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Question Number 198156 by Erico last updated on 12/Oct/23

Prove that

\frac{2 t - 1}{lnt - \ln (1 - t)} = {\int_{0}}^{1} t^{x} {(1 - t)}^{1 - x} dx

and {\int_{0}}^{1} \frac{2 t - 1}{lnt - \ln (1 - t)} dt = \frac{π}{2} {\int_{0}}^{1} \frac{x (1 - x)}{\sin (π x)} dx

Answered by witcher3 last updated on 12/Oct/23

\int_{0}^{1} e^{(1 - x) \ln (1 - t) + xln (t)} dx

= \int_{0}^{1} (1 - t) e^{x (\ln (\frac{t}{1 - t})} dx

= \frac{(1 - t) (\frac{t}{1 - t} - 1)}{\ln (t) - \ln (1 - t)} = \frac{2 t - 1}{\ln (t) - \ln (1 - t)}

\int_{0}^{1} \frac{2 t - 1}{\ln (t) - \ln (1 - t)} dt = \int_{0}^{1} \int_{0}^{1} t^{x} {(1 - t)}^{1 - x} dxdt

= \int_{0}^{1} \int_{0}^{1} t^{x} {(1 - t)}^{1 - x} dxdt = \int_{0}^{1} \int_{0}^{1} t^{x} {(1 - t)}^{1 - x} dtdx

= \int_{0}^{1} β (x + 1, 2 - x) dx

= \int_{0}^{1} \frac{x (1 - x)}{Γ (3)} . Γ (x) Γ (1 - x) dx = \frac{1}{2} \int_{0}^{1} \frac{x (1 - x) π}{\sin (π x)}

= \frac{π}{2} \int_{0}^{1} \frac{x (1 - x)}{\sin (π x)} dx = \frac{π}{2} . 7 \frac{ζ (3)}{π^{3}} = \frac{7 ζ (3)}{2 π^{2}}

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