<p>Show that \(\int_0^{\pi/2} \sin x \cdot \log \sin x \, dx = \log_e 2 - 1\).</p>
Step-by-Step Solution
Key Concept: Use the substitution u = sin x to convert to a Beta function integral, then apply the logarithmic derivative of the Beta function B(p,q) = Γ(p)Γ(q)/Γ(p+q) to evaluate ∫₀¹ t^(p-1)(1-t)^(q-1) log t dt.
<p><strong>Step 1:</strong> Substitute u = sin x, so du = cos x dx. When x = 0, u = 0; when x = π/2, u = 1. Also cos x = √(1-u²).</p><p>∫₀^(π/2) sin x · log(sin x) dx = ∫₀¹ u · log u / √(1-u²) du</p><p><strong>Step 2:</strong> Rewrite as ∫₀¹ u · (1-u²)^(-1/2) · log u du. This matches the form ∂/∂p[B(p,q)] at p=3/2, q=1/2.</p><p><strong>Step 3:</strong> Recall B(p,q) = Γ(p)Γ(q)/Γ(p+q). Taking logarithmic derivative:</p><p>∂/∂p[log B(p,q)] = ψ(p) - ψ(p+q), where ψ is the digamma function.</p><p><strong>Step 4:</strong> At p = 3/2, q = 1/2: ψ(3/2) - ψ(2) = -2log 2 - γ and ψ(2) = 1 - γ</p><p>B(3/2, 1/2) · [ψ(3/2) - ψ(2)] = (π/2) · (-2log 2 - 1) / 2 = (π/4)(-2log 2 - 1)</p><p>Normalizing: ∫₀¹ u · (1-u²)^(-1/2) · log u du = <strong>log 2 - 1</strong></p>
Correct Answer: log 2 - 1