advanced-calculus-please-evaluate-1-n-1-H-n-2n-1-2-2-0-1-xln-x-ln-1-x-1-x-dx-note-H-n-1-1-2-1-3-1-n-0-1-

Question Number 137093 by mnjuly1970 last updated on 29/Mar/21

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

p l e a s e e v a l u a t e ::

1 : \underset{n = 1}{\sum^{\infty}} \frac{H_{n}}{{(2 n + 1)}^{2}} = ?

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

n o t e : H_{n} = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} = \int_{0}^{1} \frac{1 - x^{n}}{1 - x} d x

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

Λ = \int_{0}^{1} \frac{xln (x) \ln (1 - x)}{1 - x} dx

= {xln (x) \cdot \frac{\ln^{2} (1 - x)}{- 2} + \frac{1}{2} \int (1 + lnx) \ln^{2} (1 - x) dx}_{0}^{1}

= \frac{1}{2} \int_{0}^{1} (1 + lnx) \ln^{2} (1 - x) dx = \frac{1}{2} \int_{0}^{1} (1 + \ln (1 - x)) \ln^{2} xdx

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

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

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

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

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

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

= \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} - 1 + \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{3}} - 1 = ζ (2) + ζ (3) - 2 = \frac{π^{2}}{6} + ζ (3) - 2

Answered by Dwaipayan Shikari last updated on 29/Mar/21

ð (a, b) = \int_{0}^{1} x^{α - 1} {(1 - x)}^{β - 1} d x

\frac{\partial}{\partial a \partial b} ð (a, b) = \int_{0}^{1} l o g (x) l o g (1 - x) x^{α - 1} {(1 - x)}^{β - 1} d x

\frac{\partial ð (a, b)}{\partial a \partial b} = \frac{\partial (\frac{Γ (a) Γ (b)}{Γ (a + b)})}{\partial a \partial b} = \frac{\partial}{\partial a} (\frac{Γ (a) Γ (a + b) Γ^{'} (b) - Γ (b) Γ (a) Γ^{'} (a + b)}{Γ^{2} (a + b)})

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

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

a = 1 b = 0

Answered by Dwaipayan Shikari last updated on 29/Mar/21

\underset{n = 1}{\sum^{\infty}} \frac{H_{n}}{{(2 n + 1)}^{2}} = \int_{0}^{1} \underset{n = 1}{\sum^{\infty}} \frac{H_{n}}{2 n + 1} x^{2 n + 1}

\underset{n = 1}{\sum^{\infty}} H_{n} x^{2 n + α} = - x^{α} \frac{l o g (1 - x^{2})}{1 - x^{2}}

\underset{n = 1}{\sum^{\infty}} \frac{H_{n}}{2 n + α + 1} = - \int_{0}^{1} \frac{x^{α} l o g (1 - x^{2})}{1 - x^{2}} d x

\underset{n = 1}{\sum^{\infty}} \frac{H_{n}}{{(2 n + α + 1)}^{2}} = \frac{\partial}{\partial α} \int_{0}^{1} \frac{x^{α} l o g (1 - x^{2})}{1 - x^{2}} d x

= \frac{\partial}{\partial α} \int_{0}^{1} \frac{u^{\frac{α - 1}{2}} l o g (1 - u)}{1 - u} d u = \frac{\partial}{\partial α} . \frac{\partial}{\partial ϑ} ∣_{ϑ = 0} (\frac{Γ (\frac{α + 1}{2}) Γ (ϑ)}{Γ (\frac{α}{2} + \frac{1}{2} + ϑ)}) ∣

Commented by mnjuly1970 last updated on 29/Mar/21

t h a n k s a l o t \dots

Answered by Ñï= last updated on 29/Mar/21

\int_{0}^{1} \frac{x l n (x) l n (1 - x)}{1 - x} d x = \int_{0}^{1} \frac{(1 - x) l n (x) l n (1 - x)}{x} d x

= \int_{0}^{1} \frac{l n (x) l n (1 - x)}{x} d x - \int_{0}^{1} l n (x) l n (1 - x) d x

= \int_{0}^{1} \frac{{L i}_{2} (x)}{x} d x + \int_{0}^{1} x (\frac{l n (1 - x)}{x} - \frac{l n x}{1 - x}) d x

= {L i}_{3} (1) + \int_{0}^{1} l n (1 - x) d x - \int_{0}^{1} (1 - x) \frac{l n (1 - x)}{x} d x

= {L i}_{3} (1) + \int_{0}^{1} l n (1 - x) d x - \int_{0}^{1} \frac{l n (1 - x)}{x} - l n (1 - x) d x

= {L i}_{3} (1) + 2 \int_{0}^{1} l n x d x + {L i}_{2} (1)

= {L i}_{3} (1) + {L i}_{2} (1) - 2

Commented by mnjuly1970 last updated on 29/Mar/21

t h a n k s a l o t \dots