Improper integral (√x+x²)/(x²+x³+5) from 1 to ∞

This problem asks to investigate the convergence of an improper integral of the first kind with a rational integrand.

Problem

$\int_{1}^{+ \infty} \frac{\sqrt{x} + x^{2}}{x^{2} + x^{3} + 5} d x$

Solution

Step 1

The integrand f(x) = (√x+x²)/(x²+x³+5) is continuous on [1, +∞): the numerator is positive, and the denominator does not vanish because all its coefficients are positive. Hence the integral is improper of the first kind. We apply the comparison test in limit form.

Why does the denominator not vanish?

The polynomial x² + x³ + 5 takes only positive values for x ≥ 1: all its coefficients are positive, so each term is positive. Hence the denominator is never zero on [1, +∞), and the integrand is continuous.

Step 2

Let us find a function g(x) equivalent to f(x) as x → +∞. To do this, we examine the behaviour of the numerator and denominator for large x.

Step 3

As x → +∞: the numerator √x + x² behaves like x² (highest degree), and the denominator x² + x³ + 5 behaves like x³. Hence:

\sqrt{x} + x^{2} \sim x^{2}, x^{2} + x^{3} + 5 \sim x^{3}

Step 4

The ratio of leading terms gives the equivalent function:

f (x) \sim \frac{x^{2}}{x^{3}} = \frac{1}{x}

Step 5

Let us verify the equivalence using the limit comparison test:

lim_{x \to + \infty} \frac{f (x)}{1 / x} = lim_{x \to + \infty} \frac{(\sqrt{x} + x^{2}) \cdot x}{x^{2} + x^{3} + 5}

How do we choose g(x)?

The equivalence is determined by the leading terms of the numerator and denominator as x → +∞. The leading term of the numerator is x², and of the denominator is x³. Their ratio x²/x³ = 1/x is the comparison function.

Step 6

Divide the numerator and denominator by x³ and pass to the limit:

lim_{x \to + \infty} \frac{\frac{1}{x^{3 / 2}} + 1}{1 + \frac{1}{x} + \frac{5}{x^{3}}} = 1

Step 7

The harmonic integral ∫₁⁺∞ dx/x diverges (reference integral 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.

Why does the harmonic integral diverge?

The harmonic integral ∫₁⁺∞ dx/x diverges, even though the integrand tends to zero. Proof: ∫₁ᴺ dx/x = ln N → +∞ as N → +∞. This is the reference divergent improper integral of the first kind.

Answer

Diverges.

Frequently asked questions

How do we 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 by at least 2, the integral converges. If the difference is exactly 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?

The harmonic integral is ∫₁⁺∞ dx/x. It diverges even though the integrand tends to zero. This is the reference divergent improper integral of the first kind.

What if the integrand is a fraction with roots?

If roots are present, consider their asymptotics as x → +∞. For example, √x = x^1/2, which is smaller than any positive power of x. In our case √x does not affect the leading term of the numerator (x² dominates).

What is the difference between the limit comparison test and the comparison test in inequality form?

In inequality form, one 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 with 0 < c < ∞. The limit form is more convenient and applies to a more complex structure of the integrand.