En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : Suites d'intégrales 2Intégration en mathématiques/Exercices/Suites d'intégrales 2 », n'a pu être restituée correctement ci-dessus.
Pour
a
,
n
∈
N
{\displaystyle a,n\in \mathbb {N} }
I
(
a
,
n
)
=
∫
0
1
x
n
(
1
−
x
)
a
d
x
{\displaystyle I(a,n)=\int _{0}^{1}x^{n}(1-x)^{a}\,\mathrm {d} x}
1° En intégrant par parties, montrer que :
I
(
a
+
1
,
n
)
=
a
+
1
n
+
1
I
(
a
,
n
+
1
)
{\displaystyle I(a+1,n)={\frac {a+1}{n+1}}I(a,n+1)}
2° Établir que :
I
(
a
,
n
)
−
I
(
a
,
n
+
1
)
=
I
(
a
+
1
,
n
)
{\displaystyle I(a,n)-I(a,n+1)=I(a+1,n)}
En déduire que :
I
(
a
,
n
+
1
)
=
n
+
1
n
+
a
+
2
I
(
a
,
n
)
{\displaystyle I(a,n+1)={\frac {n+1}{n+a+2}}I(a,n)}
3° L'entier
a
{\displaystyle a}
n
{\displaystyle n}
∀
n
∈
N
I
(
a
,
n
)
=
1
×
2
×
3
×
⋯
×
(
n
−
1
)
×
n
(
a
+
1
)
×
(
a
+
2
)
×
⋯
×
(
a
+
n
+
1
)
{\displaystyle \forall n\in \mathbb {N} \quad I(a,n)={\frac {1\times 2\times 3\times \cdots \times (n-1)\times n}{(a+1)\times (a+2)\times \cdots \times (a+n+1)}}}
Solution
Wikipedia-logo-v2.svg
(
n
+
1
)
I
(
a
+
1
,
n
)
=
∫
0
1
(
n
+
1
)
x
n
(
1
−
x
)
a
+
1
d
x
=
[
x
n
+
1
(
1
−
x
)
a
+
1
]
0
1
+
(
a
+
1
)
∫
0
1
x
n
+
1
(
1
−
x
)
a
d
x
=
0
+
(
a
+
1
)
I
(
a
,
n
+
1
)
{\displaystyle (n+1)I(a+1,n)=\int _{0}^{1}(n+1)x^{n}(1-x)^{a+1}\,\mathrm {d} x=\left[x^{n+1}(1-x)^{a+1}\right]_{0}^{1}+(a+1)\int _{0}^{1}x^{n+1}(1-x)^{a}\,\mathrm {d} x=0+(a+1)I(a,n+1)}
I
(
a
,
n
)
−
I
(
a
,
n
+
1
)
=
∫
0
1
(
x
n
−
x
n
+
1
)
(
1
−
x
)
a
d
x
=
I
(
a
+
1
,
n
)
{\displaystyle I(a,n)-I(a,n+1)=\int _{0}^{1}\left(x^{n}-x^{n+1}\right)(1-x)^{a}\,\mathrm {d} x=I(a+1,n)}
Grâce à la question 1, on en déduit :
I
(
a
,
n
)
=
(
1
+
a
+
1
n
+
1
)
I
(
a
,
n
+
1
)
=
a
+
n
+
2
n
+
1
I
(
a
,
n
+
1
)
{\displaystyle I(a,n)=\left(1+{\frac {a+1}{n+1}}\right)I(a,n+1)={\frac {a+n+2}{n+1}}I(a,n+1)}
I
(
a
,
0
)
=
∫
0
1
(
1
−
x
)
a
d
x
{\displaystyle I(a,0)=\int _{0}^{1}(1-x)^{a}\,\mathrm {d} x}
1
a
+
1
{\displaystyle {\frac {1}{a+1}}}
1° Soient
p
∈
Q
+
{\displaystyle p\in \mathbb {Q} _{+}}
q
∈
Q
{\displaystyle q\in \mathbb {Q} }
x
∈
[
0
,
1
[
{\displaystyle x\in \left[0,1\right[}
f
p
,
q
(
x
)
=
∫
0
x
t
p
(
1
−
t
)
q
d
t
{\displaystyle f_{p,q}(x)=\int _{0}^{x}t^{p}(1-t)^{q}\,\mathrm {d} t}
Justifier cette notation.
