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The Limit Comparison Theorem for Improper Integrals
Limit Comparison Theorem (Type I): If f and g are continuous, positive functions for all values of x, and
lim xโโ
f (x) g(x)
= k
Then:
- if 0 < k < โ, then โซ (^) โ
a
g(x) dx converges โโ
a
f (x) dx converges
- if k = 0, then โซ (^) โ
a
g(x) dx converges =โ
a
f (x) dx converges
- if k = โ, then โซ (^) โ
a
g(x) dx converges โ=
a
f (x) dx converges
All of the corresponding statements for improper integrals of type II are also true:
Limit Comparison Theorem (Type II): If f and g are continuous, positive functions for all values of x, and
lim xโ 0 +
f (x) g(x)
= k
Then:
- if 0 < k < โ, then โซ (^) a
0
g(x) dx converges โโ
โซ (^) a
0
f (x) dx converges
- if k = 0, then โซ (^) a
0
g(x) dx converges =โ
โซ (^) a
0
f (x) dx converges
- if k = โ, then โซ (^) a
0
g(x) dx converges โ=
โซ (^) a
0
f (x) dx converges
These theorems offer an alternative to using the Comparison Theorem (CT) discussed in the book when trying to determine whether an improper integral converges or diverges. The proofs of these three statements use CT, so we can conclude that in some sense, any problem the Limit Comparison Theorem (LCT) can solve could also be solved by CT, just by following the arguments in those proofs; however, sometimes the solution is easier using LCT.
(It should be noted however that there do exist some examples of convergence questions where LCT fails, but CT does not! So strictly speaking, CT is more powerful than LCT.)
First, letโs prove part (1.); the proofs of parts (2.) and (3.) are similar (you might want to try working those out for yourself!).
Proof of LCT(I-1.):
We assume that k exists and is a positive finite number, and that the limit from a to โ of g converges; we will show that the limit from a to โ of f converges. Proving the other direction can be done similarly, or simply by observing that if lim fg = k exists and is positive, then
lim gf = (^1) k must also exist and be positive...
The definition of the limit tells us that there exists some N such that
(k โ 1) <
f (x) g(x)
< (k + 1) whenever x > N.
So, for those values of x, we have
f (x) g(x)
< (k + 1) =โ f (x) < (k + 1)g(x)
We now break the integral in question into two pieces:
a
f (x) dx =
โซ N
a
f (x) dx +
N
f (x) dx
The first integral is of a continuous function on a closed, bounded interval, so we know that is finite. The convergence of the second integral is concluded by the following, which we can do because of the inequality determined above:
N
f (x) dx <
N
(k + 1)g(x) dx = (k + 1)
N
g(x) dx
(the last integral in the equation above is given to converge; therefore, by the Comparison Theorem, the integral on the left converges.)
We conclude, as desired, that the integral of f converges.
