Utilisateur:Xzapro4/IntégralesImpropres

Intégrales impropres

1)1) Cas [a,b[ où -inf < a < b < +inf

Définition

Soit $f:[a,b[\to K$ continue par morceaux

On dit que l'intégrale impropre $\int _{a}^{b}f(t){\rm {d}}t$ existe, a un sens ou converge lorsque $x\to \int _{a}^{x}f(t){\rm {d}}t$ admet une limite $I\in K$ quand $x{\overset {<}{\to }}b$

Dans ce cas, on note $\int _{a}^{b}f(t){\rm {d}}t$ cette limite I.

Lorsque $x\to \int _{a}^{x}f(t){\rm {d}}t$ n'a pas de limite (dans K), on dit que l'intégrale impropre $\int _{a}^{b}f(t){\rm {d}}t$ n'existe pas, ou diverge.

• Nature de $\int _{a}^{b}f(t){\rm {d}}t$ : convergence ou divergence

Remarque

Soit $c\in [a;b[$

$\forall x\in [a,b[,\int _{c}^{x}f(t){\rm {d}}t=\int _{c}^{a}f(t){rmd}t+\int _{a}^{x}f(t){\rm {d}}t$

D'où $\int _{c}^{b}f(t){\rm {d}}t$ converge ssi $\int _{a}^{b}f(t){\rm {d}}t$ converge

Dans les théorèmes suivants portant sur la nature de $\int _{a}^{b}f(t){\rm {d}}t$, les hypothèses faite sur f sur $[a,b[$ pourront être remplacées par les mêmes hypothèses sur le seul intervalle $[c,b[$

Début d’un théorème

Théorème

Soit $\alpha \in \mathbb {R}$

$\int _{1}^{+\infty }{\frac {1}{t^{\alpha }}}{\rm {d}}t$ converge ssi $\alpha >1$

Pour $\alpha >1,~\int _{1}^{+\infty }{\frac {1}{t^{\alpha }}}{\rm {d}}t={\frac {1}{\alpha -1}}$

Fin du théorème

Début d'une démonstration

Démonstration

$t\to {\frac {1}{t^{\alpha }}}$ est de classe ${\mathcal {C}}^{0}$ sur $[1,+\infty [$

$1\leq x:\int _{1}^{x}{\frac {1}{t^{\alpha }}}{\rm {d}}t={\begin{cases}\ln x{\rm {~si~}}\alpha =1\\{\frac {x^{-\alpha +1}-1}{-\alpha +1}}{\rm {~sinon}}\end{cases}}$

$\ln x\to +\infty (x\to +\infty )$

$x^{1-\alpha }\to {\begin{cases}0{\rm {~si~}}\alpha >1\\+\infty {\rm {~si~}}\alpha <1\\\end{cases}}x\to +\infty$

Pour $\alpha >1,\int _{1}^{+\infty }{\frac {1}{t^{\alpha }}}{\rm {d}}t=\lim _{x\to +\infty }{\frac {1-x^{1-\alpha }}{\alpha -1}}={\frac {1}{\alpha -1}}$

Fin de la démonstration

Début d’un théorème

Théorème

Soit $\alpha \in \mathbb {R}$

$\int _{0}^{+\infty }e^{-\alpha t}{\rm {d}}t$ converge ssi $\alpha >0$

Dans ce cas, $\int _{0}^{+\infty }e^{-\alpha t}{\rm {d}}t={\frac {1}{\alpha }}$

Fin du théorème

Début d'une démonstration

Démonstration

$t\to e^{-\alpha t}\in {\mathcal {C}}^{0}(\mathbb {R} ^{+})$

$\forall x\in R^{+},\int _{0}^{x}e^{-\alpha t}{\rm {d}}t={\begin{cases}x{\rm {~si~}}\alpha =0\\{\frac {e^{-\alpha x}-1}{-\alpha }}{\rm {~sinon}}\end{cases}}$

$x\to \int _{0}^{x}e^{-\alpha t}{\rm {d}}t$ admet une limite finie en $+\infty$ ssi $\alpha >0$

Pour $\alpha >0,\int _{0}^{+\infty }e^{-\alpha t}{\rm {d}}t=\lim _{x\to +\infty }{\frac {e^{-\alpha x}-1}{-\alpha }}={\frac {1}{\alpha }}$

Fin de la démonstration

& \underline{Th}\ \left( a,b \right)\in R{}^\text{2};\ a<b \\ 
& \alpha \in R \\ 
&  \\ 
& \int_{a}^{b}{\frac{1}{\left( b-t \right)^{\alpha }}dt}\ converge\ ssi\ \alpha <1 \\ 

\end{align}</math>

${\begin{aligned}&{\underline {d{\acute {e}}m}}\\&t\to {\frac {1}{\left(b-t\right)^{\alpha }}}\ \ C{}^{\circ }\ sur\ [a,b[\\&\forall x\in [a,b[\ ,\ \int _{a}^{x}{{\frac {1}{\left(b-t\right)^{\alpha }}}dt}=\left\{{\begin{aligned}&-\ln \left(b-x\right)+\ln \left(b-a\right)\ si\ \alpha =1\\&{\frac {\left(b-x\right)^{-\alpha +1}-\left(b-a\right)^{-\alpha +1}}{-\alpha +1}},\ si\ \alpha \neq 1\\\end{aligned}}\right.\\&\\&x\to \int _{a}^{x}{{\frac {1}{\left(b-t\right)^{\alpha }}}dt}\ admet\ une\ \lim ite\ finie\ en\ b\ ssi\ \alpha <1\\&\\&{\underline {Th}}\ \left(coh{\acute {e}}rence\ de\ la\ notation\right)\\&-\infty <a<b<+\infty \\&\\&Soit\ f:\ [a,b[\to K\ continue\ par\ morceaux,\ se\ prolongeant\ en\ une\ fonction\ {\tilde {f}}\ continue\ par\ morceaux\ sur\ [a,b]\\&\\&\int _{a}^{b}{f\left(t\right)dt}\ converge\ et\ \int _{a}^{b}{f\left(t\right)dt}=\int _{[a,b]}^{}{{\tilde {f}}\left(t\right)dt}\\&{\underline {d{\acute {e}}m}}\ \ {\tilde {f}}\ continue\ par\ moreaux\ sur\ [a,b]\\&soit\ F:{\begin{matrix}[a,b]\to K\\x\to \int _{a}^{x}{{\tilde {f}}\left(t\right)dt}\\\end{matrix}}\\\end{aligned}}$

