambda x}}{e^{x}+2\operatorname {Cos} .\theta +e^{-x}}}\operatorname {Sin} .(\mu x).\operatorname {d} x$

$={\frac {\varpi }{\operatorname {Sin} .\theta }}.{\frac {\left[e^{\nu (\pi +\theta )}-e^{-\nu (\pi +\theta )}\right]\operatorname {Sin} .\lambda (\varpi +\theta )-\left[e^{\nu (\pi -\theta )}-e^{-\nu (\varpi -\theta )}\right]\operatorname {Sin} .\lambda (\varpi +\theta )}{e^{2\nu \varpi }-2\operatorname {Cos} .(2\lambda \varpi )+e^{-2\nu \varpi }}}.$

(118)$\int _{0}^{\infty }{\frac {e^{ax}+(-1)^{n}e^{-ax}}{e^{x}+e^{-x}}}.x^{n}\operatorname {d} x=(-1)^{n}.{\frac {\pi }{2}}.{\frac {\operatorname {d} ^{n}.\operatorname {Sec} .{\frac {a\varpi }{2}}}{\operatorname {d} a^{n}}}.$

(119)$\int _{0}^{\infty }{\frac {e^{ax}-(-1)^{n}e^{-ax}}{e^{x}-e^{-x}}}.x^{n}\operatorname {d} x=(-1)^{n}.{\frac {\pi }{2}}.{\frac {\operatorname {d} ^{n}.\operatorname {Tang} .{\frac {a\varpi }{2}}}{\operatorname {d} a^{n}}}.$

(120)$\int _{0}^{\infty }\operatorname {Sin} .{\frac {b}{2}}\left(e^{x}+e^{-x}\right).\operatorname {Sin} .{\frac {b}{2}}\left(e^{x}-e^{-x}\right){\frac {\operatorname {d} x}{e^{x}-e^{-x}}}=+{\frac {\varpi }{4}}\operatorname {Sin} .b.$

(121)$\int _{0}^{\infty }{\frac {e^{x}\operatorname {Arc} .\left(\operatorname {Tang} .e^{-x}\right)-e^{-x}\operatorname {Arc} .\left(\operatorname {Tang} .e^{x}\right)}{e^{x}-e^{-x}}}\operatorname {d} x={\frac {\pi }{4}}\operatorname {l} (2)\,;$

et ainsi du reste.

On pourrait, au surplus, déduire directement la plupart des équations précédentes de la formule (4), combinée avec celles qui résultent de la méthode indiquée dans le 13.^e paragraphe du Mé-