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Improper Integrals
Let f : (a, +∞) → R is a function integrable on [a, A] for all A > a. We say that integral
a f^ (x)dx^ converges if limA→+∞
a f^ (x)dx^ exist.
Examples: Determine if the following integral is convergent or divergent.
1
1 x^2 dx.
A→lim+∞
1
x^2
dx = (^) A→lim+∞[−(^1 x
|A 1 )] = − (^) A→lim+∞(^1 A
Thus integral
1
1 x^2 dx^ converges.
1 √^1 x dx.
A→lim+∞
1
x
dx = (^) A→lim+∞ 2(
x|A 1 ) = (^) A→lim+∞ 2(
Thus integral
1 √^1 x dx^ diverges.
0
1 1+x^2 dx.
A→lim+∞
0
1 + x^2 dx^ =^ A→lim+∞(arctan^ |
A 0 ) =^ A→lim+∞(arctan^ A^ −^ arctan 0) =^
π
Thus integral
0
1 1+x^2 dx^ converge.
1
1 x dx.
lim A→+∞
1
x
dx = lim A→+∞
(ln |A 1 ) = lim A→+∞
(ln A − ln 1) = +∞.
Thus integral
1
1 x dx^ diverge. 1
2
Theorem 1Let f and g be continuous on [a, +∞) with 0 < f (x) ≤ g(x) for all x ≥ a. Then
a g(x)dx^ converges, then^
a f^ (x)dx^ converges
a f^ (x)dx^ diverges, then^
a g(x)dx^ diverges
Theorem 2 Let positive functions f and g are continuous on [a, +∞) and
x→lim+∞
f (x) g(x) =^ L,^0 < L <^ +∞. Then
a f^ (x)dx^ and^
a g(x)dx^ both converges or both diverges.
Integral (^) ∫ +∞ a
xp^
dx, a > 0
converges for p > 1 and diverges for p ≤ 1.
Examples: Determine if the following integral is convergent or divergent.
1 e−x
(^2) dx.
We have 0 < e−x^2 < e−x. Since
lim A→+∞
1
e−x^ = lim A→+∞
[−(e−x|A 1 ] = lim A→+∞
[−(e−A^ − e^1 )] = e.
Then
1 e
−xdx converges and from Theorem 1, ∫^ +∞ 1 e
−x^2 dx also con-
verges.
1
sin^2 x x^2 dx.^ Since^
sin^2 x x^2 ≤^
1 x^2 and^
1
1 x^2 dx,^ p^ = 2 converges, then
from Theorem 1
1
sin^2 x x^2 dx^ converges.