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TSI2
no2
t7→ t
(t+ 2)(t2+ 1) [0,+∞[ +∞
t
(t+ 2)(t2+ 1) ∼
t→+∞
t
t×t2∼
t→+∞
1
t2
Z+∞
1
1
t2tZ+∞
1
t
(t+ 2)(t2+ 1) t
Z+∞
0
t
(t+ 2)(t2+ 1) t
α, β, γ
t∈[0,+∞[
t
(t+ 2)(t2+ 1) =α t+βt +γ t2+ 1 =α(t2+ 1) + (βt +γ)(t+ 2) (t+ 2)(t2+ 1)
=(α+β)t2+ (2β+γ)t+ (α+ 2γ)
(t+ 2)(t2+ 1)
α+β= 0
2β+γ= 1
α+ 2γ= 0 
β=−α
γ=−α/2
−2α−α/2 = −5α/2=1
α=−2
5β=2 5γ=1 5x > 0
Zx
0
t
(t+ 2)(t2+ 1) t=−2 5Zx
0
t
t+ 2 +1 5Zx
0
2t t
t2+ 1 +1 5Zx
0
t
t2+ 1
=−2
5ln(x+ 2) + 2 5ln 2+1 5ln(x2+ 1) + 1 5arctan x
=1
5ln x2+ 1 (x+ 2)2+1 5arctan x+2 5ln 2
x+∞
Z+∞
0
t
(t+ 2)(t2+ 1) t=π 10 +2 ln 2 5
t= tan u t = (1 + tan2u)u u = arctan t
t7→ arctan tC1[0,+∞[ [0, π/2[
Zπ/2
0
sin u
2 cos u+ sin uu=Zπ/2
0
sin u
cos u
1
2 + sin u
cos u
u
=Zπ/2
0
tan u×1
2 + tan u×(1 + tan2u)u
1 + tan2u
=
t=tan uZ+∞
0
t
(t+ 2)(t2+ 1) t
Zπ/2
0
sin u
2 cos u+ sin uu=π 10 +2 ln 2 5

a b 0< a < b
h:t7→ e−at −e−bt
t]0,+∞[a<b ⇒e−at > e−bt
?0
h(t) = e−at −e−bt
t=1−at +o(t)−(1 −bt +o(t)) t=(b−a)t+o(t) t=b−a+o(1)
lim
t→0h(t) = b−a
h0Z1
0
e−at −e−bt
tt
? t >1
06e−at −e−bt
t6e−at t6e−at
Z+∞
1
e−at tZ+∞
1
e−at −e−bt
tt
Z+∞
0
e−at −e−bt
tt
(x, y)∈R20< x < y
u=at u =bt
Zy
x
e−at −e−bt
tt=Zy
x
e−at
tt−Zy
x
e−bt
tt
Zy
x
e−at −e−bt
tt=Zay
ax
e−u
u/a
u
a−Zby
bx
e−u
u/b
u
b
=Zay
ax
e−u
uu−Zby
bx
e−u
uu
Zy
x
e−at −e−bt
tt="Zbx
ax
e−u
uu+Zay
bx
e−u
uu#−"Zay
bx
e−u
uu−Zby
ay
e−u
uu#
Zy
x
e−at −e−bt
tt=Zbx
ax
e−t
tt−Zby
ay
e−t
tt.
z > 0t∈[az, bz]
e−bz
t6e−t t6e−az t
e−bz Zbz
az
t
t6Zbz
az
e−t
tt6e−az Zbz
az
t
t
Zbz
az
t
t= ln(bz)−ln(az) = ln b a
e−bz ln b
a6Zbz
az
e−t
tt6e−az ln b a.

TSI2
x, y ∈]0,+∞[x<y
e−bx ln b
a−e−ay ln b a6Zy
x
e−at −e−bt
tt=Zbx
ax
e−t
tt−Zby
ay
e−t
tt6e−ax ln b a−e−by ln b a
y→+∞
∀x∈]0,+∞[, e−bx ln b
a6Z+∞
x
e−at −e−bt
tt6e−ax ln b a
lim
x→0e−bx ln b
a= lim
x→0e−ax ln b
a= ln b a
Z+∞
0
e−at −e−bt
tt= ln b a.
y→+∞x→0
ζ(2)
I0, I1J0
I0=Zπ/2
0
t=π
2;I1=Zπ/2
0
cos t t = [sin t]π/2
0= 1; J0=Zπ/2
0
t2t=t3
3π/2
0
=π3
24
J1u(t) = t2v0(t) = cos t
u0(t)=2t v(t) = sin t
Zπ/2
0
t2cos t t =t2sin tπ/2
0−Zπ/2
0
2tsin t t
u(t)=2t v0(t) = sin t
u0(t)=2 v(t) = −cos t
J1=π2
4− [−2tcos t]π/2
0+ 2 Zπ/2
0
cos t t!=π2
4−2 [sin t]π/2
0=π2
4−2
I0=π
2;I1= 1; J0=π3 24 ;J1=π2 4−2.
t7→ cosnt n 0,π
2Zπ/2
0
cosnt t > 0
∀n∈NIn>0.
n∈NIn−2
u(t) = cosn+1 t v0(t) = cos t
u0(t) = −(n+ 1) sin tcosnt v(t) = sin t
In+2 =sin tcosn+1 tπ/2
0+ (n+ 1) Zπ/2
0
sin2tcosnt t