${\begin{aligned}&{\tilde {f}}\ est\ born{\acute {e}}e\ sur\ [a,b]\\&\forall \left(x,y\right)\in [a,b]{}^{\text{2}},\ \left|F\left(x\right)-F\left(y\right)\right|=\left|\int _{y}^{x}{{\tilde {f}}\left(t\right)dt}\right|\leq \left|\int _{y}^{x}{\left|{\tilde {f}}\left(t\right)\right|dt}\right|\leq \left|\int _{y}^{x}{\left|\left|f\right|\right|_{\infty }}\right|\leq \left|x-y\right|||f||_{\infty }\\&F\ lipschitzienne,\ donc\ continue\ sur\ [a,b]\\&\\&\int _{a}^{x}{f\left(t\right)dt}=\int _{a}^{x}{{\tilde {f}}\left(t\right)dt}=F\left(x\right)\to F\left(b\right)\ \left(x{\overset {<}{\mathop {\to } }}\,b\right)\\&\int _{a}^{b}{f\left(t\right)dt}\ converge\ et\ \int _{a}^{b}{f\left(t\right)dt}=F\left(b\right)=\int _{[a,b]}^{}{\tilde {f}}\\&\\&{\underline {exemple}}\ \int _{-\pi }^{0}{{\frac {s\operatorname {int} }{t}}dt}\ converge\\&\\&En\ effet,\ t\to {\frac {\sin t}{t}}\ continue\ sur\ [-\pi ,0[\ ,\ {\frac {\sin t}{t}}\to 1\ \left(t{\overset {<}{\mathop {\to } }}\,0\right)\\\end{aligned}}$

2)Cas ]a,b] où -inf < a < b < +inf

${\begin{aligned}&Soit\ f:]a,b]\to K\ continue\ par\ morceaux.\\&\\&{\underline {d{\acute {e}}f}}:\ On\ dit\ que\ l'\operatorname {int} {\acute {e}}grale\ impropre\ \int _{a}^{b}{f\left(t\right)dt}\ existe\ ,\ a\ un\ sens,\ converge\\&lors\ que\ {\begin{matrix}]a,b]\to K\\x\to \int _{x}^{b}{f\left(t\right)dt}\\\end{matrix}}\ admet\ une\ \lim ite\ I\ dans\ K\ \left(x{\overset {>}{\mathop {\to } }}\,a\right)\\&\\&I\ est\ not{\acute {e}}e\ \int _{a}^{b}{f\left(t\right)dt}\\&\\&Sinon,\ on\ dit\ que\ \int _{a}^{b}{f\left(t\right)dt}\ n'existe\ pas,\ diverge.\\&\\&{\underline {Th}}\ \alpha \in R\\&\int _{0}^{1}{{\frac {1}{t^{\alpha }}}dt}\ converge\ ssi\ \alpha <1\\&\\&{\underline {d{\acute {e}}m}}\ \\&{\begin{matrix}t\to {\frac {1}{t^{\alpha }}}\ \ C{}^{\circ }\ sur\ ]0,1]\\\forall x\in ]0,1],\ \int _{x}^{1}{\frac {dt}{t^{\alpha }}}=\left\{{\begin{aligned}&-\ln x\ si\ \alpha =1\\&{\frac {1-x^{-\alpha +1}}{-\alpha +1}}\ si\ \alpha \neq 1\\\end{aligned}}\right.\\\end{matrix}}\\\end{aligned}}$

${\begin{aligned}&x\to \int _{x}^{1}{{\frac {1}{t^{\alpha }}}dt}\ admet\ une\ \lim ite\ \left(x{\overset {>}{\mathop {\to } }}\,0\right)\ ssi\ \alpha <1\\&\\&{\underline {Th}}\ \ {\begin{matrix}\alpha \in R\\\left(a,b\right)\in R{}^{\text{2}}\ \ a<b\\\end{matrix}}\\&\int _{a}^{b}{{\frac {1}{\left(t-a\right)^{\alpha }}}dt}\ converge\ ssi\ \alpha <1\\&\\&{\underline {d{\acute {e}}m}}\ ana\log ue\\&{\underline {Th}}\int _{0}^{1}{\ln t}dt\ converge\\&\\&{\underline {d{\acute {e}}m}}\ \ln \ C{}^{\circ }\ sur\ ]0,1]\\&\forall x\in ]0,1],\ \int _{x}^{1}{\ln tdt}=\left(t\ln t-t\right)_{x}^{1}=-1-x\ln x+x\\&\int _{x}^{1}{\ln tdt}\to -1\ \left(x\to 0\right);\ \int _{0}^{1}{\ln t\ dt}\ converge\ \left({\begin{aligned}&et\ \int _{0}^{1}{\ln tdt}=-1\ \\&\left(savoir\ qu'elle\ converge,\ mais\ vers\ o{\grave {u}}\ n'est\ pas\ impor\tan t\right)\\\end{aligned}}\right)\\\end{aligned}}$

