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Question Number 160895 by HongKing last updated on 08/Dec/21

Prove that :

\underset{0}{\int^{1}} \frac{\ln (x)}{x^{n} + x^{n - 1} + \dots + 1} dx = \frac{1}{n^{2}} [ψ^{(1)} (\frac{2}{n}) - ψ^{(1)} (\frac{1}{n})]

Answered by Kamel last updated on 08/Dec/21

Ω_{n} = \int_{0}^{1} \frac{L n (x) (1 - x)}{1 - x^{n}} d x = {[\frac{d}{d s} \int_{0}^{1} \frac{x^{s} - x^{s + 1}}{1 - x^{n}} d x]}_{s = 0}

\overset{t = x^{n}}{=} \frac{1}{n} {[\frac{d}{d \underset{}{s}} \int_{0}^{1} \frac{t^{\frac{s + 1}{n} - 1} - 1 + 1 - t^{\frac{s + 2}{n} - 1}}{1 - t} d t]}_{s = 0} = \frac{1}{n^{2}} (Ψ^{(1)} (\frac{2}{n}) - Ψ^{(1)} (\frac{1}{n}))

Commented by HongKing last updated on 10/Dec/21

thank you so much dear Sir