Izan bedi
f
:
[
a
,
b
)
→
ℜ
{\displaystyle f:[a,b)\rightarrow \Re }
f
{\displaystyle f}
[
a
,
t
]
−
n
{\displaystyle [a,t]-n}
∃
lim
t
→
∞
∫
a
∞
f
(
x
)
d
x
{\displaystyle \exists \lim _{t\to \infty }\int _{a}^{\infty }f(x)dx}
∫
a
∞
f
(
x
)
d
x
{\displaystyle \int _{a}^{\infty }f(x)dx}
∫
a
∞
f
(
x
)
d
x
=
lim
t
→
∞
∫
a
t
f
(
x
)
d
x
{\displaystyle \int _{a}^{\infty }f(x)dx=\lim _{t\to \infty }\int _{a}^{t}f(x)dx}
f
:
(
−
∞
,
∞
)
→
ℜ
{\displaystyle f:(-\infty ,\infty )\rightarrow \Re }
f
{\displaystyle f}
[
t
1
,
t
2
]
−
n
{\displaystyle [t_{1},t_{2}]-n}
∫
−
∞
∞
f
(
x
)
d
x
{\displaystyle \int _{-\infty }^{\infty }f(x)dx}
∫
−
∞
a
f
(
x
)
d
x
{\displaystyle \int _{-\infty }^{a}f(x)dx}
∫
a
∞
f
(
x
)
d
x
{\displaystyle \int _{a}^{\infty }f(x)dx}
∫
−
∞
∞
f
(
x
)
d
x
=
∫
−
∞
a
f
(
x
)
d
x
+
∫
a
∞
f
(
x
)
d
x
{\displaystyle \int _{-\infty }^{\infty }f(x)dx=\int _{-\infty }^{a}f(x)dx+\int _{a}^{\infty }f(x)dx}
Orokorrean,
∫
1
∞
1
x
α
d
x
=
{
k
o
n
b
e
r
g
e
n
t
e
a
,
α
>
1
d
i
b
e
r
g
e
n
t
e
a
,
α
≤
1
{\displaystyle \int _{1}^{\infty }{1 \over x^{\alpha }}dx={\begin{cases}konbergentea,&\alpha >1\\dibergentea,&\alpha \leq 1\end{cases}}}
Konbergentzia irizpidea
aldatu
Izan bitez
f
,
g
:
[
a
,
∞
)
→
ℜ
{\displaystyle f,g:[a,\infty )\rightarrow \Re }
[
a
,
t
]
−
n
{\displaystyle [a,t]-n}
0
≤
f
(
x
)
≤
g
(
x
)
{\displaystyle 0\leq f(x)\leq g(x)}
∀
x
≥
a
{\displaystyle \forall x\geq a}
∫
a
∞
g
(
x
)
d
x
{\displaystyle \int _{a}^{\infty }g(x)dx}
⇒
∫
a
∞
f
(
x
)
d
x
{\displaystyle \Rightarrow \int _{a}^{\infty }f(x)dx}
∫
a
∞
f
(
x
)
d
x
{\displaystyle \int _{a}^{\infty }f(x)dx}
⇒
∫
a
∞
g
(
x
)
d
x
{\displaystyle \Rightarrow \int _{a}^{\infty }g(x)dx}
0
≤
f
(
x
)
≤
g
(
x
)
{\displaystyle 0\leq f(x)\leq g(x)}
∀
x
≥
a
{\displaystyle \forall x\geq a}
⇒
0
≤
∫
a
t
f
(
x
)
d
x
≤
∫
a
t
g
(
x
)
d
x
{\displaystyle \Rightarrow 0\leq \int _{a}^{t}f(x)dx\leq \int _{a}^{t}g(x)dx}
0
≤
lim
t
→
∞
∫
a
t
f
(
x
)
d
x
≤
lim
t
→
∞
∫
a
t
g
(
x
)
d
x
{\displaystyle 0\leq \lim _{t\to \infty }\int _{a}^{t}f(x)dx\leq \lim _{t\to \infty }\int _{a}^{t}g(x)dx}
⊠
{\displaystyle \boxtimes }
aldatu
Izan bedi
∫
1
∞
1
x
5
+
7
d
x
{\displaystyle \int _{1}^{\infty }{1 \over x^{5}+7}dx}
0
≤
1
x
5
+
7
≤
1
x
5
{\displaystyle 0\leq {1 \over x^{5}+7}\leq {1 \over x^{5}}}
x
≥
1
{\displaystyle x\geq 1}
∫
1
∞
1
x
5
d
x
{\displaystyle \int _{1}^{\infty }{1 \over x^{5}}dx}
α
=
5
>
1
⟹
{\displaystyle \alpha =5>1\Longrightarrow }
→
k
o
n
p
.
