Intégrale de Gauss par trois méthodes

I=Z+∞
0
e−t2dt =√π
2
g:
[0,+∞[→R
x7→ Z1
0
e−(t2+1)x2
t2+ 1 dt
(t, x)7→ Z1
0
e−(t2+1)x2
t2+ 1 dt x
g
∀x≥0, g0(x) = −2xZ1
0
e−(t2+1)x2dt =−2xe−x2Z1
0
e−(tx)2dt
u=tx
g0(x) = −2e−x2Zx
0
e−u2du =−2f0(x)f(x)f:x7→ Zx
0
e−u2du
∀x≥0, g(x)−g(0) = −¡f2(x)−f2(0)¢g(x) = π
4−f2(x)
0≤g(x)≤e−x2Z1
0
dt
1 + t2lim
x→+∞g(x)=0 f
lim
x→+∞f(x) = rπ
4
I=Z+∞
0
e−t2dt =√π
2
1
ln(1 + x)≤x∀x∈]−1,+∞[n∈N∗
∀t∈]0,√n],ln µ1−t2
n¶≤ −t2 nln µ1 + t2 n¶≤t2 n
n−n
∀t∈]0,√n],µ1−t2
n¶n
≤e−t2 ≤µ1 + t2
n¶−n
∀n∈N∗,Z√n
0µ1−t2
n¶n
dt ≤Z√n
0
e−t2dt ≤Z√n
0µ1 + t2
n¶−n
dt
t=√ncos u t =√n u
∀n∈N∗,√nZπ
2
0
sin2n+1 udu ≤Z√n
0
e−t2dt ≤√nZπ
2
0
sin2n−2udu
Zπ
2
0
sinnudu ∼
n∞rπ
2n
√π
2
n→+∞n
I=Z+∞
0
e−t2dt = lim
x→+∞Z√x
0
e−t2dt =√π
2

a > 0
Da=©(x, y)∈R2/ x2+y2≤a2ª, Ca= [−a, a]2Ia=Z ZDa e−(x2+y2)dx dy
∀a > 0, Ia=Z Z[0,a]×[0,2π] e−r2rdrdθ =ÃZ[0,a] e−r2rdr!ÃZ[0,2π] dθ!=π(1 −e−a2)
Ja=Z ZCa e−(x2+y2)dx dy
Ja=ÃZ[−a,a] e−x2dx!ÃZ[−a,a] e−y2dy!=µZa
−a
e−x2dx¶2
Da⊂Ca⊂D√2aIa≤Ja≤J√2a
∀a > 0, π(1 −e−a2)≤µZa
−a
e−x2dx¶2 ≤π(1 −e−2x2)
a+∞Z+∞
−∞
e−x2dx =√π I