38 Intégrale impropre d`une fonction continue sur un intervalle de R

R.
I
I= [a, b[−∞ < a < b ≤+∞
f: [a, b[→R C
I.
f I
a0=a < a1<··· < ap< ap+1 =b
f]ak, ak+1[ 0 ≤k≤p
a ak1≤k≤p
f]ap, b[
[ap, b[k0p−1,
f]ak, ak+1[
[ak, ak+1].
I
[α, β]⊂I.
f I, F a,
[a, b[
∀x∈[a, b[, F (x) = x
a
f(t)dt.
x∈[a0, a1[,
F(x) = x
a0
f(t)dt

R.
x∈[ak, ak+1[k1p,
F(x) =
k−1
j=0 aj+1
aj
f(t)dt +x
ak
f(t)dt.
f[a, b[
F x b I.
b
a
f(t)dt
f[a, b[.
F b f [a, b[
b
a
f(t)dt = lim
x→bx
a
f(t)dt.
f: [a, b[→R C c∈[a, b[
f[a, b[f
[c, b[b
b
a
f(t)dt =c
a
f(t)dt +b
c
f(t)dt.
∀x∈]a, b[,x
a
f(t)dt =c
a
f(t)dt +x
c
f(t)dt.
f
]a, b]−∞ ≤ a < b < +∞
b
a
f(t)dt = lim
x→ab
x
f(t)dt
f]a, b[−∞ ≤ a<b≤+∞
f]a, b[
c]a, b[c
a
f(t)dt b
c
f(t)dt
c,
f]a, b[
b
a
f(t)dt =c
a
f(t)dt +b
c
f(t)dt
= lim
x→ab
x
f(t)dt + lim
y→by
a
f(t)dt.
c
a
f(t)dt b
c
f(t)dt c.

c∈]a, b[
c
a
f(t)dt b
c
f(t)dt f ]a, b[
d∈]a, b[,
d
a
f(t)dt +b
d
f(t)dt =c
a
f(t)dt +b
c
f(t)dt.
b
a
f(t)dt
f
]a, b[
c
a
f(t)dt
b
c
f(t)dt b
a
f(t)dt.
a=−∞ b= +∞lim
x→+∞x
−x
f(t)dt
f]−∞,+∞[. f (t) = t
x
−x
f(t)dt = 0 x > 0 lim
x→+∞x
0
f(t)dt = +∞.
f]−∞,+∞[
lim
x→+∞0
−x
f(t)dt lim
x→+∞x
0
f(t)dt.
f:t7→ sin (t) [0,+∞[.
f:t7→ 1
1−e−t+ ln (t)−1 te−t
]0,+∞[+∞
0
f(t)dt = 0.
f, [a, b[b ℓ b,
f(b) = ℓ
f b F a
f[a, b], F b
lim
x→bx
a
f(t)dt = lim
x→bF(x)−F(a) = F(b)−F(a) = b
a
f(t)dt

R.
α f ]0,+∞[
f:t7→ 1
tα.
f[1,+∞[α > 1
∀α > 1,+∞
1
dt
tα=1 α−1.
f]0,1] α < 1
∀α < 1,1
0
dt
tα=1 1−α.
+∞
0
dt
tαα.
u=b−t
u=t−a a < b α Rf:t7→ 1
(b−t)αf:t7→
1
(t−a)α[a, b[α < 1
∀α < 1,b
a
dt
(b−t)α=b
a
dt
(t−a)α=1 1−α
1
(b−a)α−1.
a= 0, b = 1 α=1
2,
1
0
dt
√1−t= 2.
I= [a, b[f g
f g I
f f +λg λ
b
a
f(x)dx =b
a
f(x)dx
b
a
(f(x) + λg (x)) dx =b
a
f(x)dx +λb
a
g(x)dx.
b
a
f(x)dx b
a
g(x)dx b
a
(f(x) + g(x)) dx

f(x) = 1
x2, g (x) = −1 x2f(x) = 1 x2, g (x) = 1 x2
]0,1] .
f g
f(x) = 1, g (x) = 1
√xf(x) = 1 √x,
g(x) = 1
√x]0,1] .
fb
a
f(x)dx
b
aℜ(f) (x)dx b aℑ(f) (x)dx
b
a
f(x)dx =b
aℜ(f) (x)dx +ib aℑ(f) (x)dx.
a, b +∞
0
eat cos (bt)dt
f, g C1Ilim
x→bf(x)g(x)
b
a
f′(x)g(x)dx b
a
f(x)g′(x)dx
b
a
f(x)g′(x)dx = lim
x→bf(x)g(x)−f(a)g(a)−b
a
f′(x)g(x)dx.
[a, x]
+∞
0
tne−tdt In
n∈N.+∞
0
arctan (t2)
t2dt
φC1
J= [α, β[I= [a, b[f I
β
α
f(φ(t)) φ′(t)dt b
a
f(x)dx
b
a
f(x)dx =β
α
f(φ(t)) φ′(t)dt.
x
a
f(t)dt
π
2
0
ln (sin (t)) dt.