integrale generalisee[1]

TSI2
+∞
X
k=1
1
k2=π2 6I=Z1
0
ln t
t−1t
k∈NIk=Z1
0
tkln(t)t Ik
n
X
k=0
Ik=Z1
0
tn+1 ln(t)
t−1t−I
tln(t)
t−1t0t1
M > 0t∈]0,1[ 
tln(t)
t−16M
lim
n→+∞Z1
0
tn+1 ln(t)
t−1t= 0 I=
+∞
X
k=0
1
(k+ 1)2
H
x > 0Z+∞
0
e−t
x+tt
f:]0,+∞[→Rx∈]0,+∞[f(x) = Z+∞
0
e−t
x+tt
∀x∈]0,+∞[f(x)>Z1
0
e−1
x+tt f(x)−→
x→0+ +∞
∀x∈]0,+∞[ 0 6f(x)61
xf(x)−→
x→+∞0
Z+∞
0
te−tt
∀x∈]0,+∞[,f(x)−1
x61 x2Z+∞
0
te−tt.
f(x)∼
x→+∞
1
x
f
f
(x, h)∈]0,+∞[×R∗h>−x
2
Z+∞
0
e−t
(x+t)2t
∀t∈[0,+∞[,
1
h1 x+h+t−1 x+t+1 (x+t)262|h| x3
f(x+h)−f(x)
h+Z+∞
0
e−t
(x+t)2t62|h| x3

f]0,+∞[∀x∈]0,+∞[f0(x) = −Z+∞
0
e−t
(x+t)2t
x∈]0,+∞[ (ε, A)∈]0,1] ×[1,+∞[
ZA
ε
e−t
(x+t)2t=−e−A x+A+e−ε x+ε−ZA
ε
e−t
x+tt.
∀x∈]0,+∞[, f0(x) = −1
x+f(x)
f f C2]0,+∞[∀x∈]0,+∞[, f00(x) = 1
x2+f0(x)
g
g:]0,+∞[→Rx > 0g(x) = e−xf(x)
g]0,+∞[∀x∈]0,+∞[, g0(x) = −e−x
x
x∈]0,+∞[Z+∞
x
e−u
uu
∀x∈]0,+∞[, g(x) = Z+∞
x
e−u
uu,
∀x∈]0,+∞[, f(x) = exZ+∞
x
e−u
uu
Z+∞
x
e−u
uu∼
x→+∞
e−x
x
H
a b 0< a < b
Z+∞
0
e−at −e−bt
tt
(x, y)∈R20< x < y
Zy
x
e−at −e−bt
tt=Zbx
ax
e−t
tt−Zby
ay
e−t
tt.
z > 0
e−bz ln b
a6Zbz
az
e−t
tt6e−az ln b a.
Z+∞
0
e−at −e−bt
tt= ln b a.
H
Z+∞
0
t
(t+ 2)(t2+ 1) t
Zπ/2
0
sin u
2 cos u+ sin uu
t= tan u

TSI2
H
B(u, v) = Z1
0
tu−1(1 −t)v−1t
(u, v)∈R2D
B1
2,1 2
∀(u, v)∈D, B(u, v) = B(v, u)
∀(u, v)∈D, B(u, v) = B(u+ 1, v) + B(u, v + 1)
∀(u, v)∈D, B(u+ 1, v) = u
u+vB(u, v)
B(n+ 1, p)B(1, p) (n, p)∈N×N∗
n
B(n, p) (n, p)∈N∗×N∗
H
a0 1
ln 1
x∼
x→11−x u = 1 −x
p q −1Z1
a
xpln 1
xq
x
α r lim
x→+∞
(ln x)r
xα
0< a < 1r s 1
Z+∞
1
a
(ln x)r
xsx
s= 1 + 2α(ln x)r
xs=o1 x1+α+∞
p q −1
I(p, q) = Z1
0
xpln 1
xq
x
p q −1
I(p, q) = Z+∞
0
e−(p+1)xxqx
I(p, q) = 1
(p+ 1)q+1 I(0, q)
q I(0, q) = qI(0, q −1)
I(p, q)p−1q
H
I=Z+∞
0
x
√x(1 + x);J=Z+∞
0
ln(x)
√x(1 + x)x;K=Z+∞
0
√xln(x)
(1 + x)2x
I

I
I u =√x
J
ln(x)
√x(1 + x)=
x→0o1
x3/4
J1=Z1
0
ln(x)
√x(1 + x)x
J2=Z+∞
1
ln(x)
√x(1 + x)x
J
u=1
xJ1=−J2
J
K
0 +∞x7→ √xln(x)
1 + xK
K=1
2J+I K
H
Z+∞
0
sin(t)
tdt
Z+∞
0
sin2(t)
t2dt
t7→ u= 2t
Z+∞
0
sin2(t)
t2dt=Z+∞
0
2 sin(t) cos(t)
tdt=Z+∞
0
sin(t)
tdt .
n∈N
In=Zπ
2
0
sin2(nt)
t2dt
t7→ u=nt
lim
n→+∞
In
n=Z+∞
0
sin(t)
tdt .
n∈N
An=Zπ
2
0
sin2(nt)
sin2(t)dt .
A0A1
a, b ∈R
sin(a+b) sin(a−b) = sin2a−sin2b
n∈N
sin2(nt)−2 sin2((n+ 1)t) + sin2((n+ 2)t) = 2 sin2(t) cos(2(n+ 1)t).
An−2An+1 +An+2 = 0
n∈NAn=nπ
2
n∈N
Bn=Zπ
2
0
sin2(nt)
tan2(t)dt .
An−Bn=π
4

TSI2
x∈[0,π
2[ sin(x)6x6tan(x)
∀n∈N, Bn6In6An.
Z+∞
0
sin(t)
tdt=π 2.
H
f(x) = Z+∞
0
e−2tp1 + x2e2tt
f
a > 0Z+∞
0
e−at t
xZ+∞
0
e−2tp1 + x2e2tt
fR+
f
∀x > 0,∀t>0, xet6√1 + x2e2t6xet+e−t
2x
∀x > 0x6f(x)6x+1
6x
f+∞
f
g]0,+∞[Z+∞
1
g(t)t
H:x7→ Z+∞
x
g(t)t
C1]0,+∞[H0(x) = −g(x)
x > 0
f(x) = x2Z+∞
x
√1 + u2
u3u
fC1]0; +∞[
f
x
2f(x) = p1 + x2+x2Z+∞
x
1
u√1 + u2u
f]0,+∞[
f f00
x
Z+∞
x
1
u√1 + u2u=−ln x
√1 + x2+Z+∞
x
uln u
(1 + u2)3/2u
Z+∞
0
uln u
(1 + u2)3/2u