Improper integral (1+√x+x+x²)/(x²+x³+14) from 1 to ∞
This problem asks to investigate the convergence of an improper integral of the first kind with an infinite upper limit.
Problem
Solution
Step 1
The integrand is continuous on [1, +∞): the polynomial x² + x³ + 14 has only positive coefficients, so it does not vanish for x ≥ 1. The integral is improper of the first kind.
Why does the denominator not vanish?
The polynomial x² + x³ + 14 takes only positive values for x ≥ 1. All its coefficients are positive (1, 1, 0, 14), so for x ≥ 1 each term is positive. Hence the denominator never vanishes on [1, +∞), and the integrand is continuous.
Step 2
As x → +∞, the numerator 1 + √x + x + x² behaves like x² (the highest degree), and the denominator x² + x³ + 14 behaves like x³. Therefore f(x) is equivalent to 1/x:
How do we find the equivalent function?
Equivalence as x → +∞ is determined by the leading terms of the numerator and denominator. The leading term of the numerator is the highest power of x, which is x². The leading term of the denominator is x³. The ratio of leading terms x²/x³ = 1/x determines the behavior of the function at infinity.
Step 3
Verify with the limit comparison test:
Step 4
The integral of the comparison function g(x) = 1/x:
Step 5
The harmonic integral ∫₁⁺∞ dx/x diverges (the reference integral ∫₁⁺∞ dx/xᵅ with α = 1). By the comparison test in limit form, if lim f(x)/g(x) = c with 0 < c < ∞, then the integrals ∫ f and ∫ g converge or diverge simultaneously. Hence, the original integral diverges.
Answer
Diverges.
Frequently asked questions
How do I determine the convergence of an improper integral of a rational function?
Compare the degrees of the numerator and denominator. If the degree of the numerator is smaller than the degree of the denominator by at least 2, the integral converges. If the degree of the numerator is smaller by exactly 1 (difference 1), the integral behaves like the harmonic ∫ dx/x and diverges. If the degree of the numerator is ≥ the degree of the denominator, the integral diverges.
What is the harmonic integral and why does it diverge?
The harmonic integral is ∫₁⁺∞ dx/x. It diverges, although the integrand tends to zero. Proof: ∫₁ᴺ dx/x = ln N → +∞ as N → +∞. This is the reference divergent improper integral of the first kind.
What to do if the integrand is a fraction with roots?
When roots are present, consider their asymptotic behavior as x → +∞. For example, √x = x^(1/2), which is less than any positive power of x. In our case, √x does not affect the leading term of the numerator (x² dominates), but in other problems it may be the opposite — then the equivalent function changes.
What is the difference between the limit comparison test and the comparison test in inequality form?
In inequality form, you must prove that f(x) ≤ C·g(x) for all sufficiently large x. In limit form, it suffices to compute lim f(x)/g(x) = c (0 < c < ∞). The limit form is more convenient since it does not require an estimate on the whole interval. It applies even for a more complex structure of the integrand.
