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 .