<p>Prove that \(\displaystyle\int_0^\infty \frac{dx}{1+x^4} = \int_0^\infty \frac{x^2\,dx}{1+x^4} = \frac{\pi}{2\sqrt{2}}\).</p>
Step-by-Step Solution
Key Concept: Use the substitution x = 1/t to relate the two integrals, then add them to simplify the integrand. The key is recognizing that 1/(1+x⁴) + x²/(1+x⁴) = (1+x²)/(1+x⁴), which factors nicely as 1/(x²-√2x+1) after dividing by x².
<p><strong>Step 1: Prove the two integrals are equal</strong></p><p>Let I₁ = ∫₀^∞ dx/(1+x⁴) and I₂ = ∫₀^∞ x²dx/(1+x⁴).</p><p>In I₁, substitute x = 1/t, so dx = -dt/t²:</p><p>I₁ = ∫₀^∞ (1/t²)/(1+1/t⁴) · dt/t² = ∫₀^∞ t²/(t⁴+1) dt = I₂</p><p><strong>Step 2: Add the integrals</strong></p><p>I₁ + I₂ = ∫₀^∞ (1+x²)/(1+x⁴) dx</p><p>Since I₁ = I₂, we have: 2I₁ = ∫₀^∞ (1+x²)/(1+x⁴) dx</p><p><strong>Step 3: Simplify the integrand</strong></p><p>Divide numerator and denominator by x²:</p><p>∫₀^∞ (1+x²)/(1+x⁴) dx = ∫₀^∞ (1/x² + 1)/(x² + 1/x²) dx</p><p>Let u = x - 1/x, then du = (1 + 1/x²)dx</p><p>Note: x² + 1/x² = (x - 1/x)² + 2 = u² + 2</p><p><strong>Step 4: Evaluate the integral</strong></p><p>2I₁ = ∫₋∞^∞ du/(u² + 2) = [1/√2 · arctan(u/√2)]₋∞^∞</p><p>= 1/√2 · [π/2 - (-π/2)] = π/√2</p><p><strong>Step 5: Final answer</strong></p><p>I₁ = π/(2√2) ∴ ∫₀^∞ dx/(1+x⁴) = ∫₀^∞ x²dx/(1+x⁴) = <strong>π/(2√2)</strong></p>
Correct Answer: π/(2√2)