i
r
i
z
p
∫
1
∞
1
x
5
d
x
≥
∫
1
∞
1
x
5
+
7
d
x
{\displaystyle {\xrightarrow[{}]{konp.irizp}}\int _{1}^{\infty }{1 \over x^{5}}dx\geq \int _{1}^{\infty }{1 \over x^{5}+7}dx}
Izan bitez
f
,
g
:
[
a
,
∞
)
→
ℜ
{\displaystyle f,g:[a,\infty )\rightarrow \Re }
[
a
,
t
]
−
n
{\displaystyle [a,t]-n}
0
≤
f
(
x
)
{\displaystyle 0\leq f(x)}
0
<
g
(
x
)
{\displaystyle 0<g(x)}
∀
x
≥
a
{\displaystyle \forall x\geq a}
lim
x
→
∞
(
f
(
x
)
g
(
x
)
)
=
l
{\displaystyle \lim _{x\to \infty }{\biggl (}{f(x) \over g(x)}{\biggr )}=l}
(i)
l
≠
0
,
∞
{\displaystyle l\neq 0,\infty }
∫
a
∞
f
(
x
)
d
x
{\displaystyle \int _{a}^{\infty }f(x)dx}
⇒
∫
a
∞
g
(
x
)
d
x
{\displaystyle \Rightarrow \int _{a}^{\infty }g(x)dx}
l
=
0
{\displaystyle l=0}
∫
a
∞
g
(
x
)
d
x
{\displaystyle \int _{a}^{\infty }g(x)dx}
⇒
∫
a
∞
f
(
x
)
d
x
{\displaystyle \Rightarrow \int _{a}^{\infty }f(x)dx}
l
=
∞
{\displaystyle l=\infty }
∫
a
∞
g
(
x
)
d
x
{\displaystyle \int _{a}^{\infty }g(x)dx}
⇒
∫
a
∞
f
(
x
)
d
x
{\displaystyle \Rightarrow \int _{a}^{\infty }f(x)dx}
(i)
l
≠
0
,
∞
{\displaystyle l\neq 0,\infty }
lim
x
→
∞
(
f
(
x
)
g
(
x
)
)
=
l
⟺
[
∀
ε
>
0
:
∃
M
>
0
,
x
>
M
⇒
|
f
(
x
)
g
(
x
)
−
l
|
<
ε
]
{\displaystyle \lim _{x\to \infty }\left({\frac {f(x)}{g(x)}}\right)=l\Longleftrightarrow \left[\forall \varepsilon >0:\exists M>0,x>M\Rightarrow \left\vert {\frac {f(x)}{g(x)}}-l\right\vert <\varepsilon \right]}
Kasu partikularra,
ε
=
l
2
{\displaystyle \varepsilon ={l \over 2}}
|
f
(
x
)
g
(
x
)
−
l
|
<
l
2
⟺
−
l
2
<
f
(
x
)
g
(
x
)
−
l
<
l
2
⟺
l
2
<
f
(
x
)
g
(
x
)
<
3
l
2
⟺
0
≤
l
2
g
(
x
)
≤
f
(
x
)
≤
3
l
2
f
(
x
)
{\displaystyle \left\vert {\frac {f(x)}{g(x)}}-l\right\vert <{l \over 2}\Longleftrightarrow -{l \over 2}<{\frac {f(x)}{g(x)}}-l<{l \over 2}\Longleftrightarrow {l \over 2}<{\frac {f(x)}{g(x)}}<{3l \over 2}\Longleftrightarrow 0\leq {l \over 2}g(x)\leq f(x)\leq {3l \over 2}f(x)}
l
=
0
{\displaystyle l=0}
lim
x
→
∞
(
f
(
x
)
g
(
x
)
)
=
∞
⟺
[
∀
k
>
0
:
∃
M
>
0
,
x
>
M
⇒
|
f
(
x
)
g
(
x
)
|
<
k
]
{\displaystyle \lim _{x\to \infty }\left({\frac {f(x)}{g(x)}}\right)=\infty \Longleftrightarrow \left[\forall k>0:\exists M>0,x>M\Rightarrow \left\vert {\frac {f(x)}{g(x)}}\right\vert <k\right]}
Kasu partikularra, k=1:
|
f
(
