Let
𝑓
be a positive, continuous function with domain
𝑅
.

Assume
∫
1
∞
𝑓
(
𝑥
)
𝑑
𝑥
is divergent.

Which of the following improper integrals must also be divergent?

Select all the correct answers.

Question

Let 
𝑓
  be a positive, continuous function with domain 
𝑅
 .

Assume 
∫
1
∞
𝑓
(
𝑥
)
𝑑
𝑥
  is divergent.

Which of the following improper integrals must also be divergent?

Select all the correct answers.

BlackTom AI · Accepted Answer

Question restatement and options:
- Given a positive, continuous function f with domain R, and knowing that ∫ from 1 to ∞ of f(x) dx diverges, determine which of the following improper integrals must also be divergent. Select all that apply.
Options:
1) ∫_0^∞ f(x) dx
2) ∫_1^∞ f(x) /(1+x) dx
3) ∫_1^∞ f(x)/(1+sin^2 x) dx
4) ∫_1^∞ f(x)^(1+sin^2 x) dx
5) ∫_1^∞ f(x)^2 dx

Option-by-option analysis:
Option 1: ∫_0^∞ f(x) dx
Since f is positive and continuous on R, the integral over [0,1] is finite (a continuous function on a compact interval has a finite integral). The divergence of ∫_1^∞ f(x) dx implies that the tail integral from 1 to ∞ is infinite. Adding the finite part over [0,1] to an infinite tail cannot cancel the divergence. Therefore, ∫_0^∞ f(x) dx must diverge. This reasoning relies on the fact that the sum of a finite number and an infinite quantity is infinite.

Option 2: ∫_1^∞ f(x)/(1+x) dx
Here, for x ≥ 1, we have 1+x ≥ x, so 0 < 1/(1+x) ≤ 1/x, and in particular 1/(1+x) ≤ 1. The integrand is f(x) times a bounded factor 1/(1+x) that decays like 1/x. However, divergence of ∫_1^∞ f(x) dx does not automatically imply divergence of ∫_1^∞ f(x)/(1+x) dx, because multiplying by a decaying factor can sometimes produce a convergent integral even if the original diverges. Hence, this option need not diverge in all cases, and cannot be declared must-diverge.

Option 3: ∫_1^∞ f(x)/(1+sin^2 x) dx
Note that 1 ≤ 1+sin^2 x ≤ 2 for all x, so 1/(1+sin^2 x) satisfies 1/2 ≤ 1/(1+sin^2 x) ≤ 1. Consequently, f(x)/(1+sin^2 x) ≥ (1/2) f(x) for all x. Since ∫_1^∞ f(x) dx diverges and (1/2) f(x) is just a constant multiple of f(x), the comparison yields ∫_1^∞ f(x)/(1+sin^2 x) dx ≥ (1/2) ∫_1^∞ f(x) dx = ∞. Therefore, this integral must diverge.

Option 4: ∫_1^∞ f(x)^(1+sin^2 x) dx
The exponent 1+sin^2 x lies between 1 and 2. Depending on the size of f(x) relative to 1, raising to an exponent between 1 and 2 can either increase or decrease the value of f(x). Without additional information about the magnitude of f(x) (whether it’s consistently ≥1 or ≤1 for large x), we cannot guarantee divergence here. A counterexample can be constructed where ∫_1^∞ f(x) dx diverges but ∫_1^∞ f(x)^(1+sin^2 x) dx converges, so this option is not necessarily must-diverge.

Option 5: ∫_1^∞ f(x)^2 dx
Similar to option 4, squaring f can either amplify or dampen the integrand depending on the size of f(x). It is possible to have ∫_1^∞ f(x) dx diverge while ∫_1^∞ f(x)^2 dx converges (for example, a tail like f(x) ~ 1/x yields divergence for ∫ f but convergence for ∫ f^2). Therefore, this integral is not guaranteed to diverge.

Summary of findings:
- Must-diverge: Option 1 (∫_0^∞ f(x) dx) and Option 3 (∫_1^∞ f(x)/(1+sin^2 x) dx).
- Not necessarily must-diverge: Options 2, 4, and 5.

Therefore, the correct selections are the first and the third options.

Let 𝑓 be a positive, continuous function with domain 𝑅 . Assume ∫ 1 ∞ 𝑓 ( 𝑥 ) 𝑑 𝑥 is divergent. Which of the following improper integrals must also be divergent? Select all the correct answers.Multiple choice

Similar Questions

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