advanced-calculus-prove-that-n-1-1-n-H-n-n-2-0-1-ln-1-x-ln-1-x-x-dx-note-H-n-k-1-n-1-k-therefore-n-1-

Question Number 118924 by mnjuly1970 last updated on 21/Oct/20

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

p r o v e t h a t ::

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

n o t e :: H_{n} = \underset{k = 1}{\sum^{n}} \frac{1}{k}

t h e r e f o r e : \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n} H_{n}}{n^{2}} = \frac{- 5}{8} ζ (3) ✓

. . m . n . j u l y . 1970 \dots

Answered by mindispower last updated on 21/Oct/20

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

= - \sum_{k} \frac{1}{k} \int_{0}^{1} x^{n + k - 1} d x = - \sum_{k ⩾ 1} \frac{1}{(n + k) k}

= - \sum_{k ⩾ 1} \frac{1}{n} (\frac{1}{k} - \frac{1}{n + k}) = - \frac{1}{n} \sum_{k ⩽ n} \frac{1}{k} = - \frac{H_{n}}{n}

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

\frac{1}{n} \int x^{n - 1} l n (1 - x) d x = - \frac{H_{n}}{n^{2}}

\Rightarrow Σ \frac{{(- 1)}^{n - 1}}{n} \int_{0}^{1} x^{n - 1} l n (1 - x) d x = \sum_{n ⩾ 1} \frac{{(- 1)}^{n} H_{n}}{n^{2}}

\Rightarrow \int_{0}^{1} (\frac{1}{x} Σ \frac{{(- 1)}^{n - 1} x^{n}}{n}) l n (1 - x) d x = \sum_{n ⩾ 1} \frac{{(- 1)}^{n} H_{n}}{n^{2}}

\sum_{n ⩾ 1} \frac{{(- 1)}^{n - 1} x^{n}}{n} = l n (1 + x) \Leftrightarrow

\int_{0}^{1} \frac{l n (1 + x) l n (1 - x)}{x} d x = Σ \frac{{(- 1)}^{n} H_{n}}{n^{2}}

\int_{0}^{1} \frac{l n (1 + x) l n (1 - x)}{x} d x d o n e i n y o u r a n s w e r

Commented by mnjuly1970 last updated on 21/Oct/20

b r a v o b r a v o

m r p o w e r

g r a t e f u l s i r \dots

Commented by mindispower last updated on 21/Oct/20

w i t h e p l e a s u r