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a, b ∈R0< c < b−a
2
f:x7→ (1|x| ≤ c
0
g(t) = Zb
a
f(t−x) dx
2x+ 4x2= 3x+ 3x2
x∈R
x∈]0 ; 1[ ∪]1 ; +∞[
f(x) = Zx2
x
dt
ln t
f(x)x∈]0 ; 1[ ∪]1 ; +∞[
x > 1
Zx2
x
xdt
tln t≤f(x)≤Zx2
x
x2dt
tln t
f1+
f1−
fC1]0 ; 1[ ]1 ; +∞[
f0(x)
fC1
C∞]0 ; +∞[
R∗
f:x7→ Z2x
x
et
tdt
f
fC1R
f:R→RC1a > 0
I(x) = 1
xa+1 Zx
0
taf(t) dt
I(x)x
f∈ C([0 ; 1],R)
R1
0tnf(t) dt
R1
0tndt!n≥0
In=Z1
0
xn+1
ln(1 −x)dx

g
−c≤t−x≤c⇐⇒ t−c≤x≤t+c
t≤a−c t ≥b+c g(t) = 0
a−c≤t≤a+c g(t) = Rt+c
a1 dx=t+c−a
a+c≤t≤b−c g(t) = Rt+c
t−c1 dt= 2c
b−c≤t≤b+c g(t) = Rb
t−c1 dx=b−t+c
g
4x2−3x2= 3x−2x
3x−2x=Z1
0
x(2 + t)x−1dt4x2−3x2=Z1
0
x2(3 + t)x2−1dt
Z1
0
ϕ(t) dt= 0
ϕ
ϕ(t) = x(x(3 + t)x2−1−(2 + t)x−1)
x≤0x≥1ϕ
x∈]0 ; 1[
x(3 + t)x2−1≤(2 + t)x−1
ln x+ (x2−1) ln(3 + t)≤(x−1) ln(2 + t)
f:x7→ ln x+ (x2−1) ln(3 + t)−(x−1) ln(2 + t) ]0 ; 1]
f
f0(x) = 1
x+ 2xln(3 + t)−ln(2 + t)
x≥1/2f0(x)≥1
x+ ln(3 + t)−ln(2 + t)≥0
x≤1/2f0(x)≥2−ln(2 + t)≥2−ln 3 ≥0
f0(x)≥0f
f(1) = 0 f
Z1
0
ϕ(t) dt= 0
ϕ
ϕ
x≤0x= 0
x > 0ϕ
∀t∈[0 ; 1],1
x+ (x2−1) ln(3 + t)−(x−1) ln(2 + t)=0
t
∀t∈[0 ; 1],x2−1
3 + t=x−1 2 + t
x= 1
x= 0 x= 1
S={0,1}
x
x x2

x > 1t∈[x;x2]x≤t≤x2ln t > 0
Zx2
x
xdt
tln t≤f(x)≤Zx2
x
x2dt
tln t
Zx2
x
dt
tln t= [ln |ln t|]x2
x= ln 2
f(x)−−−−→
x→1+ln 2
x < 1x2≤xln t < 0
x2ln 2 ≤f(x)≤xln 2
f(x)−−−−→
x→1−ln 2
H t 7→ 1/ln t]0 ; 1[ ]1 ; +∞[
f(x) = H(x2)−H(x)f
C1]0 ; 1[ ]1 ; +∞[
f0(x) = x−1
ln x
fC1
f f
C1]0 ; +∞[
fC∞]0 ; 1[ ]1 ; +∞[
fln x
x−1
x= 1 + h
1
f0(x)=1 hln(1 + h) =
+∞
X
n=0
(−1)nhn
n+ 1
|h|<1
1
f0(x)C∞
f0(x)C∞
x→0+
∀t∈[x; 2x],ex≤et≤e2x
exln 2 ≤f(x)≤e2xln 2
f(x)→ln 2
x→0−f(x)→ln 2 f
f(0) = ln 2
F t 7→ et/t R∗
+FC1
f(x) = F(2x)−F(x)fC1
f0(x) = e2x−ex
x
R∗
−f0(x)−−−→
x→01
fC1Rf0(0) = 1
I(x)−f(0)
a+ 1 =1 xa+1 Zx
0
taf(t) dt−Zx
0
taf(0) dt=1
xa+1 Zx
0
ta(f(t)−f(0)) dt
ε > 0α > 0
|x| ≤ α=⇒ |f(x)−f(0)| ≤ ε
|x| ≤ α t x |f(t)−f(0)| ≤ ε
1
xa+1 Zx
0
ta(f(t)−f(0)) dt ≤ε
lim
x→0I(x) = f(0)
a+ 1

R1
0tnf(t) dt
R1
0tndt=Z1
0
(n+ 1)tnf(t) dt
u=tn+1
R1
0tnf(t) dt
R1
0tndt=Z1
0
f(u1/(n+1)) du
kfk∞
R1
0tnf(t) dt
R1
0tndt→f(1)
f: ]0 ; 1[ →R
f(x) = −ln(1 −x)
x=
+∞
X
n=0
xn
n+ 1
0
a, b ∈]0 ; 1[
−(n+ 1)In=An+Bn+Cn
An=Za
0
(n+ 1) xn
f(x)dx, Bn=Zb
a
(n+ 1) xn
f(x)dx Cn=Z1
b
(n+ 1) xn
f(x)dx
f
0≤An≤Za
0
(n+ 1)xn
f(0) =an+1
a= 1 −εnεn=ln n
n→0
ln(n)an+1 = eln(ln n)+(n+1) ln(1−εn)→0
ln(ln n)+(n+ 1) ln(1 −εn)∼ − ln n→ −∞
An= o 1
ln n
f
0≤Cn≤Z1
b
(n+ 1)xn
f(b)dx=1−bn+1 f(b)
b= 1 −ηnηn=1
n(ln n)→0
bn+1 →1f(b)∼ln n
Cn∼o1
ln n
f
bn+1 −an+1
f(b)≤Bn≤bn+1 −an+1 f(a)
bn+1 −an+1 →1f(b)∼f(a)∼ln n
−(n+ 1)In∼1
ln n
In∼ − 1
nln n
t=−ln(1 −x)x= 1 −e−t
In=−Z+∞
0
(1 −e−t)n+1
te−tdt
In=Z+∞
0
n+1
X
k=0
(−1)kn+ 1
k−e−(k+1)t tdt
0
n+1
X
k=0
(−1)kn+ 1
k= 0

0
In=
n+1
X
k=0
(−1)kn+ 1
kZ+∞
0
e−t−e−(k+1)t
tdt
Z+∞
0
e−t−e−(k+1)t
tdt= lim
ε→0Z+∞
ε
e−t−e−(k+1)t
tdt= ln(k+ 1)
n+1
X
k=0
(−1)kn+ 1
kln(k+ 1) ∼ − 1 nln n