${\begin{aligned}&{\underline {\operatorname {Re} m}}:\ -\infty \leq a<b<+\infty \\&Soit\ f:\ ]a,b]\to K\ continue\ par\ morceaux\\&\int _{x}^{b}{f\left(t\right)dt}=-\int _{-x}^{-b}{f\left(-u\right)du}=\int _{-b}^{-x}{f\left(-u\right)du}\\&Le\ probl{\grave {e}}me\ de\ la\ nature\ de\ \int _{a}^{b}{f\left(t\right)dt}\ se\ ram{\grave {e}}ne\ {\grave {a}}\ celui\ de\ la\ nature\ de\ \int _{-b}^{-a}{f\left(-u\right)du}\ avec\ -\infty <-b<-a\leq +\infty \\&\left(cf\ 1^{er}\ cas\ paragraphe\ 1\right)\\&\\&Les\ th\ {\acute {e}}nonc{\acute {e}}es\ dans\ le\ cadre\ du\ paragraphe\ 1\\&s'{\acute {e}}tendent\ au\ cas\ du\ paragraphe\ 2\\&\left(sans\ {\bmod {i}}fication\ des\ in{\acute {e}}galit{\acute {e}}s\right)\\&\\&{\underline {Th}}\ \left(coh{\acute {e}}rence\ de\ la\ d{\acute {e}}finition\right)\\&-\infty <a<b<+\infty \\&\\&Soit\ f:]a,b]\to K\ continue\ par\ morceaux.\\&\\&si\ f\ admet\ un\ prolongement\ {\tilde {f}}\ continue\ par\ moreaux\ sur\ [a,b]\\&alors\ \int _{a}^{b}{f\left(t\right)dt}\ converge\ et\ \int _{a}^{b}{f\left(t\right)dt}=\int _{[a,b]}^{}{{\tilde {f}}\left(t\right)dt}\\\end{aligned}}$

3)Cas ]a,b[ où -inf < a < b < +inf

${\begin{aligned}&{\underline {Th\ et\ d{\acute {e}}f}}:\\&-\infty \leq a<b\leq +\infty \\&Soit\ f:\ ]a,b[\to K\ continue\ par\ morceaux\\&s'il\ existe\ c\in ]a,b[\ tel\ que\ \int _{a}^{c}{f\left(t\right)dt}\ et\ \int _{c}^{b}{f\left(t\right)dt}\ convergent\\&alors\ \forall d\in ]a,b[,\ \int _{a}^{d}{f\left(t\right)dt},\ \int _{d}^{b}{f\left(t\right)dt}\ convergent,\ \\&\int _{a}^{d}{f\left(t\right)dt}+\int _{d}^{b}{f\left(t\right)dt}=\int _{a}^{c}{f\left(t\right)dt}+\int _{c}^{b}{f\left(t\right)dt}\\&on\ dit\ que\ \int _{a}^{b}{f\left(t\right)dt}\ {\underline {\ converge}},\ et\ a\ pour\ valeur\\&\int _{a}^{c}{f\left(t\right)dt}+\int _{c}^{b}{f\left(t\right)dt}\\&\\&{\underline {d{\acute {e}}m}}:\\&c,d\in ]a,b[\\&\int _{x}^{d}{f\left(t\right)dt}=\int _{x}^{c}{f\left(t\right)dt}+\int _{c}^{d}{f\left(t\right)dt}\ \ \left(x\in ]a,b[\right)\\&\int _{a}^{d}{f\left(t\right)dt}\ converge\ ssi\ \int _{a}^{c}{f\left(t\right)dt}\ converge\\&En\ cas\ de\ convergence\ \int _{a}^{d}{f\left(t\right)dt}=\int _{a}^{c}{f\left(t\right)dt}+\int _{c}^{d}{f\left(t\right)dt}\\&\int _{d}^{y}{f\left(t\right)dt}=\int _{d}^{c}{f\left(t\right)dt}+\int _{c}^{y}{f\left(t\right)dt}\ \ \left(y\in ]a,b[\right)\\\end{aligned}}$

${\begin{aligned}&\int _{d}^{b}{f\left(t\right)dt}\ converge\ ssi\ \int _{c}^{b}{f\left(t\right)dt}\ converge\\&En\ cas\ de\ convergence\ \int _{d}^{b}{f\left(t\right)dt}=\int _{d}^{c}{f\left(t\right)dt}+\int _{c}^{b}{f\left(t\right)dt}\\&\\&\int _{a}^{d}{f\left(t\right)dt}+\int _{d}^{b}{f\left(t\right)dt}=\int _{a}^{c}{f\left(t\right)dt}+++\int _{c}^{b}{f\left(t\right)dt}\\&\\&ATTENTION\ !!!{\underline {Exemple\ \alpha \in R}}\\&\int _{0}^{+\infty }{{\frac {1}{t^{\alpha }}}dt}\\&\\&\int _{0}^{1}{{\frac {1}{t^{\alpha }}}dt}\ converge\ si\ \alpha <1\\&\\&\int _{1}^{+\infty }{{\frac {1}{t^{\alpha }}}dt}\ converge\ ssi\ \alpha >1\\&\\&\forall \alpha ,\ \int _{0}^{+\infty }{\frac {1}{t^{\alpha }}}dt\ diverge\\\end{aligned}}$

