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Annales de mathématiques pures et appliquées, 1826-1827, Tome 17.djvu/110
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(87)
∫
0
∞
Cos
.
b
x
r
d
x
x
2
−
r
2
=
−
π
4
Sin
.
b
r
,
∫
0
∞
Sin
.
b
x
x
d
x
x
2
+
r
2
=
ϖ
2
Cos
.
b
r
.
{\displaystyle \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)
∫
0
∞
Cos
.
b
x
d
x
1
−
x
2
=
2
∫
0
1
Sin
.
b
2
(
x
+
1
x
)
.
Sin
.
b
2
(
x
−
1
x
)
.
d
x
x
2
−
1
=
π
2
Sin
.
b
r
.
{\displaystyle \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)
∫
0
∞
Sin
.
b
x
x
d
x
=
ϖ
2
.
∫
0
∞
Sin
.
b
x
x
.
r
2
d
x
x
2
+
r
2
=
ϖ
2
(
1
−
e
−
b
r
)
.
{\displaystyle \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)
∫
−
∞
+
∞
l
(
x
2
−
2
r
x
Cos
.
θ
+
r
2
)
d
x
1
+
x
2
=
ϖ
l
(
1
+
2
r
Sin
.
θ
+
r
2
)
.
{\displaystyle \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)
∫
−
∞
+
∞
Arc
.
Tang
.
=
r
Cos
.
θ
−
x
r
Sin
.
θ
.
d
x
1
+
x
2
=
ϖ
.
Arc
.
Tang
.
=
r
Cos
.
θ
1
+
r
Sin
.
θ
.
{\displaystyle \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)
{
∫
0
∞
l
(
1
+
s
2
x
2
)
.
r
d
x
x
2
+
r
2
=
ϖ
l
(
1
+
s
r
)
;
∫
0
∞
Arc
.
Tang
.
(
s
x
)
.
x
d
x
x
2
+
r
2
=
ϖ
2
l
(
1
+
s
x
)
.
{\displaystyle \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)
{
∫
0
∞
l
(
1
+
s
2
x
2
)
.
r
d
x
x
2
−
r
2
=
−
ϖ
Arc
.
Tang
.
=
s
r
;
∫
0
∞
Arc
.
Tang
.
(
s
x
)
.
x
d
x
x
2
−
r
2
=
ϖ
4
l
(
1
+
s
2
r
2
)
.
{\displaystyle \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)
∫
0
∞
(
Arc
.
Cot
.
x
)
2
d
x
=
2
∫
0
∞
Arc
.
Tang
.
1
x
.
x
d
x
x
2
+
1
=
ϖ
l
(
2
)
.
{\displaystyle \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)
∫
0
1
Arc
.
Tang
.
1
x
−
1
x
Arc
.
Tang
.
x
x
−
1
x
.
d
x
x
=
ϖ
2
l
(
2
)
.
{\displaystyle \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)
∫
−
∞
+
∞
(
−
x
−
1
)
a
−
1
l
(
1
+
s
x
−
1
)
r
d
x
x
2
+
r
2
=
ϖ
r
a
−
1
l
(
1
+
s
x
)
.
{\displaystyle \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).}