Advanced-Integral-Prove-that-0-1-1-x-1-x-x-2-log-x-dx-proof-0-1-1-x-2-1-x-3-log-x-dx-f-a

Question Number 142275 by mnjuly1970 last updated on 29/May/21

\dots \dots . A d v a n c e d \dots * * * * * \dots I n t e g r a l \dots \dots

P r o v e t h a t :: Φ := \int_{0}^{1} \frac{1 - x}{(1 - x + x^{2}) l o g (x)} d x =

p r o o f ::

Φ := \int_{0}^{1} \frac{1 - x^{2}}{(1 - x^{3}) l o g (x)} d x

f (a) := \int_{0}^{1} \frac{1 - x^{a}}{(1 - x^{3}) l o g (x)}

Φ := f (2) \dots \dots . . ✓

f^{'} (a) := \int_{0}^{1} \frac{- x^{a} l o g (x)}{(1 - x^{3}) l o g (x)} = \int_{0}^{1} \frac{- x^{a}}{1 - x^{3}} d x (★)

(★) :: x^{3} = y \Rightarrow f^{'} (a) := \frac{1}{3} \int_{0}^{1} \frac{y^{\frac{- 2}{3}} - y^{\frac{a}{3} - \frac{2}{3}}}{1 - y} d y

:= \frac{1}{3} \int_{0}^{1} \frac{y^{\frac{- 2}{3}} - 1 + 1 - y^{\frac{a}{3} - \frac{1}{3}}}{1 - y} d y

:= \frac{1}{3} (ψ (\frac{a}{3} + \frac{2}{3}) - ψ (\frac{2}{3}))

f (a) := l o g (Γ (\frac{a}{3} + \frac{2}{3})) - \frac{a}{3} ψ (\frac{2}{3}) + C

f (0) := 0 = l o g (Γ (\frac{2}{3})) + C

C := - l o g (Γ (\frac{2}{3}))

Φ := f (2) = l o g (Γ (\frac{4}{3})) - \frac{2}{3} ψ (\frac{2}{3}) - l o g (Γ (\frac{2}{3}))

:= l o g (\frac{Γ (\frac{4}{3})}{Γ (\frac{2}{3})}) - \frac{2}{3} ψ (\frac{2}{3}) \dots . ✓