4) propriétés

${\begin{aligned}&{\underline {Th}}\ -\infty \leq a<b\leq +\infty \\&Soit\ I\ un\ \operatorname {int} ervalle\ d'extr{\acute {e}}mit{\acute {e}}s\ a,b\\&Les\ fonctions\ continues\ par\ morceaux\ sur\ I,\ {\grave {a}}\ valeurs\ dans\ K,\ dont\ l'\operatorname {int} {\acute {e}}grale\ sur\ I\ est\ convergente,\ forment\\&un\ K-espace\ vectoriel.\\&f\to \int _{a}^{b}{f\left(t\right)dt}\ est\ une\ forme\ lin{\acute {e}}aire\ sur\ cet\ espace\ vectoriel.\\&{\underline {d{\acute {e}}m}}\ \ -\infty \leq a<b<+\infty \ \ ;\ I=]a,b]\ ,\ par\ exemple.\\&E=\left\{f:I\to K\ continue\ par\ moreaux\right\}\\&F=\left\{f:I\to K\ continue\ par\ moreaux\ tels\ que\ \int _{a}^{b}{f\left(t\right)dt}\ converge\right\}\subset E\\&\\&0\in F\\&\forall \lambda \in K,\ \forall \left(f,g\right)\in F{}^{\text{2}},\ \forall x\in ]a,b]\\&\int _{x}^{b}{\left(\lambda f+g\right)\left(t\right)dt}=\lambda \int _{x}^{b}{f\left(t\right)dt}+\int _{x}^{b}{g\left(t\right)dt}\\&d'apr{\grave {e}}s\ les\ th\ g{\acute {e}}n{\acute {e}}raux\ sur\ les\ \lim ites,\ \int _{a}^{b}{\left(\lambda f+g\right)\left(t\right)dt}\ converge\\&\lambda f+g\in F\\&F\ est\ un\ sev\ de\ E.\\&\int _{a}^{b}{\left(\lambda f+g\right)\left(t\right)dt}=\lambda \int _{a}^{b}{f\left(t\right)dt}+\int _{a}^{b}{g\left(t\right)dt}\\\end{aligned}}$

${\begin{aligned}&{\underline {Th}}\ -\infty \leq a<b\leq +\infty \\&Soit\ I\ un\ \operatorname {int} ervalle\ d'extr{\acute {e}}mit{\acute {e}}s\ a,b\\&\\&f:I\to C\ continue\ par\ morceaux\ sur\ I\\&u=\operatorname {Re} f,\ v=\operatorname {Im} f\\&\int _{a}^{b}{f\left(t\right)dt}\ converge\ ssi\ \int _{a}^{b}{u\left(t\right)dt}\ et\ \int _{a}^{b}{v\left(t\right)dt}\ convergent.\\&\\&En\ cas\ de\ convergence,\\&\int _{a}^{b}{f\left(t\right)dt}=\int _{a}^{b}{u\left(t\right)dt}+i\int _{a}^{b}{v\left(t\right)dt}\\&\int _{a}^{b}{{\overline {f\left(t\right)}}dt}\ converge\ et\ \int _{a}^{b}{{\overline {f\left(t\right)}}dt}={\overline {\int _{a}^{b}{f\left(t\right)dt}}}\\&\\&{\underline {d{\acute {e}}m}}\ \ -\infty <a<b\leq +\infty ,\ I=[a,b[\ par\ exemple\\&\forall x\in [a,b[,\ \int _{a}^{x}{f\left(t\right)dt}=\int _{a}^{x}{u\left(t\right)dt}+i\int _{a}^{x}{v\left(t\right)dt}\\&\\&th\ g{\acute {e}}n{\acute {e}}raux\ relatifs\ aux\ fonctions\ composantes.\\&\int _{a}^{x}{{\overline {f\left(t\right)}}dt}={\overline {\int _{a}^{x}{f\left(t\right)dt}}}=\int _{a}^{x}{u\left(t\right)dt}-i\int _{a}^{x}{v\left(t\right)dt}\\\end{aligned}}$

${\begin{aligned}&{\underline {Conventions}}:\\&a\in R,\ si\ f\ est\ d{\acute {e}}finie\ au\ po\operatorname {int} \ a,\ on\ note\ \int _{a}^{a}{f\left(t\right)dt}=0\\&\\&-\infty \leq a<b\leq +\infty \\&f:]a,b[\to K\ continue\ par\ moreaux\ tels\ que\ \int _{a}^{b}{f\left(t\right)dt}\ converge\\&-\int _{a}^{b}{f\left(t\right)dt}\ est\ not{\acute {e}}e\ \int _{b}^{a}{f\left(t\right)dt}\\&\\&Th\ \left(\ relation\ de\ Chasles\right)\\&Soient\ a,b,c\in {\overline {R}}\ \ \alpha =\min \left(a,b,c\right),\ \beta =max\left(a,b,c\right)\\&Soit\ f:\ ]\alpha ,\beta [\to K\ continue\ par\ morceaux.\\&\\&Si\ \int _{\alpha }^{\beta }{f\left(t\right)dt}\ converge,\ alors\ \int _{a}^{c}{f\left(t\right)dt},\int _{c}^{b}{f\left(t\right)dt},\ \int _{a}^{b}{f\left(t\right)dt}\ convergent,\\&\int _{a}^{b}{f\left(t\right)dt}=\int _{a}^{c}{f\left(t\right)dt}+\int _{c}^{b}{f\left(t\right)dt}\\&{\underline {d{\acute {e}}m}}:\ -\infty \leq c\leq a\leq b\leq +\infty \ par\ exemple\\&Par\ hypoth{\grave {e}}se,\ \int _{c}^{b}{f\left(t\right)dt}\ converge\\&D'o{\grave {u}}\ \int _{c}^{a}{f\left(t\right)dt},\ \int _{a}^{b}{f\left(t\right)dt}\ convergent\\\end{aligned}}$

