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I n = 10
x2n+1 ln x1 − x2
dx
I n
limn→+∞
I n = 0
I n+1 − I n limn→+∞
nk=1
1
k2
f [0,π]
1
n
nπ0 f
un
sin(u) du =
2
n
nk=1
(−1)k f ckn
∀k ∈ {1, . . . , n} k ∈ [(k −
1) π,kπ]
limn→+∞
1
n
nk=1
(−1)k f ckn
= 0 lim
n→+∞
1
n
nπ0 f
un
sin(u) du = 0
limn→+∞
π0 f (t)sin(nt) dt = 0
limn→+∞
π
2
0
f (t)sin(nt) dt = 0
(α,β ) ∈ R2 π
0
αt + βt2
cos(nt) dt =
1
n2
π0
αt + βt2
dt
cos t + cos (2t) + . . . + cos (nt)
= 1
2
sin (2n+1)t2
sin t2− 1
sin p − sin q = 2 cos p+q2 sin p−q2
π2
0
sin (2n + 1) t
sin t dt =
π
2
nk=1
π0
αt + βt2
cos(nt) dt
limn→+∞
nk=1
π0
αt + βt2
cos(nt) dt = −
1
2
π0
αt + βt2
dt
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limn→+∞
nk=1
1
k2 =
π2
6 = −4
10
x ln x
1 − x2dx
∀x ∈ ]0,1[ f n (x)
= x2n+1 ln x
1 − x2
f n ]0,1[ 10
x2n+1 ln x
1 − x2 dx
f n
limx→0
x ln x = 0 limx→0
x2nx ln x
1 − x2 = 0
ln x ln x = ln(1 − (1 − x)) = − (1
− x) + o (1 − x)
ln x
1 − x =
− (1 − x) + (1 − x) ε (1 − x)
1 − x = −1 + ε (1 − x) lim
x→1
ln x
1 − x = −1
f n (x) = x2n+1 ln x
(1 + x) (1 − x) limx→1
f n (x) = −1
2
f n gn : gn (x) = 0
x = 0−12 x = 1
f n (x) x ∈ ]0,1[
gn 10 f n (x) dx =
10 gn (x) dx
hn ∀x ∈ [0,1] hn (x)
=
0 x = 0−12 x = 1
x ln x
1 − x2 x ∈ ]0,1[
hn [0,1] M
∀x ∈ [0,1] hn (x) M
0 10 |f n (x)| dx = 10
x2n |hn (x)| dx 10 x2nM dx =
M x2n+1
2n + 11
0
= M
2n + 1 →n→+∞0
limn→+∞
10 |f n (x)| dx = 0
limn→+∞
10 f n (x) dx = 0 = limn→+∞
I n
I n+1 − I n = 10
x2n+1+2 ln x
1 − x2 dx −
10
x2n+1 ln x
1 − x2 dx =
10
x2n+1+2 − x2n+1
ln x
1 − x2 dx
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I n+1 − I n = 10
x2n+1
x2 − 1
ln x
1 − x2 dx = −
10 x
2n+1 ln xdx
I n+1 − I n = − 10 x2n+1
ln xdx = − x2n+22n + 2 ln x1
0+ 10 x2n+2(2n + 2) x dx = x
2n+2
(2n + 2)210
I n+1 − I n = 1
(2n + 2)2 =
1
4 (n + 1)2
nk=1
1
k2 =
nk=1
I k − I k−1 = 4 (I n −
I 0)
limn→+∞
I n = 0, limn→+∞
nk=1
1
k2 = −4I 0
+∞k=1
1
k2
1
n
nπ0 f
un
sin(u) du =
1
n
nk=1
kπ(k−1)π f
un
sin(u) du
u → sin u [(k − 1) π,kπ]
1n
nπ0 f
un
sin(u) du =
1n
nk=1
f ckn
kπ(k−1)π sin (u) du ck ∈
[(k − 1) π,kπ]
1
n
nπ0 f
un
sin(u) du =
1
n
nk=1
f ckn
[cos u]kπ(k−1)π =
2
n
nk=1
(−1)k f ckn
f [0,π] f
[0,π] |f | M
ε > 0
∃N > 0 ∀ (x,y) ∈ [0,π]2 |x − y|
1
N π ⇒ |f (x) − f (y)| ε
f ckn
− f
kπn
ε 1n nk=1 (−1)k f ckn
−
1
n
nk=1
(−1)k f kπn
ε
1
n
nk=1
(−1)k f ckn
1
n
nk=1
(−1)k f kπn
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1n nk=1 (−1)k f kπn
1n[n2
]−1k=1
f 2k+2n
π
− f 2k+1n
π + 1
n
nk=2[n
2]
f kπn
π
n N
1n nk=1 (−1)k f kπn
1n[n2 ]−1k=1
ε + 1
n
nk=2[n
2]
M ε + M
n
n 1
n ε
1n nk=1 (−1)k f kπn
ε lim
n→+∞
1
n
nk=1
(−1)k f ckn
= lim
n→+∞
1
n
nk=1
(−1)k f kπn
= 0
u = nt π0 f (t)sin(nt)
dt =
1
n
π0 f
un
sin(u) du lim
n→+∞
π0 f (t)sin(nt) dt = 0
u = nt π2
0 f (t)sin(nt) dt = 1
n
nπ2
0 f un
