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Annales de mathématiques pures et appliquées, 1826-1827, Tome 17.djvu/109
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(79)
∫
0
∞
x
a
d
x
x
2
+
2
r
x
Cos
.
θ
+
r
2
=
ϖ
r
a
−
1
Sin
.
a
ϖ
.
Sin
.
a
θ
Sin
.
θ
.
{\displaystyle \int _{0}^{\infty }{\frac {x^{a}\operatorname {d} x}{x^{2}+2rx\operatorname {Cos} .\theta +r^{2}}}={\frac {\varpi r^{a-1}}{\operatorname {Sin} .a\varpi }}.{\frac {\operatorname {Sin} .a\theta }{\operatorname {Sin} .\theta }}.}
(80)
∫
0
∞
x
λ
Cos
.
[
μ
l
(
x
)
]
x
2
+
2
x
Cos
.
θ
+
1
d
x
{\displaystyle \int _{0}^{\infty }{\frac {x\lambda \operatorname {Cos} .\left[\mu \operatorname {l} (x)\right]}{x^{2}+2x\operatorname {Cos} .\theta +1}}\operatorname {d} x}
=
ϖ
Sin
.
θ
.
[
e
μ
(
π
+
θ
)
+
e
−
μ
(
ϖ
+
θ
)
]
Cos
.
λ
(
ϖ
−
θ
)
−
[
e
μ
(
π
−
θ
)
+
e
−
μ
(
ϖ
−
θ
)
]
Cos
.
λ
(
ϖ
+
θ
)
e
2
μ
ϖ
−
2
Cos
.
(
2
λ
ϖ
)
+
e
−
2
μ
ϖ
.
{\displaystyle ={\frac {\varpi }{\operatorname {Sin} .\theta }}.{\frac {\left[e^{\mu (\pi +\theta )}+e^{-\mu (\varpi +\theta )}\right]\operatorname {Cos} .\lambda (\varpi -\theta )-\left[e^{\mu (\pi -\theta )}+e^{-\mu (\varpi -\theta )}\right]\operatorname {Cos} .\lambda (\varpi +\theta )}{e^{2\mu \varpi }-2\operatorname {Cos} .(2\lambda \varpi )+e^{-2\mu \varpi }}}.}
(81)
∫
0
∞
x
λ
Sin
.
[
μ
l
(
x
)
]
x
2
+
2
x
Cos
.
θ
+
1
d
x
{\displaystyle \int _{0}^{\infty }{\frac {x\lambda \operatorname {Sin} .\left[\mu \operatorname {l} (x)\right]}{x^{2}+2x\operatorname {Cos} .\theta +1}}\operatorname {d} x}
=
ϖ
Sin
.
θ
.
[
e
μ
(
π
+
θ
)
−
e
−
μ
(
π
+
θ
)
]
Sin
.
λ
(
ϖ
+
θ
)
−
[
e
μ
(
π
−
θ
)
−
e
−
μ
(
ϖ
−
θ
)
]
Sin
.
λ
(
ϖ
−
θ
)
e
2
μ
ϖ
−
2
Cos
.
(
2
λ
ϖ
)
+
e
−
2
μ
ϖ
.
{\displaystyle ={\frac {\varpi }{\operatorname {Sin} .\theta }}.{\frac {\left[e^{\mu (\pi +\theta )}-e^{-\mu (\pi +\theta )}\right]\operatorname {Sin} .\lambda (\varpi +\theta )-\left[e^{\mu (\pi -\theta )}-e^{-\mu (\varpi -\theta )}\right]\operatorname {Sin} .\lambda (\varpi -\theta )}{e^{2\mu \varpi }-2\operatorname {Cos} .(2\lambda \varpi )+e^{-2\mu \varpi }}}.}
(82)
∫
0
∞
x
a
1
+
x
2
[
l
(
x
)
]
n
d
x
=
∫
0
1
x
a
+
(
−
1
)
n
(
1
x
)
a
x
+
1
x
[
l
(
x
)
]
n
d
x
x
{\displaystyle \int _{0}^{\infty }{\frac {x^{a}}{1+x^{2}}}\left[\operatorname {l} (x)\right]^{n}\operatorname {d} x=\int _{0}^{1}{\frac {x^{a}+(-1)^{n}\left({\frac {1}{x}}\right)^{a}}{x+{\frac {1}{x}}}}\left[\operatorname {l} (x)\right]^{n}{\frac {\operatorname {d} x}{x}}}
=
π
2
.
d
n
.
Sec
.
(
1
2
a
ϖ
)
d
a
n
.
{\displaystyle ={\frac {\pi }{2}}.{\frac {\operatorname {d} ^{n}.\operatorname {Sec} .\left({\frac {1}{2}}a\varpi \right)}{\operatorname {d} a^{n}}}.}
(83)
∫
0
∞
x
a
x
2
−
1
[
l
(
x
)
]
n
d
x
=
∫
0
1
x
a
−
(
−
1
)
n
(
1
x
)
a
x
+
1
x
[
l
(
x
)
]
n
d
x
x
{\displaystyle \int _{0}^{\infty }{\frac {x^{a}}{x^{2}-1}}\left[\operatorname {l} (x)\right]^{n}\operatorname {d} x=\int _{0}^{1}{\frac {x^{a}-(-1)^{n}\left({\frac {1}{x}}\right)^{a}}{x+{\frac {1}{x}}}}\left[\operatorname {l} (x)\right]^{n}{\frac {\operatorname {d} x}{x}}}
=
π
2
.
d
n
.