Déterminer la fonction dérivée de
f
p
,
q
{\displaystyle f_{p,q}}
En se limitant à
p
⩾
1
{\displaystyle p\geqslant 1}
(
a
,
b
,
c
)
∈
R
3
{\displaystyle (a,b,c)\in \mathbb {R} ^{3}}
(
p
,
q
)
{\displaystyle (p,q)}
∀
x
∈
[
0
,
1
[
a
f
p
,
q
(
x
)
+
b
f
p
−
1
,
q
−
1
(
x
)
=
x
p
(
1
−
x
)
q
(
c
x
−
q
)
{\displaystyle \forall x\in \left[0,1\right[\quad af_{p,q}(x)+bf_{p-1,q-1}(x)=x^{p}(1-x)^{q}(cx-q)}
On distinguera les cas
q
=
0
{\displaystyle q=0}
q
≠
0
{\displaystyle q\neq 0}
{
a
=
(
p
+
q
)
(
1
+
p
+
q
)
b
=
−
p
q
c
=
p
+
q
.
{\displaystyle {\begin{cases}a=(p+q)(1+p+q)\\b=-pq\\c=p+q.\end{cases}}}
2° Pour
x
∈
[
0
,
1
[
{\displaystyle x\in [0,1[}
r
∈
Q
+
{\displaystyle r\in \mathbb {Q} _{+}}
F
r
(
x
)
=
∫
0
x
t
r
(
1
−
t
)
r
+
3
d
t
{\displaystyle F_{r}(x)=\int _{0}^{x}{\frac {t^{r}}{(1-t)^{r+3}}}\,\mathrm {d} t}
dans laquelle n'intervient aucun signe d'intégration.
(On mettra la fonction
F
r
{\displaystyle F_{r}}
f
p
−
1
,
q
−
1
{\displaystyle f_{p-1,q-1}}
Solution
Wikipedia-logo-v2.svg
La fonction
t
↦
t
p
(
1
−
t
)
q
{\displaystyle t\mapsto t^{p}(1-t)^{q}}
[
0
,
1
[
{\displaystyle \left[0,1\right[}
[
0
,
x
]
{\displaystyle \left[0,x\right]}
x
∈
[
0
,
1
[
{\displaystyle x\in \left[0,1\right[}
f
p
,
q
{\displaystyle f_{p,q}}
Les deux fonctions à égaler coïncident toujours en
0
{\displaystyle 0}
]
0
,
1
[
{\displaystyle \left]0,1\right[}
x
p
−
1
(
1
−
x
)
q
(
c
x
−
q
)
+
q
x
p
(
1
−
x
)
q
−
1
{\displaystyle x^{p-1}(1-x)^{q}(cx-q)+qx^{p}(1-x)^{q-1}}
a
x
(
1
−
x
)
+
b
=
p
(
1
−
x
)
(
c
x
−
q
)
−
q
x
(
c
x
−
q
)
+
c
x
(
1
−
x
)
{\displaystyle ax(1-x)+b=p(1-x)(cx-q)-qx(cx-q)+cx(1-x)}
a
=
c
(
p
+
q
+
1
)
,
a
=
p
(
c
+
q
)
+
q
2
+
c
,
b
=
−
p
q
{\displaystyle a=c(p+q+1),\quad a=p(c+q)+q^{2}+c,\quad b=-pq}
a
=
c
(
p
+
q
+
1
)
,
(
p
+
q
−
c
)
q
=
0
,
b
=
−
p
q
{\displaystyle a=c(p+q+1),\quad (p+q-c)q=0,\quad b=-pq}
si
q
≠
0
{\displaystyle q\neq 0}
(
a
,
b
,
c