x
)
g
(
x
)
|
<
1
⟹
f
(
x
)
g
(
x
)
<
1
⟹
f
(
x
)
<
g
(
x
)
>
0
{\displaystyle \left\vert {\frac {f(x)}{g(x)}}\right\vert <1\Longrightarrow {\frac {f(x)}{g(x)}}<1\Longrightarrow f(x)<g(x)>0}
∫
0
∞
f
(
x
)
d
x
{\displaystyle \int _{0}^{\infty }f(x)dx}
k
o
n
b
e
r
g
e
n
t
e
a
⟸
∫
a
∞
g
(
x
)
d
x
{\displaystyle konbergentea\Longleftarrow \int _{a}^{\infty }g(x)dx}
k
o
n
b
e
r
g
e
n
t
e
a
{\displaystyle konbergentea}
⊠
{\displaystyle \boxtimes }
aldatu
∫
2
∞
x
+
1
x
3
d
x
{\displaystyle \int _{2}^{\infty }{x+1 \over {\sqrt {x^{3}}}}dx}
lim
x
→
∞
x
+
1
x
3
x
+
1
x
=
lim
x
→
∞
(
x
+
1
)
x
x
x
=
1
⟹
{\displaystyle \lim _{x\to \infty }{{x+1 \over {\sqrt {x^{3}}}} \over {x+1 \over {\sqrt {x}}}}=\lim _{x\to \infty }{(x+1){\sqrt {x}} \over x{\sqrt {x}}}=1\Longrightarrow }
Orduan,
∫
1
∞
1
x
d
x
{\displaystyle \int _{1}^{\infty }{1 \over {\sqrt {x}}}dx}
d
i
b
e
r
g
e
n
t
e
a
(
α
=
1
/
2
<
1
)
{\displaystyle dibergentea(\alpha =\operatorname {1} /\operatorname {2} <1)}
⟹
∫
a
∞
x
+
1
x
3
d
x
{\displaystyle \Longrightarrow {\displaystyle \int _{a}^{\infty }{x+1 \over {\sqrt {x^{3}}}}dx}}
![{\displaystyle [a,t]-n}](https://wikimedia.org/api/rest_v1/media/math/render/svg/beae9f4be00558b7c831f6872b78af31a0eb3a36)
![{\displaystyle [t_{1},t_{2}]-n}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1dd72fc18bdb988ff2502d75208d790869d65b2a)
![{\displaystyle {\xrightarrow[{}]{konp.irizp}}\int _{1}^{\infty }{1 \over x^{5}}dx\geq \int _{1}^{\infty }{1 \over x^{5}+7}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d26c0716446bb2143ff560c1317875972e7e958c)
![{\displaystyle \lim _{x\to \infty }\left({\frac {f(x)}{g(x)}}\right)=l\Longleftrightarrow \left[\forall \varepsilon >0:\exists M>0,x>M\Rightarrow \left\vert {\frac {f(x)}{g(x)}}-l\right\vert <\varepsilon \right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7338dad93bb3f746d5faa1bda634f91a2ce9bb3b)
![{\displaystyle \lim _{x\to \infty }\left({\frac {f(x)}{g(x)}}\right)=\infty \Longleftrightarrow \left[\forall k>0:\exists M>0,x>M\Rightarrow \left\vert {\frac {f(x)}{g(x)}}\right\vert <k\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ddf565a5e70a591a317f02fbaec94a593615913a)
Datuak: Q464118
Multimedia: Improper integral / Q464118