Tang
.
(
1
2
a
ϖ
)
d
a
n
.
{\displaystyle ={\frac {\pi }{2}}.{\frac {\operatorname {d} ^{n}.\operatorname {Tang} .\left({\frac {1}{2}}a\varpi \right)}{\operatorname {d} a^{n}}}.}
(84)
∫
0
∞
x
a
−
1
−
x
b
−
1
l
(
x
)
.
d
x
x
2
+
1
=
+
(
l
.
Tang
.
a
ϖ
4
−
l
.
Tang
.
b
ϖ
4
)
.
{\displaystyle \int _{0}^{\infty }{\frac {x^{a-1}-x^{b-1}}{\operatorname {l} (x)}}.{\frac {\operatorname {d} x}{x^{2}+1}}=+\left(\operatorname {l} .\operatorname {Tang} .{\frac {a\varpi }{4}}-\operatorname {l} .\operatorname {Tang} .{\frac {b\varpi }{4}}\right).}
(85)
∫
0
∞
x
a
−
1
−
x
b
−
1
l
(
x
)
.
d
x
x
2
−
1
=
−
(
l
.
Sin
.
a
π
2
−
l
.
Sin
.
b
ϖ
2
)
.
{\displaystyle \int _{0}^{\infty }{\frac {x^{a-1}-x^{b-1}}{\operatorname {l} (x)}}.{\frac {\operatorname {d} x}{x^{2}-1}}=-\left(\operatorname {l} .\operatorname {Sin} .{\frac {a\pi }{2}}-\operatorname {l} .\operatorname {Sin} .{\frac {b\varpi }{2}}\right).}
(86)
∫
0
∞
Cos
.
b
x
.
r
d
x
x
2
+
r
2
=
π
2
.
e
−
b
r
,
∫
0
∞
Sin
.
b
x
.
x
d
x
x
2
+
r
2
=
π
2
.
e
−
b
r
.
{\displaystyle \int _{0}^{\infty }\operatorname {Cos} .bx.{\frac {r\operatorname {d} x}{x^{2}+r^{2}}}={\frac {\pi }{2}}.e^{-br},\quad \int _{0}^{\infty }\operatorname {Sin} .bx.{\frac {x\operatorname {d} x}{x^{2}+r^{2}}}={\frac {\pi }{2}}.e^{-br}.}
![{\displaystyle \int _{0}^{\infty }{\frac {x\lambda \operatorname {Cos} .\left[\mu \operatorname {l} (x)\right]}{x^{2}+2x\operatorname {Cos} .\theta +1}}\operatorname {d} x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5033203dd7d72d0ea86862e0dbe0c4fb9a4cd32f)
![{\displaystyle ={\frac {\varpi }{\operatorname {Sin} .\theta }}.{\frac {\left[e^{\mu (\pi +\theta )}+e^{-\mu (\varpi +\theta )}\right]\operatorname {Cos} .\lambda (\varpi -\theta )-\left[e^{\mu (\pi -\theta )}+e^{-\mu (\varpi -\theta )}\right]\operatorname {Cos} .\lambda (\varpi +\theta )}{e^{2\mu \varpi }-2\operatorname {Cos} .(2\lambda \varpi )+e^{-2\mu \varpi }}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c053446e7ab064311c8a0a90e2a99535aee2762)
![{\displaystyle \int _{0}^{\infty }{\frac {x\lambda \operatorname {Sin} .\left[\mu \operatorname {l} (x)\right]}{x^{2}+2x\operatorname {Cos} .\theta +1}}\operatorname {d} x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3c08d40b06c15e3cdac41f205a5aec0bc2c59790)
![{\displaystyle ={\frac {\varpi }{\operatorname {Sin} .\theta }}.{\frac {\left[e^{\mu (\pi +\theta )}-e^{-\mu (\pi +\theta )}\right]\operatorname {Sin} .\lambda (\varpi +\theta )-\left[e^{\mu (\pi -\theta )}-e^{-\mu (\varpi -\theta )}\right]\operatorname {Sin} .\lambda (\varpi -\theta )}{e^{2\mu \varpi }-2\operatorname {Cos} .(2\lambda \varpi )+e^{-2\mu \varpi }}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/41079089adcd84cef6e99aa32cc5a3a976c35973)
![{\displaystyle \int _{0}^{\infty }{\frac {x^{a}}{1+x^{2}}}\left[\operatorname {l} (x)\right]^{n}\operatorname {d} x=\int _{0}^{1}{\frac {x^{a}+(-1)^{n}\left({\frac {1}{x}}\right)^{a}}{x+{\frac {1}{x}}}}\left[\operatorname {l} (x)\right]^{n}{\frac {\operatorname {d} x}{x}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8ce4e0273010029eea72c4f4210bbcec7899eaec)
![{\displaystyle \int _{0}^{\infty }{\frac {x^{a}}{x^{2}-1}}\left[\operatorname {l} (x)\right]^{n}\operatorname {d} x=\int _{0}^{1}{\frac {x^{a}-(-1)^{n}\left({\frac {1}{x}}\right)^{a}}{x+{\frac {1}{x}}}}\left[\operatorname {l} (x)\right]^{n}{\frac {\operatorname {d} x}{x}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2abd79d71206410ce585f79ec4836fec8e377c5c)
