Integral inpropio

Kalkuluan, funtzio baten integral inpropio bat integral zehatz baten limitea da, integrazio-tartearen mutur bateko edo bietako puntuak bere eremuan ez dagoen zenbaki batera hurbiltzen direnean, ${\displaystyle \infty }$-ra edo ${\displaystyle -\infty }$-ra. Gainera, integral definitu bat inpropioa da integral definituaren funtzio integratua integrazio-tarte osoan jarraitua ez denean. Bi egoerak ere gerta daitezke.

Muga infinituko integral inpropioak

Izan bedi ${\displaystyle f:[a,b)\rightarrow \Re }$ eta ${\displaystyle f}$ ${\displaystyle [a,t]-n}$ integragarria (t>a zanik), ${\displaystyle \exists \lim _{t\to \infty }\int _{a}^{\infty }f(x)dx}$, ${\displaystyle \int _{a}^{\infty }f(x)dx}$ integral inpropioa konbergentea da.

${\displaystyle \int _{a}^{\infty }f(x)dx=\lim _{t\to \infty }\int _{a}^{t}f(x)dx}$

Izan bedi ${\displaystyle f:(-\infty ,\infty )\rightarrow \Re }$ eta ${\displaystyle f}$ ${\displaystyle [t_{1},t_{2}]-n}$ integragarria. Orduan, ${\displaystyle \int _{-\infty }^{\infty }f(x)dx}$ integral inpropioa konbergentea da${\displaystyle \int _{-\infty }^{a}f(x)dx}$ eta ${\displaystyle \int _{a}^{\infty }f(x)dx}$ konbergenteak direnean. Kasu horretan, ${\displaystyle \int _{-\infty }^{\infty }f(x)dx=\int _{-\infty }^{a}f(x)dx+\int _{a}^{\infty }f(x)dx}$.

Orokorrean,

${\displaystyle \int _{1}^{\infty }{1 \over x^{\alpha }}dx={\begin{cases}konbergentea,&\alpha >1\\dibergentea,&\alpha \leq 1\end{cases}}}$

Funtzio ez-negatiboen integral inpropioen konbergentzia irizpideak

Konbergentzia irizpidea

Izan bitez ${\displaystyle f,g:[a,\infty )\rightarrow \Re }$ funtzioa integragarriak ${\displaystyle [a,t]-n}$ (t>a zanik). Suposatu ${\displaystyle 0\leq f(x)\leq g(x)}$ ${\displaystyle \forall x\geq a}$. Orduan,

${\displaystyle \int _{a}^{\infty }g(x)dx}$ konbergentea ${\displaystyle \Rightarrow \int _{a}^{\infty }f(x)dx}$ konbergentea

${\displaystyle \int _{a}^{\infty }f(x)dx}$ dibergentea ${\displaystyle \Rightarrow \int _{a}^{\infty }g(x)dx}$ dibergentea

Froga

${\displaystyle 0\leq f(x)\leq g(x)}$ bada ${\displaystyle \forall x\geq a}$

${\displaystyle \Rightarrow 0\leq \int _{a}^{t}f(x)dx\leq \int _{a}^{t}g(x)dx}$. Beraz, ${\displaystyle 0\leq \lim _{t\to \infty }\int _{a}^{t}f(x)dx\leq \lim _{t\to \infty }\int _{a}^{t}g(x)dx}$

${\displaystyle \boxtimes }$

Adibideak

Izan bedi ${\displaystyle \int _{1}^{\infty }{1 \over x^{5}+7}dx}$ integral inpropioa.

${\displaystyle 0\leq {1 \over x^{5}+7}\leq {1 \over x^{5}}}$, ${\displaystyle x\geq 1}$.

${\displaystyle \int _{1}^{\infty }{1 \over x^{5}}dx}$, ${\displaystyle \alpha =5>1\Longrightarrow }$ konbergentea ${\displaystyle {\xrightarrow[{}]{konp.irizp}}\int _{1}^{\infty }{1 \over x^{5}}dx\geq \int _{1}^{\infty }{1 \over x^{5}+7}dx}$ konbergentea.

Limitearen irizpidea

Izan bitez ${\displaystyle f,g:[a,\infty )\rightarrow \Re }$ funtzioa integragarriak ${\displaystyle [a,t]-n}$. Suposatu ${\displaystyle 0\leq f(x)}$ eta ${\displaystyle 0<g(x)}$ ${\displaystyle \forall x\geq a}$ eta ${\displaystyle \lim _{x\to \infty }{\biggl (}{f(x) \over g(x)}{\biggr )}=l}$. Orduan,

(i) ${\displaystyle l\neq 0,\infty }$

${\displaystyle \int _{a}^{\infty }f(x)dx}$ konbergentea ${\displaystyle \Rightarrow \int _{a}^{\infty }g(x)dx}$ konbergentea