${\begin{aligned}&et\ \int _{c}^{b}{f\left(t\right)dt}=\int _{c}^{a}{f\left(t\right)dt}+\int _{a}^{b}{f\left(t\right)dt}\\&\int _{a}^{b}{f\left(t\right)dt}=-\int _{c}^{a}{f\left(t\right)dt}+\int _{c}^{b}{f\left(t\right)dt}\\&\\&{\underline {Extension}}:\ -\infty \leq a<b<c\leq +\infty \\&Soit\ f:]a,b[\cup ]b,c[\to K\ \ continue\ par\ morceaux\ sur\ ]a,b[\ et\ ]b,c[.\\&Si\ \int _{a}^{b}{f\left(t\right)dt}\ et\ \int _{b}^{c}{f\left(t\right)dt}\ convergent,\ alors\ par\ convention\\&on\ dit\ que\ \int _{a}^{c}{f\left(t\right)dt}\ converge\ et\ \int _{a}^{c}{f\left(t\right)dt}=\int _{a}^{b}{f\left(t\right)dt}+\int _{b}^{c}{f\left(t\right)dt}\\&\\&{\underline {Ex:}}\ f\left(t\right)=\left\{{\begin{aligned}&e^{-{\frac {1}{t{}^{\text{2}}}}}\ si\ -1\leq t<0\\&{\frac {1}{\sqrt {t}}}\ si\ 0<t\leq 1\\\end{aligned}}\right.\\&f\ continue\ par\ moreaux\ sur\ [-1,0[,\ sur\ ]0,1]\\&f\left(t\right)\to 0\ \left(t{\overset {<}{\mathop {\to } }}\,0\right)\\\end{aligned}}$

${\begin{aligned}&f\ se\ prolonge\ en\ une\ fonction\ continue\ sur\ [-1,0]\\&Donc\ \int _{-1}^{0}{f\left(t\right)dt}\ converge.\\&\\&{\frac {1}{2}}<1:\ \int _{0}^{1}{f\left(t\right)dt}=\int _{0}^{1}{{\frac {1}{\sqrt {t}}}dt}\ converge\\&D'o{\grave {u}}\ \int _{-1}^{1}{f\left(t\right)dt}\ converge\ \left(alors\ que\ f\left(t\right){\underset {t{\overset {>}{\mathop {\to } }}\,0}{\mathop {\to } }}\,0\right)\\&\\&{\underline {Th\ \left(\operatorname {int} {\acute {e}}gration\ par\ parties\right)}}\\&-\infty <a<b\leq +\infty \\&Soient\ u,v:\ [a,b[\to K\ continues\ sur\ [a,b[\ et\ de\ classe\ C^{1}\ par\ moreaux\ sur\ [a,b[\\&\forall x\in [a,b[,\ \int _{a}^{x}{u'\left(t\right)v\left(t\right)dt}=\left[u\left(t\right)v\left(t\right)\right]_{a}^{x}-\int _{a}^{x}{u\left(t\right)v'\left(t\right)dt}\\&si\ deux\ des\ \exp ressions\ ont\ une\ \lim ite\ dans\ K\ lorsque\ x{\overset {<}{\mathop {\to } }}\,b\\&alors\ la\ 3^{e}\ aussi\ et\\&\int _{a}^{b}{u'\left(t\right)v\left(t\right)dt}={\underset {x{\overset {<}{\mathop {\to } }}\,b}{\mathop {\lim } }}\,\left(u\left(x\right)v\left(x\right)\right)-u\left(a\right)v\left(a\right)-\int _{a}^{b}{u\left(t\right)v'\left(t\right)dt}\\&\\&{\underline {d{\acute {e}}m}}\ th\ g{\acute {e}}n{\acute {e}}raux\ sur\ les\ \lim ites\\\end{aligned}}$

${\begin{aligned}&ATTENTION\ a\in R,\ f:[a,+\infty [\to K\ continue\ par\ morceaux\\&{\frac {1}{t}}\to 0\ \left(t\to +\infty \right)\ mais\ \int _{1}^{+\infty }{{\frac {1}{t}}dt}\ diverge\\\end{aligned}}$

${\begin{aligned}&Si\ \int _{a}^{+\infty }{f\left(t\right)dt}\ converge\ et\ si\ \exists \ell \in {\overline {R}},\ f\left(t\right){\underset {\left(t\to +\infty \right)}{\mathop {\to } }}\,\ell \\&alors\ \ell =0\\&\\&{\underline {d{\acute {e}}m}}\\&si\ \ell \neq 0,\ on\ peut\ \sup poser\ \ell \in R*_{+}\cup \left\{+\infty \right\}\ \left(quitte\ {\grave {a}}\ changer\ f\ en\ -f\right)\\\end{aligned}}$

${\begin{aligned}&\exists k>0,\ \exists A\geq a,\ \forall x\in [A,+\infty [,\ f\left(x\right)\geq k\\&\\&\forall x\geq A,\ \int _{A}^{x}{f\left(t\right)dt}\geq \int _{A}^{x}{k}=k\left(x-a\right)\\&\int _{A}^{x}{f\left(t\right)dt}\to +\infty \ \left(x\to +\infty \right),\ or\ \int _{A}^{+\infty }{f\left(t\right)dt}\ converge\\&D'o{\grave {u}}\ \ell =0\\\end{aligned}}$

