1-1-ln-x-dx-

Question Number 153164 by alisiao last updated on 05/Sep/21

\int_{- 1}^{1} l n (Γ (x)) d x

Answered by Ar Brandon last updated on 05/Sep/21

Ω = \int_{- 1}^{1} \ln (Γ (x)) d x = \underset{Ω_{1}}{\underset{⏟}{\int_{- 1}^{0} \ln (Γ (x)) d x}} + \underset{Ω_{2}}{\underset{⏟}{\int_{0}^{1} \ln (Γ (x)) d x}}

Ω_{2} = \int_{0}^{1} \ln (Γ (x)) d x = \int_{0}^{1} \ln (Γ (1 - x)) d x

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

= \frac{1}{2} \int_{0}^{1} \ln (Γ (x) Γ (1 - x)) d x = \frac{1}{2} \int_{0}^{1} \ln (\frac{π}{\sin (π x)}) d x

= \frac{\ln π}{2} - \frac{1}{2} \int_{0}^{1} \ln (\sin (π x)) d x = \frac{\ln π}{2} - \frac{1}{2 π} \underset{= 0}{\underset{⏟}{\int_{0}^{π} \ln (\sin (u)) d u}}