1)1) Cas [a,b[ où -inf < a < b < +inf
Nature de
∫
a
b
f
(
t
)
d
t
{\displaystyle \int _{a}^{b}f(t){\rm {d}}t}
Début d’un théorème
Théorème
Soit
α
∈
R
{\displaystyle \alpha \in \mathbb {R} }
∫
1
+
∞
1
t
α
d
t
{\displaystyle \int _{1}^{+\infty }{\frac {1}{t^{\alpha }}}{\rm {d}}t}
α
>
1
{\displaystyle \alpha >1}
Pour
α
>
1
,
∫
1
+
∞
1
t
α
d
t
=
1
α
−
1
{\displaystyle \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
→
1
t
α
{\displaystyle t\to {\frac {1}{t^{\alpha }}}}
C
0
{\displaystyle {\mathcal {C}}^{0}}
[
1
,
+
∞
[
{\displaystyle [1,+\infty [}
1
≤
x
:
∫
1
x
1
t
α
d
t
=
{
ln
x
s
i
α
=
1
x
−
α
+
1
−
1
−
α
+
1
s
i
n
o
n
{\displaystyle 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
→
+
∞
(
x
→
+
∞
)
{\displaystyle \ln x\to +\infty (x\to +\infty )}
x
1
−
α
→
{
0
s
i
α
>
1
+
∞
s
i
α
<
1
x
→
+
∞
{\displaystyle x^{1-\alpha }\to {\begin{cases}0{\rm {~si~}}\alpha >1\\+\infty {\rm {~si~}}\alpha <1\\\end{cases}}x\to +\infty }
Pour
α
>
1
,
∫
1
+
∞
1
t
α
d
t
=
lim
x
→
+
∞
1
−
x
1
−
α
α
−
1
=
1
α
−
1
{\displaystyle \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
α
∈
R
{\displaystyle \alpha \in \mathbb {R} }
∫
0
+
∞
e
−
α
t
d
t
{\displaystyle \int _{0}^{+\infty }e^{-\alpha t}{\rm {d}}t}
α
>
0
{\displaystyle \alpha >0}
Dans ce cas,
∫
0
+
∞
e
−
α
t
d
t
=
1
α
{\displaystyle \int _{0}^{+\infty }e^{-\alpha t}{\rm {d}}t={\frac {1}{\alpha }}}
Fin du théorème
Début d'une démonstration
Démonstration
t
→
e
−
α
t
∈
C
0
(
R
+
)
{\displaystyle t\to e^{-\alpha t}\in {\mathcal {C}}^{0}(\mathbb {R} ^{+})}
∀
x
∈
R
+
,
∫
0
x
e
−
α
t
d
t
=
{
x
s
i
α
=
0
e
−
α
x
−
1
−
α
s
i
n
o
n
{\displaystyle \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
→
∫
0
x
e
−
α
t
d
t
{\displaystyle x\to \int _{0}^{x}e^{-\alpha t}{\rm {d}}t}
+
∞
{\displaystyle +\infty }
α
>
0
{\displaystyle \alpha >0}
Pour
α
>
0
,
∫
0
+
∞
e
−
α
t
d
t
=
lim
x
→
+
∞
e
−
α
x
−
1
−
α
=
1
α
{\displaystyle \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>
d
e
´
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→
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−
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C
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s
o
i
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F
:
[
a
,
b
]
→
K
x
→
∫
a
x
f
~
(
t
)
d
t
{\displaystyle {\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}}}
f
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(
x
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(
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x
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(
t
)
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t
=
∫
a
x
f
~
(
t
)
d
t
=
F
(
x
)
→
F
(
b
)
(
x
→
<
b
)
∫
a
b
f
(
t
)
d
t
c
o
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e
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∫
a
b
f
(
t
)
d
t
=
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(
b
)
=
∫
[
a
,
b
]
f
~
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x
e
m
p
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_
∫
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π
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s
int
t
d
t
c
o
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E
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e
f
f
e
t
,
t
→
sin
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t
c
o
n
t
i
n
u
e
s
u
r
[
−
π
,
0
[
,
sin
t
t
→
1
(
t
→
<
0
)
{\displaystyle {\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
S
o
i
t
f
:
]
a
,
b
]
→
K
c
o
n
t
i
n
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e
p
a
r
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o
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a
u
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.
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→
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(
t
)
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t
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_
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∫
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C
∘
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0
,
1
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∀
x
∈
]
0
,
1
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,
∫
x
1
d
t
t
α
=
{
−
ln
x
s
i
α
=
1
1
−
x
−
α
+
1
−
α
+
1
s
i
α
≠
1
{\displaystyle {\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}}}
x
→
∫
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(
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)
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x
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1
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d
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(
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;
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{\displaystyle {\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}}}
Re
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{\displaystyle {\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}}}
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{\displaystyle {\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}}}
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{\displaystyle {\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}}}
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{\displaystyle {\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}}}
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{\displaystyle {\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}}}
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{\displaystyle {\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}}}
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{\displaystyle {\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}}}
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{\displaystyle {\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}}}
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{\displaystyle {\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}}}
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{\displaystyle {\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}}}
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{\displaystyle {\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}}}
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{\displaystyle {\begin{aligned}&F\ d{\acute {e}}croissante\\&th.\ \lim ite\ d'une\ fonction\ monotone\\\end{aligned}}}
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{\displaystyle {\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}}}
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{\displaystyle {\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}}}
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{\displaystyle {\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}}}
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{\displaystyle {\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}}}
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{\displaystyle {\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}}}
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{\displaystyle {\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}}}
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{\displaystyle {\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}}}
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b2faf120a116a7ee20d774fb64d66b1331d34c27)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2f2e263b100fd54420d4e1a8ec64f15a72f0982e)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e0548acf89df8907578c86b58a51fe1f936f83d5)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aa260651efcb8c2ef74fc48f79901442c0939267)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f0e8a72a9d07e9773a718da200cea54f8c7b03b1)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a93614f1f0cabd398e25a7e1133ff6aae19a1e2b)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8e224ec6041f4bde9fdaaa71213fecf4238cc8f8)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/39fb4a3690696a8c7cfa8bd0be8328bebffbb4ef)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5c6f95cc4f9b208a8858c578bc24856ec5c2a3f0)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0cd79ef38124ac6b701a250a45316f8814a97edf)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/69d7f0ca6d604a01c0885a2477e580d307f7a021)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8b9aa4fc30c9e42e494c8d266ecf3812dec8f043)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b68069a3acd327ba4e24ebd64b1a412c06876a53)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7dbbf05c561bf3722b6d3b604e2aa99c44c1337b)
