correction

2014 −2015
◦2
F(x) = Z2π
0
ln(1 + xcos t)
cos tdt.
F(x)|x|>1 ] −1,1[
1−1
x∈[−1,1] F(x)=2Zπ
0
ln(1 + xcos t)
cos tdt
x∈[−1,1] F(x) = 2 Zπ/2
0
ln(1 + xcos t)−ln(1 −xcos t)
cos tdt
F
−1<u<1
ln(1 + u)
u
≤1
1− |u|
x∈]−1,1[ |f(x, t)| ≤ |x|
1−|x|F]−1,1[
FC1]−1,1[ F0
t∈[0, π] cos t=1−tan2(t/2)
1+tan2(t/2)
u= tan(t/2) F0(x) = 2π
√1−x2
F(0) F(x)x∈]−1,1[
x > 1t π x < −1
0 (1 + xcos t)<0
f(x, t) = ln(1+xcos t)
cos tF
x∈]−1,1[ t→f(x, t) [0,2π] cos(t) = 0 π/2 3π/2
ln(1 + u)∼u u →0t→f(x, t)π/2 3π/2
x F
x= 1 π/2 3π/2
1 + cos t= 0 t=π−∞
πcos t∼ −1 + (t−π)2/2 ln(1 + cos t)∼
2 ln |t−π|f(1, t)∼2 ln |t−π|R2π
0ln |t−π|dt
Rπ
0ln s ds ln s s(ln s−1) 0 0
F(1)
x=−1 1 −cos t= 0 t= 0 t= 2π
cos t t = 0 1 −cos t∼t2/2
f(−1, t)∼2 ln tRπ
0ln t dt
F(−1)
x∈[−1,1] F(x) = Rπ
0f(x, t)dt +R2π πf(x, t)dt
u= 2π−tcos(2π−u) = cos(u)R2π
πf(x, t)dt =−R0 πf(x, u)du =
Rπ
0f(x, t)dt

Rπ
0f(x, t)dt =Rπ/2 0f(x, t)dt +Rπ π/2f(x, t)dt t0=π−t
cos(π−t0) = −cos(t0)
F(−x) = −F(x)F
y→ln(1 + y)u0
ln(1 + u)
u
≤
sup
y∈[0,u]
1
1 + y
y→1/(1 + y)−|u|
u=xcos t]−1,1[ x∈]−1,1[
ln(1 + xcos t)
xcos t
≤1
1− |xcos t||cos t| ≤ 1
a∈[0,1[ x∈[−a, a]|f(x, t)| ≤ a/(1 −a)t∈[0, π]
f x
F[−a, a]
a∈[0,1[ F]−1,1[
∂f
∂x (x, t) = 1 1+xcos t]−1,1[×[0, π]x∈[−a, a] 0 < a < 1
∂f
∂x (x, t)
≤1
1−aRπ
0
1
1−adt < ∞f x ∈[−a, a]
FC1
[−a, a] 0 < a < 1FC1]−1,1[
t∈R1−tan2(t/2)
1+tan2(t/2) =cos2(t/2)−sin2(t/2) cos2(t)+sin2(t)= cos(2 ∗t/2) = cos(t)
F0(x)=2Rπ
0
1+tan2(t/2)
1+x+(1−x) tan2(t/2) dt u = tan(t/2) du =
1
2
1
1+tan2(t/2) dt =1
2
1
1+u2dt F 0(x)=4R∞
0
1
1+x+(1−x)u2du =4
1+xR∞
0
1
1+ 1−x
1+xu2du v =
uq1−x
1+xF0(x) = 4
√1−x2R∞
0
1
1+v2dv =2π
√1−x2
F(0) = 0
F F 002π
√1−x2
2πarcsin(x) + C F (0) = 0 F(x) = 2πarcsin(x)x∈]−1,1[
(E) ln(x)y0(x) + y(x)
x= 1.
x→ln(ln(x))
(EH)
(E)
(E)
g(x) = x
ln(x+1) x∈]−1,0[∪]0,+∞[g(0) = 1 g
C1]−1,+∞[
(E) ]0,∞[
(E)y0(x)+(xln x)−1y(x) = (ln x)−1
]0,1[ ]1,∞[ (E)
∂
∂x ln(ln(x)) = (xln x)−1x > 1
(EH) (xln x)−1
(EH)y(x) = Cexp(−ln(ln x)) = C(ln x)−1x > 1x∈]0,1[
y(x) = Cexp(−ln |ln x|) = C0(ln x)−1(EH)
x∈]0,1[7→ C(ln x)−1x∈]1,∞[7→ C(ln x)−1C∈R
(E)y0(x) =
C(x) (ln x)−1C0(x) (ln x)−1= (ln x)−1C0(x) = 1 C(x) = x
(E)x∈]0,1[7→ (x+C)(ln x)−1
x∈]1,∞[7→ (x+C)(ln x)−1C∈R

ln(1 + x)∼x x →0g(x)∼xx−1→1x→0g
0
g0 0
g(x)−1
x=x−ln(x+1)
x22 ln(x+ 1) = x−x2/2 + o(x2)x→0
x−ln(x+1)
x2∼x2/2 x2→1/2x→0g0gC1
]−1,0[∪]0,+∞[C1
gC1]−1,∞[
x→x−1 (x−1)(ln x)−1
C1]0,∞[ (E) ]0,∞[
(x−1)(ln x)−1