${\begin{aligned}&Si\ \int _{a}^{+\infty }{f\left(t\right)dt}\ converge,\ {\underline {ne\ pas\ en\ d{\acute {e}}duire}}\\&que\ f\left(t\right)\to 0\ \left(t\to +\infty \right)\\\end{aligned}}$

Fonctrions positives

1) th généraux

${\begin{aligned}&{\underline {Th}}\ -\infty <a<b\leq +\infty \\&\\&Soit\ f:[a,b[\to R\ continue\ par\ morceaux,\ positive\\&\\&Alors\ F:{\begin{matrix}[a,b[\to R\\x\to \int _{a}^{x}{f\left(t\right)dt}\\\end{matrix}}\ est\ croissante\\&\int _{a}^{b}{f\left(t\right)dt}\ converge\ ssi\ F\ est\ major{\acute {e}}e\\&Dans\ ce\ cas,\ \int _{a}^{b}{f\left(t\right)dt}={\underset {x{\overset {<}{\mathop {\to } }}\,b}{\mathop {\lim } }}\,F\left(x\right)={\underset {x\in [a,b[}{\mathop {Sup} }}\,\int _{a}^{x}{f\left(t\right)dt}\\&\\&{\underline {d{\acute {e}}m}}\\&a\leq x<y<b\ =>\ F\left(y\right)-F\left(x\right)=\int _{x}^{y}{f\left(t\right)dt}\geq 0\\&F\ croissante\ sur\ [a,b[\\&\int _{a}^{b}{f\left(t\right)dt}\ converge\ ssi\ F\ admet\ une\ \lim ite\ finie\ en\ b\\&\int _{a}^{b}{f\left(t\right)dt}\ converge\ ssi\ F\ est\ major{\acute {e}}e\ sur\ [a,b[\\&si\ F\ major{\acute {e}}e,\ {\underset {x{\overset {<}{\mathop {\to } }}\,b}{\mathop {\lim } }}\,F\left(x\right)={\underset {[a,b[}{\mathop {Sup} }}\,F\\\end{aligned}}$

${\begin{aligned}&{\underline {Th}}\ -\infty \leq a<b<+\infty \\&Soit\ f:]a,b]\to R\ continue\ par\ morceaux,\ positive.\\&Alors\ F:{\begin{matrix}]a,b]\to R\\x\to \int _{x}^{b}{f\left(t\right)dt}\\\end{matrix}}\ est\ d{\acute {e}}croissante\\\end{aligned}}$

${\begin{aligned}&\int _{a}^{b}{f\left(t\right)dt}\ converge\ ssi\ F\ est\ major{\acute {e}}e\ sur\ ]a,b]\\&\\&Da,s\ ce\ cas,\ \int _{a}^{b}{f\left(t\right)dt}={\underset {x{\overset {>}{\mathop {\to } }}\,a}{\mathop {\lim } }}\,F\left(x\right)={\underset {x\in ]a,b]}{\mathop {Sup} }}\,F\left(x\right)\\&\\&{\underline {d{\acute {e}}m}}:\\&a<x<y\leq b=>F\left(y\right)-F\left(x\right)=\int _{y}^{x}{f\left(t\right)dt}\leq 0\\\end{aligned}}$

${\begin{aligned}&F\ d{\acute {e}}croissante\\&th.\ \lim ite\ d'une\ fonction\ monotone\\\end{aligned}}$

${\begin{aligned}&{\underline {Cons{\acute {e}}quence}}:\\&Dans\ les\ th\ qui\ suivent,\ on\ pourra\ remplacer\ [a,b[\\&par\ un\ \operatorname {int} ervalle\ ]c,d]\ sans\ {\bmod {i}}fier\ les\ in{\acute {e}}galit{\acute {e}}s\ dans\ les\ conclusions.\\&\\&{\underline {Th}}\ -\infty <a<b\leq +\infty \\&\\&Si\ f:[a,b[\to R\ continue\ par\ morceaux,\ est\ positive\\&si\ \int _{a}^{b}{f\left(t\right)dt}\ converge,\\&alors\ \int _{a}^{b}{f\left(t\right)dt}\geq 0\\&\\&{\underline {d{\acute {e}}m}}\\&F\left(x\right)=\int _{a}^{x}{f}\geq 0\ \left(x\in [a,b[\right)\\&\int _{a}^{b}{f\left(t\right)dt}={\underset {{\overset {<}{\mathop {x\to b} }}\,}{\mathop {\lim } }}\,\\\end{aligned}}$

${\begin{aligned}&{\underline {Th}}\ -\infty <a<b\leq +\infty \\&Si\ f:[a,b[\to R\ est\ continue\ et\ positive,\\&Si\ \int _{a}^{b}{f\left(t\right)dt}\ converge,\ si\ \int _{a}^{b}{f\left(t\right)dt}=0\\&alors\ f=0\\&\\&{\underline {d{\acute {e}}m}}\\&\forall x\in [a,b[,\ F\left(x\right)=\int _{a}^{x}{f\left(t\right)dt}\geq 0\\&\forall x\in [a,b[,\ 0\leq F\left(x\right)\leq {\underset {[a,b[}{\mathop {Sup} }}\,F=\int _{a}^{b}{f\left(t\right)dt}=0\\&F=0\\&\forall t\in [a,b[\ ,\ f=F'=0\\&\\&{\underline {Th}}\ de\ comparaison\ des\ fonctions\ positives\\&-\infty <a<b\leq +\infty \\&Soient\ f,g:[a,b[\to R\ continues\ par\ morceaux,\ telles\ que\\&0\leq f\leq g\\&si\ \int _{a}^{b}{g\left(t\right)dt}\ converge,\ \int _{a}^{b}{f\left(t\right)dt}\ converge\ et\ \int _{a}^{b}{f}\leq \int _{a}^{b}{g}\\&Si\ \int _{a}^{b}{f\left(t\right)dt}\ diverge,\ \int _{a}^{b}{g\left(t\right)dt}\ diverge.\\\end{aligned}}$

