Let f : [0, ∞) → [0,∞) be a continuous function with the property that

10.3.8. Let f : [0, ∞) → [0,∞) be a continuous function with the property that

lim

x → ∞ f (x)

f (t)dt = ℓ > 0 , ℓ finite.

√ Prove that lim x → ∞ f (x) x exists and is finite, and evaluate it in terms of ℓ .

394 10 Applications of the Integral Calculus Solution. Define F # (x) = x

0 f (t)dt, for any x ≥ 0. Since f takes on nonnegative values, F is increasing, and by our hypothesis,

lim F

∞ → (x) = sup {F(x) : x ≥ 0} > 0.

We first prove that

lim F (x) = +∞ . x → ∞

Assuming that this limit is some finite L > 0, it follows that

lim f (x) = ℓ/L . x → ∞

Hence f (x) ≥ ℓ/(2L) for x ≥ x 0 , for some x 0 ≥ 0, so " x

F (x) = F(x 0 )+

f (t)dt ≥ ℓ(x − x 0 )/(2L) , x ≥x 0 ,

which would contradict the assumption that lim x →∞ F (x) is finite. This establishes that lim x →∞ F (x) = +∞. Therefore, l’Hˆopital’s rule yields

F 2 (x)

lim ∞ = 2 lim ∞ f (x)F (x) = 2ℓ . x →

It follows that

Finally, writing

f (x)F(x)

f (x) x =

F √ (x)

√ for x large enough, we conclude that lim x → ∞ f (x) x exists, is finite, and equal to

10.3.9. (Barb˘alat’s lemma, [5]) . Let f : [0, ∞)→R be uniformly continuous and Riemann integrable. Prove that f (x)→0 as x →∞ .

Solution. Arguing by contradiction, there exist ε > 0 and a sequence (x n ) n ≥0 ⊂ [0, ∞) such that x n →∞ as n→∞ and | f (x n )| ≥ ε for all integers n ≥ 0. By the uniform continuity of f , there exists δ > 0 such that for any n ∈ N and for all x ∈ [0,∞),

|x n − x| ≤ δ =⇒ | f (x n ) − f (x)| ≤ . 2

Therefore, for all x ∈ [x n ,x n + δ ] and for any n ∈ N, f (x) ≥ | f (x n )|−| f (x n )− f (x)| ≥ ε /2. This yields, for all n ∈ N,

f (x)dx −

f (x)dx =

f (x)dx =

| f (x)|dx ≥ >0.

10.3 Improper Integrals 395 By hypothesis, the improper Riemann integral # ∞

0 f (x)dx exists. Thus, the left-hand side of the inequality converges to 0 as n →∞. This contradicts the above

relation, and the proof is concluded. ⊓ ⊔

x → ∞ f (x) = 0 . Prove that ∞ 0 f (x)dx converges.

10.3.10. Let f : [0, ∞)→R

be a decreasing function such that lim

Solution. We first observe that our assumptions imply that f > 0 on (0, ∞) and

f is integrable on [0, T ] for any T > 0. Fix T and choose n ∈ N such that n π <T≤ (n + 1) π . Then

n −1 " (k+1)π

f (x) sin x dx =

0 k ∑ =0 k π

∑ k (−1)

| f (x)sin x|dx +

f (x) sin x dx .

k =0

We consider the limit as T →∞, directly for last term and by Leibniz’s alternating series test for the first term. First,

" T " (n+1)π " (n+1)π

f (x) sin x dx ≤

| f (x)||sin x|dx ≤ f (n π )

|sinx|dx = 2 f (n π ),

and the f (x)→0 hypothesis shows that the last term above has limit 0 as n→∞, that is, as T →∞.

Next, the limit of the first term exists if and only if ∑ ∞ k =0 (−1) k a k converges, # where a (k+1)π k = k π

f (x)|sin x|dx. But " (k+1)π

" (k+2)π

f (x)|sin x|dx ≥

f (x)|sin x|dx = a k +1 ,

(k+1)π

because f is decreasing, and for the same reason,

" (k+1)π

a k ≤ f (k π ) |sinx|dx = 2 f (k π )→0 as k→∞.

The alternating series thus converges, and the proof is complete. ⊓ ⊔ This exercise may be used to deduce the convergence of many improper integrals.

# For instance, ∞

0 (sin x)/x dx converges (we observe that 0 is not a “bad” point, since sin x /x→1 as x→0. An alternative proof of this result follows from the following

exercise:

Exercise. Prove that if 0 <a<R<∞ , then

" R sin x

2 dx < .

396 10 Applications of the Integral Calculus Next, we are concerned with the Gaussian integral (or probability integral)

# +∞ −∞ e −x 2 dx .