How to Determine Convergence of Improper Integrals with Endpoint Singularities

What We're Solving:

We need to analyze the improper integral ∫[0 to π] ((1 + cos(2 cot(θ)))/2) * (1/sin²(θ)) dθ and determine whether it converges or diverges. The tricky part is that our integrand has singularities (blows up) at both endpoints θ = 0 and θ = π where sin(θ) = 0.

The Approach:

When dealing with improper integrals, we need to:

• 1. Identify the problematic points (where the function becomes infinite)
• 2. Split the integral at any interior points if needed
• 3. Use limits to approach the singularities
• 4. Analyze the behavior near each singularity using comparison tests or direct evaluation

Think of it like approaching a cliff edge - we can't just walk to the edge, but we can carefully approach it and see what happens!

Step-by-Step Solution:

Step 1: Identify the singularities The function f(θ) = ((1 + cos(2 cot(θ)))/2) * (1/sin²(θ)) has problems where sin(θ) = 0, which occurs at θ = 0 and θ = π.

Step 2: Split the integral Since we have singularities at both endpoints, let's split at θ = π/2: ∫[0 to π] f(θ) dθ = ∫[0 to π/2] f(θ) dθ + ∫[π/2 to π] f(θ) dθ

Step 3: Analyze behavior near θ = 0 As θ → 0⁺:

• sin(θ) ≈ θ, so 1/sin²(θ) ≈ 1/θ²
• cot(θ) = cos(θ)/sin(θ) → ∞
• cos(2 cot(θ)) oscillates between -1 and 1
• So (1 + cos(2 cot(θ)))/2 is bounded between 0 and 1

Near θ = 0, our integrand behaves like: (bounded function) × (1/θ²)

Step 4: Test convergence near θ = 0 We need to check if ∫[0 to ε] (1/θ²) dθ converges for small ε > 0. ∫[0 to ε] (1/θ²) dθ = lim[a→0⁺] [-1/θ]ₐᵋ = lim[a→0⁺] (-1/ε + 1/a) = +∞

This diverges! Since our integrand is comparable to 1/θ², the integral diverges near θ = 0.

Step 5: Analyze behavior near θ = π As θ → π⁻:

• θ = π - u where u → 0⁺
• sin(π - u) = sin(u) ≈ u
• Similar analysis shows 1/sin²(θ) ≈ 1/(π-θ)²

By the same reasoning, the integral diverges near θ = π.

Step 6: Apply comparison test Since (1 + cos(2 cot(θ)))/2 ≥ 0 for all θ, and we have:

• Near θ = 0: integrand ≥ C₁/θ² for some constant C₁ > 0
• Near θ = π: integrand ≥ C₂/(π-θ)² for some constant C₂ > 0

Both ∫[0 to ε] 1/θ² dθ and ∫[π-ε to π] 1/(π-θ)² dθ diverge.

The Answer:

The improper integral diverges. It has non-integrable singularities at both θ = 0 and θ = π, where the integrand behaves like 1/θ² and 1/(π-θ)² respectively. Since ∫ 1/x² dx diverges near x = 0, our original integral cannot converge.

Memory Tip:

Remember the "p-test" for integrals: ∫ 1/x^p dx near x = 0 converges only if p < 1. Since we have p = 2 > 1, we get divergence! It's like trying to fit an infinitely tall spike into a finite box - it just won't work! 📈

Great job working through this challenging problem - improper integrals with oscillating functions can be tricky, but breaking them down systematically always helps!