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Question Number 164103 by mnjuly1970 last updated on 14/Jan/22

p r o v e t h a t

Ω = \int_{0}^{1} \ln (\frac{1 + x}{1 - x}) . \frac{d x}{x \sqrt{1 - x^{2}}} = \frac{π^{2}}{2}

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Answered by Lordose last updated on 14/Jan/22

Ω \overset{x = \frac{1 - x}{1 + x}}{=} 2 \int_{0}^{1} \frac{\ln (\frac{1}{x})}{{(1 + x)}^{2} (\frac{1 - x}{1 + x} \sqrt{1 - {(\frac{1 - x}{1 + x})}^{2}}} dx

Ω = \int_{0}^{1} \frac{\ln (\frac{1}{x})}{\sqrt{x} (1 - x)} dx \overset{x = x^{2}}{=} 4 \int_{0}^{1} \frac{\ln (\frac{1}{x})}{(1 - x^{2})} dx

Ω = - 4 \underset{k = 1}{\sum^{\infty}} \int_{0}^{1} x^{2 k} \ln (x) dx \overset{IBP}{=} 4 \underset{k = 1}{\sum^{\infty}} \frac{1}{{(2 k + 1)}^{2}}

Ω = \underset{k = 1}{\sum^{\infty}} \frac{1}{{(k + \frac{1}{2})}^{2}} dx = ψ^{(1)} (\frac{1}{2})

Ω = \frac{π^{2}}{2} ▴ ▴ ▴

Commented by mnjuly1970 last updated on 14/Jan/22

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