sin(u) du =
1
n
[n2
]π0 f
un
sin(u) du +
1
n
nπ2
[n2 ]π f un
sin(u) du
f 0,π2 M
π20 f (t)sin(nt) dt 1n
[n2 ]π0 f un sin(u) du
+ 1n nπ2[n2 ]π M du
n
2
−1 [n
2]π
0f u
n
sin(u) du
→
n→+∞0
+ M π
2n
limn→+∞
π2
0 f (t)sin(nt) dt = 0
π0
αt + βt2
cos(nt) dt =
αt + βt2
sin(nt)n
π0
=0
− π0 (α + 2βt)
sin(nt)
n dt
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π0
αt + βt2
cos(nt) dt = −
− (α + 2βt)
cos(nt)
n2
π0
− π0 2β
cos(nt)
n2 dt
π0αt + βt2 cos(nt) dt = (α +
2βπ) (−1)n
n2 − α
n2 − 2β sin(nt)
n3π0
=0
= (α + 2βπ) (−1)n
n2 − α
n2
π0
αt + βt2
cos(nt) dt =
1
n2
α = −1α + 2βπ = 0
⇐⇒
α = −1
β = 1
2π
π0 αt + βt2 dt = α t2
2 + β
t3
3 π
0 = α
π2
2 + β
π3
3
α = −1
β = 1
2π
π0
−t +
t2
2π
dt = −
π2
2 +
π3
6π = −
3π2
6 +
π2
6 = −
π2
3
π0
−t +
t2
2π
dt = −
π2
3
cos t + cos (2t) + . . . + cos (nt) =n
k=1cos(kt) = R n
k=1eikt = Reit n−1
k=0eikt = Reiteint − 1
eit − 1
cos t + cos (2t) + . . . + cos (nt) = R
eite
int
2
eint
2 − e−int
2
eit
2
eit
2 − e−it
2
= Re it2 (n+1) sin nt2
sin t2
=
cos (n+1)t2 sin nt2
sin t2
sin p − sin q = 2 cos p+q2
sin
p−q2
p = nt + t2+
q =
t2 cos
(n+1)t2 sin
nt2 =
12
sin
nt + t2
− sin t2
cos t + cos (2t) + . . . + cos (nt) =
sin
nt + t2
− sin t2
2sin t2
= 1
2
sin (2n+1)t2
sin t2− 1
cos t + cos (2t) + . . . + cos (nt)
= 1
2
sin (2n+1)t2
sin t2− 1
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1
2
π0
1
sin (2n+1)t2sin t2
− 1
dt =
π0 [cos t + cos (2t) + . . . + cos
(nt)] dt =
nk=1
π0 cos(kt) dt =
nk=1
− sin(kt)
k
π0
=0
∀k>0
12
π0
1 sin
(2n+1)t2
sin t2− 1
dt = 0 = 1
2
π0
sin (2n+1)t2sin t2
dt − 12
π0 dt =
12
π0
sin (2n+1)t2sin t2
dt − π
2
π0
sin (2n+1)t
2
sin t2dt = π
u = t2 π0
sin (2n+1)t2sin t2
dt = π2
0 2sin(2n + 1) u
sin u du
π2
0
sin(2n + 1) u
sin u du =
π
2
nk=1
π0
αt + βt2
cos(nt) dt =
π0
αt + βt2
nk=1
cos(nt)
dt =
π0
αt + βt2
12
sin (2n+1)t2
sin t2− 1
dt
u = t2n
k=1
π0
αt + βt2
cos(nt) dt =
π2
0
2αu + 4βu2
sin(2n + 1) usin u
du − 1
2
π0
αt + βt2
dt
nk=1
π0
αt + βt2
cos(nt) dt =
π2
0 2
αu + 2βu2 sin (2n + 1) u
sin u du −
1
2
π0
αt + βt2
dt
f (t) = αu
+ 2βu2
sin u
sin u sin u = u + o (u)
limt→0
f (t) = α
f π2
0 2
αu + 2βu2 sin(2n + 1) u
sin u du =
π2
0 2αu + 2βu2
sin u sin((2n + 1) u) du = 2
π2
0 f (u)sin((2n + 1) u) du
limn→+∞
π2
0 f (u)sin((2n + 1) u) du = 0
n
k=1 π
0
αt + βt2
cos(nt) dt =
π2
02
αu + 2βu2
sin(2n + 1) u
sin u du
→n→+∞0−
1
2
π
0
αt + βt2
dt
limn→+∞
nk=1
π0
αt + βt2
cos(nt) dt = −
1
2
π0
αt + βt2
dt
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α = −1
β = 1
2π
π
0 −t +
t2
2πcos(nt) dt =
1
n2
nk=1
π0
−t +
t2
2π
cos(nt) dt =
nk=1
1
k2
limn→+∞
nk=1
π0
−t +
t2
2π
cos(nt) dt = lim
n→+∞
nk=1
1
k2 =
1
2
π0
−t +
t2
2π
dt =
π2
6
limn→+∞
nk=1
1
k2 = −4I 0 = −4
10
x2n+1 ln x
1 − x2 dx
limn→+∞
nk=1
1
k2 =
π2
6 = −4
10
x ln x
1 − x2dx
