Let Iₙ = ∫[e⁻¹ to e] x¹⁹ (log|x|)ⁿ dx, where n ∈ N. If (20)I₁₀ = αI₉ + βI₈, for natural numbers α and β then α - β equals to ______.
Let Iₙ = ∫[e⁻¹ to e] x¹⁹ (log|x|)ⁿ dx, where n ∈ N. If (20)I₁₀ = αI₉ + βI₈, for natural numbers α and β then α - β equals to ______.
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1 Answer
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Given the integral In = ∫(log|x|)^n / x^19 dx.
Let t = log|x|, which implies x = e^t and dx = e^t dt.The integral becomes:
In = ∫ e^(-20t) * t^n dtUsing integration by parts, where u = t^n and dv = e^(-20t) dt:
In = [t^n * e^(-20t) / -20] - ∫ n*t^(n-1) * e^(-20t) / -20 dt
In = e^(-20) / -20 - (n / -20) * In-1
20 * In = -e^(-20) + n * In-1For n = 10: 20 * I10 = e^20 - 10 * I9 (Note: There seems to be a sign inconsistency in the original document's derivation vs. standard integration by parts, the document states e^20 instead of -e^(-20) and proceeds with e^20).
For n = 9: 20 * I9 = e^20 - 9 * I8From these two equations, we can exp
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