0-1-log-x-log-x-1-x-x-1-x-dx-Any-help-

Question Number 142690 by Rankut last updated on 04/Jun/21

\int_{0}^{1} \frac{l o g (x) l o g (\frac{x}{1 - x})}{\sqrt{\frac{x}{1 - x}}} d x

Any help

Answered by Dwaipayan Shikari last updated on 04/Jun/21

\int_{0}^{1} x^{a - 1} {(\frac{x}{1 - x})}^{b - 1} d x = \int_{0}^{1} x^{a + b - 2} {(1 - x)}^{1 - b} d x = \frac{Γ (a + b - 1) Γ (2 - b)}{Γ (a + 1)}

H e r e a = 1

\int_{0}^{1} l o g (x) x^{a - 1} {(\frac{x}{1 - x})}^{b - 1} d x = \frac{\partial}{\partial a} . \frac{Γ (a + b - 1)}{Γ (a + 1)} Γ (2 - b)

= \frac{Γ (a + 1) Γ^{'} (a + b - 1) - Γ^{'} (a + 1) Γ (a + b - 1)}{Γ^{2} (a + 1)} Γ (2 - b)

= Γ^{'} (b) Γ (2 - b) - ψ (2) Γ (b) Γ (2 - b)

\int_{0}^{1} l o g (x) l o g (\frac{x}{1 - x}) {(\frac{x}{1 - x})}^{b - 1} d x

= Γ^{″} (b) Γ (2 - b) - Γ^{'} (2 - b) Γ^{'} (b) - ψ (2) Γ^{'} (b) Γ (2 - b) + ψ^{'} (2) Γ^{'} (2 - b) Γ (b)

H e r e

p u t b = \frac{1}{2}