)
{\displaystyle (a,b,c)}
si
q
=
0
{\displaystyle q=0}
c
(
p
+
1
,
0
,
1
)
{\displaystyle c(p+1,0,1)}
c
∈
R
{\displaystyle c\in \mathbb {R} }
c
=
p
{\displaystyle c=p}
F
r
=
f
r
,
−
r
−
3
=
f
p
−
1
,
q
−
1
{\displaystyle F_{r}=f_{r,-r-3}=f_{p-1,q-1}}
(
p
,
q
)
=
(
r
+
1
,
−
r
−
2
)
{\displaystyle (p,q)=(r+1,-r-2)}
0
×
f
r
+
1
,
−
r
−
2
(
x
)
+
(
r
+
1
)
(
r
+
2
)
F
r
(
x
)
=
x
r
+
1
(
1
−
x
)
−
r
−
2
(
−
x
+
r
+
2
)
{\displaystyle 0\times f_{r+1,-r-2}(x)+(r+1)(r+2)F_{r}(x)=x^{r+1}(1-x)^{-r-2}(-x+r+2)}
F
r
(
x
)
=
(
r
+
2
−
x
)
x
r
+
1
(
r
+
1
)
(
r
+
2
)
(
1
−
x
)
r
+
2
{\displaystyle F_{r}(x)={\frac {(r+2-x)\,x^{r+1}}{(r+1)(r+2)(1-x)^{r+2}}}}
On pose, pour
h
{\displaystyle h}
k
{\displaystyle k}
H
h
,
k
=
∫
0
1
x
h
ln
k
x
d
x
{\displaystyle H_{h,k}=\int _{0}^{1}x^{h}\ln ^{k}x\,\mathrm {d} x}
1° Calculer
H
h
,
0
{\displaystyle H_{h,0}}
2° Justifier l'existence de
H
h
,
k
{\displaystyle H_{h,k}}
h
>
0
{\displaystyle h>0}
h
=
0
{\displaystyle h=0}
k
>
0
{\displaystyle k>0}
3° Prouver que si
k
>
0
{\displaystyle k>0}
H
h
,
k
=
−
k
h
+
1
H
h
,
k
−
1
{\displaystyle H_{h,k}=-{\frac {k}{h+1}}H_{h,k-1}}
4° En déduire
H
h
,
k
{\displaystyle H_{h,k}}
Solution
H
h
,
0
=
1
h
+
1
{\displaystyle H_{h,0}={\frac {1}{h+1}}}
La fonction
x
↦
x
h
ln
k
x
{\displaystyle x\mapsto x^{h}\ln ^{k}x}
]
0
,
1
]
{\displaystyle \left]0,1\right]}
0
{\displaystyle 0}
h
>
0
{\displaystyle h>0}
lim
x
→
0
x
(
ln
x
)
k
=
0
{\displaystyle \lim _{x\to 0}x\left(\ln x\right)^{k}=0}
Fonction logarithme/Croissances comparées#Comparaison entre ln(x) et x en 0⁺ ).
H
h
,
k
=
1
h
+
1
[
x
h
+
1
ln
k
x
]
0
1
−
k
h
+
1
∫
0
1
x
h
+
1
ln
k
−
1
x
x
d
x
=
−
k
h
+
1
H
h
,
k
−
1
{\displaystyle H_{h,k}={\frac {1}{h+1}}\left[x^{h+1}\ln ^{k}x\right]_{0}^{1}-{\frac {k}{h+1}}\int _{0}^{1}x^{h+1}{\frac {\ln ^{k-1}x}{x}}\,\mathrm {d} x=-{\frac {k}{h+1}}H_{h,k-1}}
h
=
0
{\displaystyle h=0}
H
h
,
k
=
(
−
1
)
k
k
!
