nice-integral-0-1-li-2-1-x-2-x-dx-m-n-

Question Number 142773 by mnjuly1970 last updated on 05/Jun/21

\dots \dots . n i c e \dots \dots i n t e g r a l \dots \dots

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

\dots \dots . m . n \dots

Answered by mnjuly1970 last updated on 05/Jun/21

ϕ := \int_{0}^{1} \frac{{l i}_{2} (x)}{1 + x} d x = {[{l i}_{2} (x) . l n (1 + x)]}_{0}^{1} + \int_{0}^{1} \frac{l n (1 - x) . l n (1 + x)}{x} d x

= \frac{π^{2}}{6} l n (2) - \frac{5}{8} ζ (3) \dots

d e r i v e d e a r l i e r :

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

Answered by qaz last updated on 05/Jun/21

ϕ = \int_{0}^{1} \frac{{Li}_{2} (1 - x)}{2 - x} dx = \int_{0}^{1} \frac{{Li}_{2} (x)}{1 + x} dx = \frac{π^{2}}{6} \ln 2 + \int_{0}^{1} \frac{\ln (1 + x) \ln (1 - x)}{x} dx

\int_{0}^{1} \frac{\ln (1 + x) \ln (1 - x)}{x} dx

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

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

= - \int_{0}^{1} \frac{\ln (1 - x^{2}) lnx}{1 - x^{2}} (- 2 x) dx - \frac{1}{2} \cdot \frac{ζ (3)}{4} + \int_{0}^{1} \frac{\ln (1 - x) lnx}{1 - x} (- 1) dx

= 2 \int_{0}^{1} \frac{\ln (1 - x^{2}) lnx}{1 - x^{2}} xdx - \frac{ζ (3)}{8} - \int_{0}^{1} \frac{\ln (1 - x) lnx}{1 - x} dx

= \frac{1}{2} \int_{0}^{1} \frac{\ln (1 - u) lnu}{1 - u} du - \frac{ζ (3)}{8} - \int_{0}^{1} \frac{\ln (1 - x) lnx}{1 - x} dx

= - \frac{1}{2} \int_{0}^{1} \frac{\ln (1 - x) lnx}{x} dx - \frac{ζ (3)}{8}

= - \frac{1}{2} \int_{0}^{1} \frac{{Li}_{2} (x)}{x} dx - \frac{ζ (3)}{8}

= - \frac{1}{2} ζ (3) - \frac{ζ (3)}{8}

= - \frac{5}{8} ζ (3)

\Rightarrow ϕ = \frac{π^{2}}{6} \ln 2 - \frac{5}{8} ζ (3)

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\int_{0}^{1} \frac{\ln (1 - x) \ln (1 + x)}{x} dx

= \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{n} \int_{0}^{1} x^{n - 1} \ln (1 - x) dx

= \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{n} (- \frac{H_{n}}{n})

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

= - \frac{5}{8} ζ (3)

Commented by mnjuly1970 last updated on 05/Jun/21

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