0-1-Li-2-x-log-1-x-dx-Li-2-x-polylogaritm-function-

Question Number 152019 by mathdanisur last updated on 25/Aug/21

Ω = \underset{0}{\int^{1}} {L i}_{2} (x) l o g (1 + x) d x = ?

{L i}_{2} (x) - p o l y l o g a r i t m f u n c t i o n

Answered by mnjuly1970 last updated on 25/Aug/21

Answered by Olaf_Thorendsen last updated on 25/Aug/21

Ω = \int_{0}^{1} {Li}_{2} (x) \log (1 + x) d x

{Li}_{2} (x) = \underset{k = 1}{\sum^{\infty}} \frac{x^{k}}{k^{2}}

\int {Li}_{2} (x) d x = \underset{k = 1}{\sum^{\infty}} \frac{x^{k + 1}}{k^{2} (k + 1)}

Ω = {[\underset{k = 1}{\sum^{\infty}} \frac{x^{k + 1}}{k^{2} (k + 1)} . \log (1 + x)]}_{0}^{1} - \int_{0}^{1} \underset{k = 1}{\sum^{\infty}} \frac{x^{k + 1}}{k^{2} (k + 1)} . \frac{d x}{1 + x}

Ω = \log 2 \underset{k = 1}{\sum^{\infty}} \frac{1}{k^{2} (k + 1)} - \underset{k = 1}{\sum^{\infty}} \frac{1}{k^{2} (k + 1)} \int_{0}^{1} \frac{x^{k + 1}}{1 + x} d x

\int_{0}^{1} \frac{x^{k + 1}}{1 + x} d x = - \int_{0}^{1} \frac{1 - x^{k + 1}}{1 + x} d x + \int_{0}^{1} \frac{d x}{1 + x}

\int_{0}^{1} \frac{x^{k + 1}}{1 + x} d x = - \int_{0}^{1} \underset{p = 0}{\sum^{k}} x^{p} d x + \int_{0}^{1} \frac{d x}{1 + x}

\int_{0}^{1} \frac{x^{k + 1}}{1 + x} d x = {[- \underset{p = 0}{\sum^{k}} \frac{x^{p + 1}}{p + 1} d x + \log (1 + x)]}_{0}^{1}

\int_{0}^{1} \frac{x^{k + 1}}{1 + x} d x = - \underset{p = 0}{\sum^{k}} \frac{1}{p + 1} d x + \log 2 = \log 2 - H_{k + 1}

Ω = \log 2 \underset{k = 1}{\sum^{\infty}} \frac{1}{k^{2} (k + 1)} - \underset{k = 1}{\sum^{\infty}} \frac{\log 2 - H_{k + 1}}{k^{2} (k + 1)}

Ω = \underset{k = 1}{\sum^{\infty}} \frac{H_{k + 1}}{k^{2} (k + 1)}

\dots b r e a k \dots I c o m e b a c k

Commented by mathdanisur last updated on 25/Aug/21

Alot cool S er, thank you, please

Commented by mathdanisur last updated on 26/Aug/21

Answer Ser

Answered by Kamel last updated on 25/Aug/21

Commented by mathdanisur last updated on 26/Aug/21

The answer is wrong

Answered by qaz last updated on 25/Aug/21

Ω = [(1 + x) \ln (1 + x) - (1 + x)] {Li}_{2} (x) ∣_{0}^{1} + \int_{0}^{1} (1 + x) [\ln (1 + x) - 1] \frac{\ln (1 - x)}{x} dx

= \frac{π^{2}}{3} \ln 2 - \frac{π^{2}}{3} + \int_{0}^{1} (\frac{\ln (1 + x) \ln (1 - x)}{x} - \frac{\ln (1 - x)}{x} + \ln (1 + x) \ln (1 - x) - \ln (1 - x)) dx

= \frac{π^{2}}{3} \ln 2 - \frac{π^{2}}{3} - \frac{5}{8} ζ (3) + \frac{π^{2}}{6} + (- \frac{π^{2}}{6} + \ln^{2} 2 - 2 \ln 2 + 2) + 1

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

Commented by mathdanisur last updated on 26/Aug/21

The answer is wrong Ser