(
h
+
1
)
k
+
1
{\displaystyle H_{h,k}={\frac {(-1)^{k}\,k!}{(h+1)^{k+1}}}}
h
=
0
{\displaystyle h=0}
Soit
n
{\displaystyle n}
k
⩽
n
{\displaystyle k\leqslant n}
I
k
,
n
=
∫
0
1
(
n
k
)
x
k
(
1
−
x
)
n
−
k
d
x
{\displaystyle I_{k,n}=\int _{0}^{1}{n \choose k}x^{k}(1-x)^{n-k}\,\mathrm {d} x}
Pour
0
⩽
k
<
n
{\displaystyle 0\leqslant k<n}
I
k
,
n
{\displaystyle I_{k,n}}
I
k
+
1
,
n
{\displaystyle I_{k+1,n}}
En déduire
I
k
,
n
{\displaystyle I_{k,n}}
n
{\displaystyle n}
Solution
En intégrant par parties, on obtient :
∫
0
1
(
n
k
)
x
k
(
1
−
x
)
n
−
k
d
x
=
[
(
n
k
)
x
k
+
1
k
+
1
(
1
−
x
)
n
−
k
]
0
1
+
∫
0
1
(
n
k
+
1
)
x
k
+
1
(
1
−
x
)
n
−
(
k
+
1
)
d
x
{\displaystyle \int _{0}^{1}{n \choose k}x^{k}(1-x)^{n-k}\,\mathrm {d} x=\left[{n \choose k}{\frac {x^{k+1}}{k+1}}(1-x)^{n-k}\right]_{0}^{1}+\int _{0}^{1}{n \choose {k+1}}x^{k+1}(1-x)^{n-(k+1)}\,\mathrm {d} x}
ce qui se traduit par :
I
k
,
n
=
I
k
+
1
,
n
{\displaystyle I_{k,n}=I_{k+1,n}}
On a donc :
I
k
,
n
=
I
n
,
n
=
∫
0
1
(
n
n
)
x
n
(
1
−
x
)
n
−
n
d
x
=
∫
0
1
x
n
d
x
=
[
x
n
+
1
n
+
1
]
0
1
=
1
n
+
1
{\displaystyle I_{k,n}=I_{n,n}=\int _{0}^{1}{n \choose n}x^{n}(1-x)^{n-n}\,\mathrm {d} x=\int _{0}^{1}x^{n}\,\mathrm {d} x=\left[{\frac {x^{n+1}}{n+1}}\right]_{0}^{1}={\frac {1}{n+1}}}
![{\displaystyle (n+1)I(a+1,n)=\int _{0}^{1}(n+1)x^{n}(1-x)^{a+1}\,\mathrm {d} x=\left[x^{n+1}(1-x)^{a+1}\right]_{0}^{1}+(a+1)\int _{0}^{1}x^{n+1}(1-x)^{a}\,\mathrm {d} x=0+(a+1)I(a,n+1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0ad480ceddb9841c5aad3b444e87d0b1be71b8ef)
![{\displaystyle \left[0,x\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c3267511c133b4e330dd893690d32e04773a5170)
![{\displaystyle \left]0,1\right[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9a5471595b2e7b5c1ecc10ea08df3ea3fa5d14a7)
![{\displaystyle I_{n}=-\underbrace {\left[\operatorname {e} ^{-t}\sin ^{2n}t\right]_{0}^{+\infty }} _{0}+2n\int _{0}^{+\infty }\operatorname {e} ^{-t}\sin ^{2n-1}t\cos t\,\mathrm {d} t=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bfc5720d80bb8ddcf60420ca9799be49e09db5b2)
![{\displaystyle -2n\underbrace {\left[\operatorname {e} ^{-t}\sin ^{2n-1}t\cos t\right]_{0}^{+\infty }} _{0}+2n\int _{0}^{+\infty }\operatorname {e} ^{-t}\left((2n-1)\sin ^{2n-2}t\cos ^{2}t-\sin ^{2n}t\right)\,\mathrm {d} t\,\mathrm {d} t=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e568459d64eeae35d0eafecf80e4ea784bb7b4d2)
