0-1-ln-2-x-Li-2-x-x-dx-

Question Number 162177 by mnjuly1970 last updated on 27/Dec/21

ϕ = \int_{0}^{1} \frac{\ln^{2} (x) . {Li}_{2} (x)}{x^{}} d x = ?

Answered by Ar Brandon last updated on 27/Dec/21

ϕ = \int_{0}^{1} \frac{{l o g}^{2} {x L i}_{2} (x)}{x} d x = \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} \int_{0}^{1} \frac{x^{n} {l o g}^{2} x}{x} d x

= \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} ({[\frac{1}{3} x^{n} {l o g}^{3} x]}_{0}^{1} - \frac{n}{3} \int_{0}^{1} x^{n - 1} {l o g}^{3} x d x)

= - \frac{1}{3} \underset{n = 1}{\sum^{\infty}} \frac{1}{n} \cdot \frac{\partial^{3}}{\partial α^{3}} ∣_{α = n - 1} \int_{0}^{1} x^{α} d x

= - \frac{1}{3} \underset{n = 1}{\sum^{\infty}} \frac{1}{n} \cdot \frac{\partial^{3}}{\partial α^{3}} ∣_{α = n - 1} \frac{1}{α + 1}

= - \frac{1}{3} \underset{n = 1}{\sum^{\infty}} \frac{1}{n} (\frac{- 6}{n^{4}}) = 2 \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{5}} = 2 ζ (5)

Commented by mnjuly1970 last updated on 27/Dec/21

g r a t e f u l s i r b r a n d o n

Commented by Ar Brandon last updated on 27/Dec/21

My pleasure, Sir .