Page:Annales de mathématiques pures et appliquées, 1826-1827, Tome 17.djvu/96

(16) $\int _{0}^{\infty }{\frac {f(x)+f(-x)}{2}}.{\frac {r\operatorname {d} x}{x^{2}-r^{2}}}={\frac {\pi }{4}}\left[f(x)-f(-r)\right]{\sqrt {-1}}.$

(17) $\int _{0}^{\infty }{\frac {f(x)-f(-x)}{2{\sqrt {-1}}}}.{\frac {x\operatorname {d} x}{x^{2}-r^{2}}}={\frac {\pi }{4}}\left[f(x)+f(-r)\right].$

(18) $\int _{0}^{\infty }{\frac {f(x)-f(-x)}{2{\sqrt {-1}}}}.{\frac {r^{2}\operatorname {d} x}{x\left(x^{2}-r^{2}\right)}}={\frac {\pi }{2}}\left[f(0)+f(r{\sqrt {-1}})\right].$

(19) $\int _{-\infty }^{+\infty }{\frac {f(x)}{r-x{\sqrt {-1}}}}\operatorname {d} x=0.$

(20) $\int _{-\infty }^{+\infty }{\frac {f(x)}{r+x{\sqrt {-1}}}}\operatorname {d} x=2\varpi f(r{\sqrt {-1}}).$

(21) $\int _{-\infty }^{+\infty }{\frac {f(x)}{\left(r-x{\sqrt {-1}}\right)^{m}}}.\operatorname {d} x=0.$

(22) $\int _{-\infty }^{+\infty }{\frac {f(x)}{\left(r+x{\sqrt {-1}}\right)^{m}}}.\operatorname {d} x=(-1)^{m-1}.{\frac {2\varpi }{\Gamma (m)}}.{\frac {\operatorname {d} ^{m-1}.f\left(r{\sqrt {-1}}\right)}{\operatorname {d} r^{m-1}}}.$^[1]

Si l’on suppose $f(x)={\frac {\operatorname {f} (x)}{\operatorname {l} \left(r-x{\sqrt {-1}}\right)}}\,;$ on trouvera

(23)$\quad \left\{{\begin{aligned}&\int _{-\infty }^{+\infty }{\frac {\operatorname {f} (x)}{\operatorname {l} \left(r-x{\sqrt {-1}}\right)}}.\operatorname {d} x=-2\varpi .\operatorname {f} \left(1-r{\sqrt {-1}}\right),\mathrm {\ pour\ } r<1\,;\\\\&\int _{-\infty }^{+\infty }{\frac {\operatorname {f} (x)}{\operatorname {l} \left(r-x{\sqrt {-1}}\right)}}.\operatorname {d} x=-\varpi .\operatorname {f} (0),\mathrm {\ pour\ } r=1\,;\\\\&\int _{-\infty }^{+\infty }{\frac {\operatorname {f} (x)}{\operatorname {l} \left(r-x{\sqrt {-1}}\right)}}.\operatorname {d} x=0,\mathrm {\ pour\ } r>1.\end{aligned}}\right.$

1. Le $\Gamma$ est celui de M. Legendre, de sorte que $\Gamma (m)=1.2.3\ldots (m-n).$ C’est le $(m-1)!$ de M. Kramp.
   J. D. G.