![{\displaystyle I_{n}=\int _{0}^{1}{\frac {\mathrm {d} x}{(1+x^{n}){\sqrt[{n}]{1+x^{n}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00983bedf69ab7076321124a4b0c51f9038adcc1)
![{\displaystyle \int _{0}^{1}{\frac {\mathrm {d} x}{\sqrt[{n}]{1+x^{n}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/020964c12c3068865ee850c463945d2500d342ba)
![{\displaystyle I_{1}=\int _{0}^{1}{\frac {\mathrm {d} x}{(1+x)^{2}}}=\left[{\frac {-1}{1+x}}\right]_{0}^{1}={\frac {1}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4877a75bd22bf459de5fae9b9ccd52d0ac42c8ef)
![{\displaystyle \int _{0}^{1}{\frac {\mathrm {d} x}{\sqrt[{n}]{1+x^{n}}}}=\left[{\frac {x}{\sqrt[{n}]{1+x^{n}}}}\right]_{0}^{1}+\int _{0}^{1}{\frac {x^{n}\,\mathrm {d} x}{(1+x^{n}){\sqrt[{n}]{1+x^{n}}}}}={\frac {1}{\sqrt[{n}]{2}}}+\int _{0}^{1}{\frac {\left(1+x^{n}\right)\,\mathrm {d} x}{(1+x^{n}){\sqrt[{n}]{1+x^{n}}}}}-I_{n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9d3dc34b589cd612ed996c170020aae5150caa45)
![{\displaystyle I_{n}={\frac {1}{\sqrt[{n}]{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/77ad7f944e4abc0888b0155d33f96b9e31da1d41)
![{\displaystyle \left]0,1\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/091455b6e9ffc33569e2fd6f11ca6e7a4dc074f4)
![{\displaystyle H_{h,k}={\frac {1}{h+1}}\left[x^{h+1}\ln ^{k}x\right]_{0}^{1}-{\frac {k}{h+1}}\int _{0}^{1}x^{h+1}{\frac {\ln ^{k-1}x}{x}}\,\mathrm {d} x=-{\frac {k}{h+1}}H_{h,k-1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eea8f7e22b88ac48ea72c75dcd2ac04b6fb8e13f)
![{\displaystyle I_{n}(h)=\left[f_{n-1}\right]_{-n}^{-n+h}=(h-1)\operatorname {e} ^{-n+h}+\operatorname {e} ^{-n}=\operatorname {e} ^{-n}\left((h-1)\operatorname {e} ^{h}+1\right)=\operatorname {e} ^{-n}I_{0}(h)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a8e651f18560d5b73e1d696d4f93c6f372dac5b4)
![{\displaystyle \int _{0}^{1}{n \choose k}x^{k}(1-x)^{n-k}\,\mathrm {d} x=\left[{n \choose k}{\frac {x^{k+1}}{k+1}}(1-x)^{n-k}\right]_{0}^{1}+\int _{0}^{1}{n \choose {k+1}}x^{k+1}(1-x)^{n-(k+1)}\,\mathrm {d} x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a7d310b973a9908a428134311bf1ea8c5eb6096e)
![{\displaystyle I_{k,n}=I_{n,n}=\int _{0}^{1}{n \choose n}x^{n}(1-x)^{n-n}\,\mathrm {d} x=\int _{0}^{1}x^{n}\,\mathrm {d} x=\left[{\frac {x^{n+1}}{n+1}}\right]_{0}^{1}={\frac {1}{n+1}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f2fcced7a4a2f127fa08ede05912acdb723f62d3)
