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Arkansas Tech University MATH 2924: Calculus II Dr. Marcel B. Finan
47 Comparison Tests for Improper Integrals
Sometimes it is difficult to find the exact value of an improper integral by
antidifferentiation, for instance the integral
∫ (^) ∞ 0 e
−x^2 dx. However, it is still
possible to determine whether an improper integral converges or diverges.
The idea is to compare the integral to one whose behavior we already know,
such us
∫ (^) ∞ 1
1 xp^ dx^ which converges for^ p >^ 1 and diverges otherwise;
∫ (^) ∞ 0 e
cxdx which converges for c < 0 and diverges for c ≥ 0;
∫ (^1) 0
1 xp^ dx^ which converges for^ p <^ 1 and diverges otherwise.
The comparison method consists of the following:
Theorem 47.
Suppose that f and g are continuous and 0 ≤ g(x) ≤ f (x) for all x ≥ a. Then
(a) if
∫ (^) ∞ a f^ (x)dx^ is convergent, so is^
∫ (^) ∞ a g(x)dx
(b) if
∫ (^) ∞ a g(x)dx^ is divergent, so is^
∫ (^) ∞ a f^ (x)dx.
This is only common sense: if the curve y = g(x) lies below the curve y =
f (x), and the area of the region under the graph of f (x) is finite, then of
course so is the area of the region under the graph of g(x). Similar results
hold for the other types of improper integrals.
Example 47.
Determine whether
∫ (^) ∞ 1 √^1 x^3 +
dx converges.
Solution.
For x ≥ 1 we have that x^3 + 5 ≥ x^3 so that
x^3 + 5 ≥
x^3. Thus, √^1 x^3 +
√^1 x^3
. Letting f (x) = √^1 x^3 and g(x) = √^1 x^3 + then we have that
0 ≤ g(x) ≤ f (x). From the previous section we know that
∫ (^) ∞ 1
1 x
3 2
dx is con-
vergent, a p-integral with p = 3 2 >^1.^ By the comparison test,^
∫ (^) ∞ 1 √^1 x^3 + dx is
convergent.
The next question is to estimate such a convergent improper integral.
Example 47.
Estimate the value of
∫ (^) ∞ 1 √^1 x^3 + dx with an error of less than 0. 01.
Solution.
We want to find b such that
∣ ∣ ∣ ∣ ∣
∫ (^) ∞
1
x^3 + 5
dx −
∫ (^) b
1
x^3 + 5
dx
∣ ∣ ∣ ∣ ∣
But (^) ∫ ∞
1
x^3 + 5
dx =
∫ (^) b
1
x^3 + 5
dx +
∫ (^) ∞
b
x^3 + 5
dx.
Thus, the problem is to find b such that
∣ ∣ ∣ ∣ ∣
∫ (^) ∞
b
x^3 + 5
dx
∣ ∣ ∣ ∣ ∣
From the example above, we have
∫ (^) ∞
b
x^3 + 5
dx <
∫ (^) ∞
b
x^3
dx =
b
So it suffices to choose b such that √^2 b < 0 .01 or b > 40 , 000 , say for example
b = 45000. In this case,
∫ (^) ∞
1
x^3 + 5
dx ≈
∫ (^45) , 000
1
x^3 + 5
dx = 1. 69824.
Example 47.
Investigate the convergence of
∫ (^) ∞ 4
dx ln x− 1.
Solution.
For x ≥ 4 we know that ln x − 1 < ln x < x. Thus, 1 ln x− 1
x
. Let g(x) = 1 x
and f (x) = 1 ln x− 1.^ Thus, 0^ < g(x)^ ≤^ f^ (x).^ Since^
∫ (^) ∞ 4
1 x dx^ =^
∫ (^) ∞ 1
1 x dx^ −^
∫ (^4) 1
1 x dx
which is divergent since
∫ (^) ∞ 1
1 x dx^ is divergent being a p-integral with^ p^ = 1. By the comparison test
∫ (^) ∞ 4
dx ln x− 1 is divergent.
Example 47.
Investigate the convergence of the improper integral
∫ (^) ∞ 1
sin√ x+ x dx.
Solution.
We know that − 1 ≤ sin x ≤ 1. Thus 2 ≤ sin x + 3 ≤ 4. Since x ≥ 1 , we
have √^2 x ≤ sin√^ xx+3 ≤ √^4 x. Note that the two integrals
∫ (^) ∞ 1 √^2 x dx^ and^
∫ (^) ∞ 1 √^4 x dx
(a) If
∫ (^) ∞ a g(x)dx^ converges, then so does^
∫ (^) ∞ a f^ (x)dx.
(b)
∫ (^) ∞ a g(x)dx^ converges if and only if^
∫ (^) ∞ a f^ (x)dx^ does. (c) If
∫ (^) ∞ a g(x)dx^ diverges, then so does^
∫ (^) ∞ a f^ (x)dx.
Proof.
(a) Suppose that limx→∞
f (x) g(x) = 0.^ Let^ >^ 0 be given. Then there is a^ b > a
such that
f (x) g(x) < ^ for all^ x^ ≥^ b.^ Thus,^ f^ (x)^ < g(x) for all^ x^ ≥^ b.^ By the
comparison test, if
∫ (^) ∞ a g(x)dx^ is convergent so does^
∫ (^) ∞ a f^ (x)dx.
(b) Now, suppose that limx→∞ f (x) g(x) =^ L,^ where^ L^ is a finite positive constant. Let < L. Then there is a constant b > a such that for all x ≥ b we have
∣ ∣ ∣ ∣ ∣
f (x)
g(x)
− L
∣ ∣ ∣ ∣ ∣
That is,
L − <
f (x)
g(x)
< L + .
Thus, for x ≥ b we have (L − )g(x) < f (x) < (L + )g(x). Now the result
follows from the comparison test.
(c) Finally, suppose that limx→∞ f (x) g(x) =^ ∞.^ Then there is^ b > a^ such that f (x)
∫^ g (x)^ ≥^ 1 for all^ x^ ≥^ b.^ That is,^ g(x)^ ≤^ f^ (x) for all^ x^ ≥^ b.^ Therefore, if ∞ a f^ (x)dx^ converges, then^
∫ (^) ∞ a g(x)dx^ converges.
Remark 47.
The Comparison Test and Limit Comparison Test also apply, modified as
appropriate, to other types of improper integrals.
Example 47.
Show that the improper integral
∫ (^) ∞ 1
1 1+x^2 dx^ is convergent.
Solution.
Since the integral
∫ (^) ∞ 1
dx x^2 is convergent (p-integral with^ p^ = 2^ >^ 1) and since
limx→∞
1 1+x^2 1 x^2
= limx→∞ x^2 x^2 +1 = 1, by the limit comparison test (Theorem 47.
(b)) we have
∫ (^) ∞ 1
dx x^2 +1 is also convergent.
