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, \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
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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
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### 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
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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
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| Autoritate kontrola | 
* **Wikimedia proiektuak**
* Wd Datuak: Q464118
* Commonscat Multimedia: Improper integral

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* Wd Datuak: Q464118
* Commonscat Multimedia: Improper integral