Improper integral dx/ln²x from 1 to 2

This problem asks to investigate the convergence of an improper integral of the second kind with a singular point at the lower limit.

Problem

$\int_{1}^{2} \frac{d x}{{ln}^{2} x}$

Solution

Step 1

The integrand f(x) = 1/ln²x is not defined at x = 1 (the denominator vanishes) but is continuous on every segment [1+ε, 2] for ε > 0. Hence the integral is improper of the second kind with a singular point at x = 1.

Why is the integral improper of the second kind?

An improper integral of the second kind arises when the integrand is unbounded (tends to infinity) at some point of the integration interval. In our case ln x → 0 as x → 1+, so 1/ln²x → +∞. The singular point x = 1 is the left endpoint of the interval.

Step 2

To apply the comparison test, find an equivalent function near the singular point. Use L'Hôpital's rule for the ratio ln x / (x−1).

Step 3

As x → 1+, both numerator and denominator tend to zero — an indeterminacy 0/0. Applying L'Hôpital's rule:

lim_{x \to 1 +} \frac{ln x}{x - 1} = lim_{x \to 1 +} \frac{1 / x}{1} = 1

Why is L'Hôpital's rule needed?

L'Hôpital's rule allows us to resolve indeterminacies of the form 0/0 or ∞/∞. In our case the limit of ln x/(x−1) as x → 1+0 equals 1. This means ln x ~ (x−1) as x → 1+0 — a key equivalence for further analysis.

Step 4

From this equality follows the equivalence:

ln x \sim (x - 1), {ln}^{2} x \sim {(x - 1)}^{2}

Step 5

Hence the integrand is equivalent to:

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

Step 6

Verify the equivalence using the limit comparison test with g(x) = 1/(x−1)²:

lim_{x \to 1 +} \frac{f (x)}{1 / {(x - 1)}^{2}} = {(lim_{x \to 1 +} \frac{x - 1}{ln x})}^{2} = 1^{2} = 1

Step 7

The integral ∫₁² dx/(x−1)² diverges: it is the reference improper integral of the second kind ∫_a^b dx/(x−a)^α with α = 2 ≥ 1. By the limit comparison test (lim f/g = 1, 0 < 1 < ∞), the integrals ∫ f and ∫ g converge or diverge simultaneously. Hence the original integral diverges.

Why does the reference integral diverge?

The generalized reference integral of the second kind ∫_a^b dx/(x−a)^α converges for α < 1 and diverges for α ≥ 1. In our case α = 2 ≥ 1, so the integral of g(x) diverges. Proof: ∫_{1+ε}^{2} dx/(x−1)² = [−1/(x−1)]_{1+ε}^{2} = −1 + 1/ε → +∞ as ε → 0+.

Answer

Diverges.

Frequently asked questions

How do I investigate an improper integral of the second kind for convergence?

Find an equivalent function g(x) near the singular point. Usually g(x) has the form 1/(x−a)^α (for a singular point at x = a). Apply the limit comparison test: if lim f(x)/g(x) = c with 0 < c < ∞, the integrals converge or diverge simultaneously. Reference: ∫_a^b dx/(x−a)^α converges for α < 1 and diverges for α ≥ 1.

How do I find the equivalent function for 1/ln²x as x → 1?

Use L'Hôpital's rule or a Taylor series expansion. By L'Hôpital's rule, lim ln x/(x−1) = lim (1/x)/1 = 1 as x → 1. This means ln x ~ (x−1), so ln²x ~ (x−1)². Then 1/ln²x ~ 1/(x−1)².

What is the generalized reference integral of the second kind?

This is the integral ∫_a^b dx/(x−a)^α with a singular point at the lower limit. For α < 1 it converges, for α ≥ 1 it diverges. It is the key tool for investigating convergence of integrals of the second kind.

What is the difference between improper integrals of the first and second kind?

An integral of the first kind has an infinite limit of integration. An integral of the second kind has a singular point inside the interval or at its endpoint where the integrand is unbounded. Mixed cases also occur: for example, an integral with an infinite upper limit and a singular point at zero (as in the next problem).