cos2x+ sin2x= 1
In+2 = (n+ 1) Zπ/2
0
(1 −cos2t) cosnt t = (n+ 1)In−(n+ 1)In+2
(n+ 2)In+2 = (n+ 1)In
In+2 =n+ 1
n+ 2In.
t∈[0, π/2] h(t) = π
2sin t−t
h0(t) = π
2cos t−1h00(t) = −π 2sin t < 0∀t∈0,π 2
h00,π
2h0(0) = π 2−1>0h0(π/2) = −1<0
α= arccos 2
π
h(0) = 0 h[0, α]h>0
h[α, π/2] h(π/2) = 0 h>0
h[0, π/2] tπ
2sin t>t
∀t∈[0, π/2],06t6π
2sin(t).
t∈0,π
206t26π2
4sin2(t) = π2 4(1 −cos2t) cosnt>0
06t2cosnt6π2
4(cosnt−cosn+2 t)
0π/2
06Zπ/2
0
t2cosnt6Zπ/2
0
π2
4(cosnt−cosn+2 t)
06Jn6π2
4(In−In+2).
In>0
06Jn
In
6π2
41−In+2 In
06Jn
In
6π2
41−n+ 1 n+ 2=π2 4×1 n+ 2
π2
4×1 n+ 2 −→
n→+∞0
lim
n→+∞
In
Jn
= 0
In+2 JnJn+2
u(t) = cosn+2 t v0(t) = 1
u0(t) = −(n+ 2) sin tcosn+1 t v(t) = t
In+2 =Zπ/2
0
cosn+2 t t =tcosn+2 tπ/2
0+ (n+ 2) Zπ/2
0
tsin tcosn+1 t t

TSI2
u(t) = sin tcosn+1 t v0(t) = t
u0(t) = cosn+2 t−(n+ 1) sin2tcosnt v(t) = t2
2
In+2 = (n+ 2) ht2
2sin tcosn+1 tiπ/2
0−1
2Zπ/2
0
t2cosn+2 t t +n+ 1
2Zπ/2
0
t2sin2tcosnt t!
= (n+ 2) −1
2Jn+2 +n+ 1 2Zπ/2
0
t2(1 −cos2t) cosnt t!
JnJn+2
In+2 = (n+ 2) −1
2Jn+2 +n+ 1 2(Jn−Jn+2)
=n+ 2
2(n+ 1)Jn−(n+ 2)Jn+2
In+2 =(n+ 1)(n+ 2)Jn−(n+ 2)2Jn+2
2.
In
In+2
In
=n+ 1
n+ 2 =(n+ 1)(n+ 2) 2
Jn
In
−(n+ 2)2
2
Jn+2
In
=(n+ 1)(n+ 2)
2
Jn
In
−(n+ 2)2
2
n+ 1
(n+ 2)In+2
Jn+2
n+ 1
n+ 2 =(n+ 1)(n+ 2) 2Jn In
−Jn+2
In+2 
(n+ 1)(n+ 2)
Jn
In
−Jn+2
In+2
=2
(n+ 2)2.
n= 2k
J2k
I2k
−J2k+2
I2k+2
=2
(2k+ 2)2=1 2(k+ 1)2.
k0n−1
n−1
X
k=0 J2k
I2k
−J2k+2
I2k+2 =J0 I0
−J2n
I2n
=1
2
n−1
X
k=0
1
(k+ 1)2=1 2
n
X
k=1
1
k2=1 2Sn
lim
n→+∞
J2n
I2n
= 0 (Sn)n∈N∗
ζ(2) = lim
n→+∞Sn= lim n→+∞2J0
I0
−J2n
I2n= 2 ×J0 I0
=π2
6
ζ(2) = π2
6