0-1-log-2-x-dx-

Question Number 135127 by Dwaipayan Shikari last updated on 10/Mar/21

\int_{0}^{1} {l o g}^{2} (Γ (x)) d x

Answered by mathmax by abdo last updated on 11/Mar/21

we have Γ (x) . Γ (1 - x) = \frac{π}{\sin (π x)} \Rightarrow

\log (Γ (x)) + \log (Γ (1 - x)) = \log (π) - \log (\sin (π x)) \Rightarrow

\log {(Γ (x))}^{2} + 2 \log (Γ (x)) \log (Γ (1 - x)) + \log^{2} (Γ (1 - x))

= \log^{2} (π) - 2 \log π \log (\sin (π x)) + \log^{2} (\sin (π x)) \Rightarrow

\int_{0}^{1} \log^{2} (Γ (x)) dx + \int_{0}^{1} \log^{2} (Γ (1 - x)) dx + 2 \int_{0}^{1} \log (Γ (x)) \log (Γ (1 - x)) dx

= \log^{2} (π) - 2 \log π \int_{0}^{1} \log (\sin (π x)) dx + \int_{0}^{1} \log^{2} (\sin (π x)) dx

\int_{0}^{1} \log^{2} (Γ (1 - x)) dx =_{1 - x = t} \int_{0}^{1} \log^{2} (Γ (t)) dt

\int_{0}^{1} \log (\sin (π x)) dx =_{π x = t} \frac{1}{π} \int_{0}^{π} \log (sint) dt

= \frac{1}{π} \int_{0}^{\frac{π}{2}} \log (sint) dt + \frac{1}{π} \int_{\frac{π}{2}}^{π} \log (sint) dt (\to t = \frac{π}{2} + u)

= \frac{1}{π} (- \frac{π}{2} \log 2) + \frac{1}{π} (- \frac{π}{2} \log 2) = - \log (2)

\Rightarrow 2 \int_{0}^{1} \log^{2} (Γ (x)) dx + 2 \int_{0}^{1} \log (Γ (x)) . \log (Γ (1 - x)) dx

= \log^{2} π + 2 \log π \log 2 + \int_{0}^{1} \log^{2} (\sin (π x)) dx \dots be continued \dots