${\begin{aligned}&{\underline {d{\acute {e}}m}}:\ \forall x\in [a,b[,\ F\left(x\right)=\int _{a}^{x}{f\left(t\right)dt},\ G\left(x\right)=\int _{a}^{x}{g\left(t\right)dt}\\&\forall x\in [a,b[,\ F\left(x\right)\leq G\left(x\right)\\&\sup posons\ que\ \int _{a}^{b}{g\left(t\right)dt}\ converge\\&\forall x\in [a,b[,\ G\left(x\right)\leq {\underset {[a,b[}{\mathop {Sup} }}\,G=\int _{a}^{b}{g\left(t\right)dt}\\&F\ major{\acute {e}}e\ sur\ [a,b[\\&\int _{a}^{b}{f\left(t\right)dt}\ converge\\&\int _{a}^{b}{f\left(t\right)dt}={\underset {[a,b[}{\mathop {Sup} }}\,F\leq \int _{a}^{b}{g\left(t\right)dt}\\&{\underline {Exemple}}\ \int _{-\infty }^{0}{e^{t}\sin {}^{\text{2}}tdt}\ converge\\&\\&f:t\to e^{t}\sin {}^{\text{2}}t\ continue\ positive\ sur\ R^{-}\\&\forall t\in R^{-},\ f\left(t\right)\leq e^{t}\\&\forall x,\ \int _{x}^{0}{e^{t}dt}=1-e^{x}\\&1-e^{x}\to 1\ \left(x\to -\infty \right)\\&\int _{-\infty }^{0}{e^{t}dt}\ converge\\&d'o{\grave {u}}\ \int _{-\infty }^{0}{f\left(t\right)dt}\ \ converge\\\end{aligned}}$

${\begin{aligned}&{\underline {Th}}\ -\infty <a<b\leq +\infty \\&Soietn\ f,g:[a,b[\to R\ continue\ par\ moreaux,\ positives\\&si\ f={}_{b}O\left(g\right)\\&alors\ la\ convergence\ de\ \int _{a}^{b}{g\left(t\right)dt}\ induit\ la\ convergence\ de\ \int _{a}^{b}{f\left(t\right)dt}\\&la\ divergence\ de\ \int _{a}^{b}{f\left(t\right)dt}\ induit\ la\ divergence\ de\ \int _{a}^{b}{g\left(t\right)dt}\\&{\underline {d{\acute {e}}m}}\ f=_{b}O\left(g\right)\\&\exists M>0,\ \exists c\in [a,b[,\ \forall t\in [c,b[,\ 0\leq f\left(t\right)\leq Mg\left(t\right)\\&th\ de\ comparaison:\ si\ \int _{c}^{b}{g\left(t\right)dt}\ converge,\ \int _{c}^{b}{f\left(t\right)dt}\ converge.\\&\\\end{aligned}}$

${\begin{aligned}&\\&si\ u\in \left\{f,g\right\}:\ \int _{c}^{b}{u\left(t\right)dt}\ converge\ ssi\ \int _{a}^{b}{u\left(t\right)dt}\ converge\\&\\&{\underline {Th}}\ \ -\infty <a<b\leq +\infty \\&Soient\ f,g:[a,b[\to R\ continues\ par\ morceaux,\ positives.\\&si\ f\left(t\right){\underset {b}{\mathop {\tilde {\ }} }}\,g\left(t\right)\\&alors\ \int _{a}^{b}{f\left(t\right)dt}\ et\ \int _{a}^{b}{g\left(t\right)dt}\ sont\ de\ m{\hat {e}}me\ nature\\&\\&{\underline {d{\acute {e}}m}}\ f\left(t\right)=_{b}O\left(g\left(t\right)\right),\ g\left(t\right)=_{b}O\left(f\left(t\right)\right)\\&cf\ th\ pr{\acute {e}}c{\acute {e}}dent\\\end{aligned}}$

${\begin{aligned}&{\underline {exemple}}\\&f\left(t\right)={\frac {\left|\ln t\right|}{t{}^{\text{2}}+1}}:\ nature\ de\ \int _{0}^{+\infty }{f\left(t\right)dt}\ ?\\&f\ C{}^{\circ }\ ,\ positive\ sur\ \ [0,+\infty [\\&f\left(t\right){\underset {0}{\mathop {\tilde {\ }} }}\,\left|\ln t\right|=-\ln t\\&\int _{0}^{1}{\ln tdt}\ converge\\&\int _{0}^{1}{f\left(t\right)dt}\ converge\ \left(th\ pr{\acute {e}}c{\acute {e}}dent\right)\\&f\left(t\right){\underset {+\infty }{\mathop {\tilde {\ }} }}\,{\frac {\ln t}{t{}^{\text{2}}}}\\\end{aligned}}$

