Let Iₙ = ∫₁ᵉ x¹⁹(log |x|)ⁿ dx, where n∈N. If (20)I₁₀ = αI₉ + βI₈, for natural numbers α and β, then α - β equals to ___________.

Solution

Maths sir · 4 steps

Get a reduction formula by parts, then use it twice to remove the e²⁰ term.

1. Integrate by parts. Take (log x)ⁿ as the first function and x¹⁹ as the second. On [1, e], x > 0, so log|x| = log x. The boundary term is e²⁰/20, because log 1 = 0 and log e = 1.

   Iₙ = [x²⁰/20(log x)ⁿ]₁ᵉ - n/20∫₁ᵉ x¹⁹(log x)ⁿ⁻¹ dx ⇒ 20Iₙ = e²⁰ - nIₙ₋₁

2. Use it twice. With n = 10: 20I₁₀ = e²⁰ - 10I₉. With n = 9: 20I₉ = e²⁰ - 9I₈, so e²⁰ = 20I₉ + 9I₈.
3. Remove the constant. Substitute this in: 20I₁₀ = 20I₉ + 9I₈ - 10I₉ = 10I₉ + 9I₈. So α = 10 and β = 9.
4. So. α - β = 10 - 9 = 1.

The idea to keep. For ∫ xᵐ(log x)ⁿ dx, integrate by parts with (log x)ⁿ as the first function. Each round lowers the power of log x by one. Use the formula again to cancel any constant left over.

Answer from the NTA final answer key. Solution by PrepWiser.