advanced-calculus-first-prove-that-1-0-1-ln-1-x-ln-1-x-x-dx-5-8-3-then-conclude-that-2-0-1-ln-2-1-

Question Number 135252 by mnjuly1970 last updated on 11/Mar/21

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

f i r s t p r o v e t h a t ::

ϕ_{1} = \int_{0}^{1} \frac{l n (1 - x) l n (1 + x)}{x} d x = \frac{- 5}{8} ζ (3)

t h e n c o n c l u d e t h a t :

ϕ_{2} = \int_{0}^{1} \frac{{l n}^{2} (1 + x)}{x} d x = \frac{ζ (3)}{4}

\dots . m . n \dots

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

Φ_{1} = \int_{0}^{1} \frac{\ln (1 - x)}{x} \ln (1 + x) dx we have \ln^{^{'}} (1 + x) = \frac{1}{1 + x} = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{n}

\Rightarrow \ln (1 + x) = \sum_{n = 0}^{\infty} {(- 1)}^{n} \frac{x^{n + 1}}{n + 1} + c (c = 0) = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \frac{x^{n}}{n} \Rightarrow

\frac{\ln (1 + x)}{x} = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \frac{x^{n - 1}}{n} \Rightarrow Φ_{1} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} \int_{0}^{1} x^{n - 1} \ln (1 - x) dx

A_{n} = \int_{0}^{1} x^{n - 1} \ln (1 - x) dx = {[\frac{x^{n} - 1}{n} \ln (1 - x)]}_{0}^{1} + \int_{0}^{1} \frac{x^{n} - 1}{n (1 - x)} dx

= - \frac{1}{n} \int_{0}^{1} \frac{x^{n} - 1}{x - 1} dx = - \frac{1}{n} \int_{0}^{1} (1 + x + x^{2} + \dots . + x^{n - 1}) dx

= - \frac{1}{n} {[x + \frac{x^{2}}{2} + \dots . + \frac{x^{n}}{n}]}_{0}^{1} = - \frac{1}{n} (1 + \frac{1}{2} + \dots . + \frac{1}{n}) = - \frac{H_{n}}{n} \Rightarrow

Φ_{1} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} H_{n} = \sum_{n = 1}^{\infty} {(- 1)}^{n} \frac{H_{n}}{n^{2}}

rest to find the value of this serie \dots . be continued \dots

Commented by mnjuly1970 last updated on 12/Mar/21

t h a n k y o u s o m u c h

Σ \frac{{(- 1)}^{n} H_{n}}{n^{2}} = \frac{- 5}{8} ζ (3)