${\begin{aligned}&t^{\frac {3}{2}}{\frac {\ln t}{t{}^{\text{2}}}}={\frac {\ln t}{\sqrt {t}}}\to 0\ \left(t\to +\infty \right)\\&{\frac {\ln t}{t{}^{\text{2}}}}{\underset {+\infty }{\mathop {=} }}\,o\left({\frac {1}{t^{\frac {3}{2}}}}\right)\\&\\&\int _{1}^{+\infty }{{\frac {1}{t^{\frac {3}{2}}}}dt}\ converge\ \left({\frac {3}{2}}>1\right)\\&th\ de\ comparaison\ des\ fonctions\ positives:\\&\int _{1}^{+\infty }{f\left(t\right)dt}\ converge\\&\int _{0}^{+\infty }{f\left(t\right)dt}\ converge\\\end{aligned}}$

gamma

${\begin{aligned}&{\underline {Th}}\ \Gamma :x\to \int _{0}^{+\infty }{t^{x-1}e^{-t}dt}\\&\Gamma \ est\ d{\acute {e}}finie\ sur\ ]0,+\infty [\\&\\&{\underline {d{\acute {e}}m}}:\\&Soit\ x\in R\\&\phi _{x}:]0,+\infty [\to R\\&t\to t^{x-1}e^{-t}\ \ \ est\ continue\ et\ {\underline {positive}}\ sur\ ]0,+\infty [\\&\\&t{}^{\text{2}}\phi _{x}\left(t\right)=t^{x+1}e^{-t}\to 0\ \left(t\to +\infty \right)\\&\phi _{x}\left(t\right){\underset {+\infty }{\mathop {=} }}\,o\left({\frac {1}{t{}^{\text{2}}}}\right)\\&\\&\int _{1}^{+\infty }{{\frac {1}{t{}^{\text{2}}}}dt}\ converge\ \left(2>1\right)\\&d'o{\grave {u}}\ \int _{1}^{+\infty }{\phi _{x}\left(t\right)dt}\ converge\ \ \left(x\in R\right)\\&\phi _{x}\left(t\right){\underset {0}{\mathop {\tilde {\ }} }}\,t^{x-1}={\frac {1}{t^{1-x}}}\\&\int _{0}^{1}{{\frac {1}{t^{1-x}}}dt}\ converge\ ssi\ 1-x<1\\\end{aligned}}$

${\begin{aligned}&\int _{0}^{1}{\phi _{x}\left(t\right)}\ converge\ ssi\ x>0\\&\Gamma \left(x\right)\ existe\ ssi\ x>0\\\end{aligned}}$

bertrand

${\begin{aligned}&{\underline {\Pr op}}\ a>1\ \ \left(\alpha ,\beta \right)\in R{}^{\text{2}}\\&\int _{a}^{+\infty }{{\frac {1}{t^{\alpha }\left(\ln t\right)^{\beta }}}dt}\ converge\ ssi\ \left\{{\begin{aligned}&\alpha >1\ et\ \beta \ r{\acute {e}}el\ quelconque\\&ou\\&\alpha =1\ et\ \beta >1\\\end{aligned}}\right.\\&\\&{\underline {d{\acute {e}}m}}:\\&f_{\alpha ,\beta }:t\to {\frac {1}{t^{\alpha }\left(\ln t\right)^{\beta }}}\ d{\acute {e}}f,\ C{}^{\circ }\ sur\ \ [a,+\infty [\\&\\&\forall \beta \in R,\ t^{\gamma }f_{\alpha ,\beta }\left(t\right)={\frac {t^{\gamma -\alpha }}{\left(\ln t\right)^{\beta }}}{\underset {t\to +\infty }{\mathop {\to } }}\,\left\{{\begin{aligned}&0\ si\ \gamma -\alpha <0\\&\\&+\infty \ si\ \gamma -\alpha >0\\\end{aligned}}\right.\\&\\&*{\underline {\alpha >1,\ \beta \in R}}:\\&choisir\ \gamma \ tel\ que\ 1<\gamma <\alpha \\&\left\{{\begin{aligned}&f_{\alpha ,\beta }\left(t\right){\underset {+\infty }{\mathop {=} }}\,o\left({\frac {1}{t^{\gamma }}}\right)\\&\int _{1}^{+\infty }{{\frac {1}{t^{\gamma }}}dt}\ converge\\\end{aligned}}\right.:\ \int _{a}^{+\infty }{f_{\alpha ,\beta }\left(t\right)dt}\ converge\\\end{aligned}}$

${\begin{aligned}&*{\underline {\alpha <1}}:\\&choisir\ \gamma \ tel\ que\ \alpha <\gamma \leq 1\\&\left\{{\begin{aligned}&{\frac {1}{t^{\gamma }}}{\underset {+\infty }{\mathop {=} }}\,o\left(f_{\alpha ,\beta }\left(t\right)\right)\\&\int _{1}^{+\infty }{{\frac {1}{t^{\gamma }}}dt}\ diverge\\\end{aligned}}\right.\ \ \ \ \int _{a}^{+\infty }{{\frac {1}{t^{\alpha }\left(\ln t\right)^{\beta }}}dt}\ diverge\\&\\&*{\underline {\alpha =1}}\\&\int _{a}^{x}{{\frac {1}{t\left(\ln t\right)^{\beta }}}dt}=\left\{{\begin{aligned}&\ln \left(\ln x\right)-\ln \left(\ln a\right)\ \ \ si\ \beta =1\\&{\frac {\left(\ln x\right)^{-\beta +1}-\left(\ln a\right)^{-b+1}}{-\beta +1}}\ \ \ si\ \beta \neq 1\\\end{aligned}}\right.\\&\ln \left(\ln x\right)\to +\infty \ \left(x\to +\infty \right)\\&\left(\ln x\right)^{1-\beta }{\underset {x\to +\infty }{\mathop {\to } }}\,\left\{{\begin{aligned}&+\infty \ \ si\ \beta <1\\&0\ \ si\ \beta >1\\\end{aligned}}\right.\\&\int _{a}^{+\infty }{{\frac {1}{t\left(\ln t\right)^{\beta }}}dt}\ converge\ ssi\ \beta >1\\\end{aligned}}$

ch13p17