(ii) ${\displaystyle l=0}$

${\displaystyle \int _{a}^{\infty }g(x)dx}$ konbergentea ${\displaystyle \Rightarrow \int _{a}^{\infty }f(x)dx}$ konbergentea

(iii) ${\displaystyle l=\infty }$

${\displaystyle \int _{a}^{\infty }g(x)dx}$ dibergentea ${\displaystyle \Rightarrow \int _{a}^{\infty }f(x)dx}$ dibergentea

Froga

(i) ${\displaystyle l\neq 0,\infty }$

${\displaystyle \lim _{x\to \infty }\left({\frac {f(x)}{g(x)}}\right)=l\Longleftrightarrow \left[\forall \varepsilon >0:\exists M>0,x>M\Rightarrow \left\vert {\frac {f(x)}{g(x)}}-l\right\vert <\varepsilon \right]}$

Kasu partikularra, ${\displaystyle \varepsilon ={l \over 2}}$. Orduan,

${\displaystyle \left\vert {\frac {f(x)}{g(x)}}-l\right\vert <{l \over 2}\Longleftrightarrow -{l \over 2}<{\frac {f(x)}{g(x)}}-l<{l \over 2}\Longleftrightarrow {l \over 2}<{\frac {f(x)}{g(x)}}<{3l \over 2}\Longleftrightarrow 0\leq {l \over 2}g(x)\leq f(x)\leq {3l \over 2}f(x)}$

(ii) ${\displaystyle l=0}$

${\displaystyle \lim _{x\to \infty }\left({\frac {f(x)}{g(x)}}\right)=\infty \Longleftrightarrow \left[\forall k>0:\exists M>0,x>M\Rightarrow \left\vert {\frac {f(x)}{g(x)}}\right\vert <k\right]}$

Kasu partikularra, k=1:

${\displaystyle \left\vert {\frac {f(x)}{g(x)}}\right\vert <1\Longrightarrow {\frac {f(x)}{g(x)}}<1\Longrightarrow f(x)<g(x)>0}$

${\displaystyle \int _{0}^{\infty }f(x)dx}$ ${\displaystyle konbergentea\Longleftarrow \int _{a}^{\infty }g(x)dx}$ ${\displaystyle konbergentea}$

${\displaystyle \boxtimes }$

Adibideak

${\displaystyle \int _{2}^{\infty }{x+1 \over {\sqrt {x^{3}}}}dx}$ integral inpropioaren konbergentzia aztertu.

${\displaystyle \lim _{x\to \infty }{{x+1 \over {\sqrt {x^{3}}}} \over {x+1 \over {\sqrt {x}}}}=\lim _{x\to \infty }{(x+1){\sqrt {x}} \over x{\sqrt {x}}}=1\Longrightarrow }$ f(x) eta g(x) bera izaera bera dute.

Orduan,

${\displaystyle \int _{1}^{\infty }{1 \over {\sqrt {x}}}dx}$ ${\displaystyle dibergentea(\alpha =\operatorname {1} /\operatorname {2} <1)}$ ${\displaystyle \Longrightarrow {\displaystyle \int _{a}^{\infty }{x+1 \over {\sqrt {x^{3}}}}dx}}$ dibergentea

Konbergentzia absolutua

Funtzioa negatiboa denean konbergentzia absolutua aztertu behar da.

${\displaystyle \int _{a}^{\infty }|f(x)|dx}$ konbergente ${\displaystyle \Longrightarrow \int _{a}^{\infty }f(x)dx}$ absolutuki konbergente ${\displaystyle \Longrightarrow \int _{a}^{\infty }f(x)dx}$ konbergente eta ${\displaystyle \left\vert {\int _{a}^{\infty }f(x)dx}\right\vert \leq \int _{a}^{\infty }|f(x)|dx}$

Adibideak

${\displaystyle \int _{1}^{\infty }{sin(x) \over x^{3}}dx}$ -ren konbergentzia aztertu:

${\displaystyle \left\vert {sin(x) \over x^{3}}\right\vert \leq {|sin(x)| \over x^{3}}={1 \over 3}}$ ${\displaystyle ,x>1}$

${\displaystyle \int _{1}^{\infty }{1 \over x^{3}}dx}$ eta ${\displaystyle \alpha =3>1}$ denez, integrala konbergentea da. Horrek inplikatzen du ${\displaystyle \int _{1}^{\infty }\left\vert {\frac {sin(x)}{x^{3}}}\right\vert {}dx}$ konbergentea izatea eta, beraz, ${\displaystyle \int _{1}^{\infty }{sin(x) \over x^{3}}dx}$ absolutuki konbergentea ${\displaystyle \Longrightarrow \int _{1}^{\infty }{sin(x) \over x^{3}}dx}$ konbergentea.

Kanpo estekak

• Datuak: Q464118
• Multimedia: Improper integral / Q464118