Advanced-calculus-0-1-x-2-ln-x-ln-1-x-dx-

Question Number 137251 by mnjuly1970 last updated on 31/Mar/21

\dots \dots A d v a n c e d \dots c a l c u l u s \dots \dots

ϕ = \int_{0}^{1} x^{2} l n (x) l n (1 - x) d x = ? ? ?

Answered by Ar Brandon last updated on 31/Mar/21

ϕ = \int_{0}^{1} x^{2} \ln (x) \ln (1 - x) dx = \frac{\partial}{\partial α} ∣_{α = 0} - \int_{0}^{1} x^{2 + α} \underset{n = 1}{\sum^{\infty}} \frac{x^{n}}{n} dx

= \frac{\partial}{\partial α} ∣_{α = 0} - \underset{n = 1}{\sum^{\infty}} \frac{1}{n} \int x^{2 + α + n} dx = \frac{\partial}{\partial α} ∣_{α = 0} - \underset{n = 1}{\sum^{\infty}} \frac{1}{n (3 + α + n)}

= \underset{n = 1}{\sum^{\infty}} \frac{1}{n {(n + 3)}^{2}} = \underset{n = 1}{\sum^{\infty}} (\frac{1}{9 n} - \frac{1}{9 (n + 3)} - \frac{1}{3 {(n + 3)}^{2}})

= \frac{1}{9} \underset{n = 1}{\sum^{\infty}} (\frac{1}{n} - \frac{1}{n + 3}) - \frac{1}{3} (\underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} - (1 + \frac{1}{2^{2}} + \frac{1}{3^{2}}))

= \frac{1}{9} (1 + \frac{1}{2} + \frac{1}{3}) - \frac{1}{3} (ζ (2) - (1 + \frac{1}{2^{2}} + \frac{1}{3^{2}}))

= \frac{11}{54} - \frac{π^{2}}{18} + \frac{49}{108} = \frac{71}{108} - \frac{π^{2}}{18}

Commented by Ar Brandon last updated on 31/Mar/21

:)

Commented by Dwaipayan Shikari last updated on 31/Mar/21

E v e n s h o r t e r :)

Commented by mnjuly1970 last updated on 31/Mar/21

t h a n k s a l o t m r b r a n d o n \dots

Commented by mnjuly1970 last updated on 31/Mar/21

g r a t e f u l \dots

Commented by Ar Brandon last updated on 31/Mar/21

{You}^{'} re welcome, Sir!

Commented by Dwaipayan Shikari last updated on 31/Mar/21

B u t, w h e n e v e r y o u a r e c a l c u l a t i n g

\int_{0}^{1} x^{π} {(1 - x)}^{e} l o g (x) l o g (1 - x) d x

y o u n e e d t h i s ν^{'} (e + 1, π + 1); -)

Commented by Ar Brandon last updated on 31/Mar/21

OK, bro. Thanks

Answered by Dwaipayan Shikari last updated on 31/Mar/21

\int_{0}^{1} x^{2} l o g (x) l o g (1 - x) d x = ν^{'} (3, 1)

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

\int_{0}^{1} x^{a - 1} {(1 - x)}^{b - 1} l o g (1 - x) l o g (x) = ν^{'} (a, b)

= \frac{\partial}{\partial a} (\frac{Γ (a) Γ^{'} (1) - ψ (a + 1) Γ (1) Γ (a)}{Γ (a + 1)})

= \frac{\partial}{\partial a} (\frac{- γ}{a} - \frac{ψ (a + 1)}{a}) = \frac{γ}{a^{2}} + \frac{ψ (1 + a)}{a^{2}} - \frac{ψ^{'} (a + 1)}{a} ∣_{a = 3}

= \frac{γ}{9} + \frac{ψ (4)}{9} - \frac{ψ^{'} (4)}{a} = \frac{γ - γ + 1 + \frac{1}{2} + \frac{1}{3}}{9} - \frac{1}{3} (\frac{π^{2}}{6} - 1 - \frac{1}{2^{2}} - \frac{1}{3^{2}})

= \frac{11}{54} + \frac{1}{3} (\frac{49}{36}) - \frac{π^{2}}{18} = \frac{71}{108} - \frac{π^{2}}{18}

Commented by mnjuly1970 last updated on 31/Mar/21

t h a n k y o u s o m u c h m r p a y a n . .

Commented by mnjuly1970 last updated on 31/Mar/21

v e r y n i c e . .