<p><strong>705.</strong> Let \(I(n) = \displaystyle\int_0^{\pi} \ln(1 - 2n\cos x + n^2)\,dx\). Find the value of \(\dfrac{I(100)}{I(10)}\).</p>
Step-by-Step Solution
Key Concept: Recognize that I(n) = ∫₀^π ln(1 - 2n cos x + n²)dx depends on whether |n| < 1 or |n| > 1. For |n| ≠ 1, use the algebraic factorization 1 - 2n cos x + n² = |e^(ix) - n|² and apply logarithm properties to find I(n) = 2π ln|n| when |n| > 1.
<p><strong>Step 1:</strong> Recognize the quadratic form. Note that 1 - 2n cos x + n² = (n - cos x)² + sin² x = |e^(ix) - n|².</p><p><strong>Step 2:</strong> Use the property that ln|a - b|² = 2ln|a - b|. For the integral, consider the function's behavior based on |n|.</p><p><strong>Step 3:</strong> For |n| > 1 (both 100 and 10 satisfy this): Apply the known result that ∫₀^π ln(1 - 2n cos x + n²)dx = 2π ln|n| when |n| > 1. This can be derived using the geometric series expansion or contour integration.</p><p><strong>Step 4:</strong> Calculate I(100) = 2π ln(100) and I(10) = 2π ln(10).</p><p><strong>Step 5:</strong> Compute the ratio: I(100)/I(10) = [2π ln(100)]/[2π ln(10)] = ln(100)/ln(10) = ln(10²)/ln(10) = 2ln(10)/ln(10) = 2.</p><p>∴ Answer: <strong>2</strong></p>
Correct Answer: 2