0-1-ln-1-x-x-2-x-dx-0-1-ln-1-x-3-ln-1-x-x-dx-pi-2-6-0-1-x-3n-1-n-dx-n-1-1-3n-2-pi-2-18-pi-2-9-

Question Number 139224 by mnjuly1970 last updated on 24/Apr/21

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

ϕ = \int_{0}^{1} \frac{l n (1 - x^{3}) - l n (1 - x)}{x} d x

= Ω + \frac{π^{2}}{6}

Ω = - Σ \int_{0}^{1} \frac{x^{3 n - 1}}{n} d x = - \underset{n = 1}{\sum^{\infty}} \frac{1}{3 n^{2}}

= \frac{- π^{2}}{18} \dots . . ≽ ϕ = \frac{π^{2}}{9}

\dots \dots . \frac{π^{2}}{9} - \frac{π^{2}}{24} = \frac{5 π^{2}}{72}

Commented by qaz last updated on 24/Apr/21

G r e a t, S i r, i w i l l c o m p l e t e t h a t i n t e g r a l .

Commented by mnjuly1970 last updated on 24/Apr/21

i a p p o l o g i z e . t h a n k y o u m a s t e r \dots

Answered by phanphuoc last updated on 24/Apr/21

π^{2} / 24 ? ? ?

Answered by mathmax by abdo last updated on 25/Apr/21

Φ = \int_{0}^{1} \frac{\log (x^{2} + x + 1)}{x} dx \Rightarrow Φ = \int_{0}^{1} \frac{\log (1 - x^{3}) - \log (1 - x)}{x} dx

= \int_{0}^{1} \frac{\log (1 - x^{3})}{x} dx - \int_{0}^{1} \frac{\log (1 - x)}{x} dx

we have \log^{^{'}} (1 - x) = - \frac{1}{1 - x} = - \sum_{n = 0}^{\infty} x^{n} \Rightarrow \log (1 - x) = - \sum_{n = 0}^{\infty} \frac{x^{n + 1}}{n + 1}

= - \sum_{n = 1}^{\infty} \frac{x^{n}}{n} \Rightarrow \frac{\log (1 - x)}{x} = - \sum_{n = 1}^{\infty} \frac{x^{n - 1}}{n} \Rightarrow

\int_{0}^{1} \frac{\log (1 - x)}{x} dx = - \sum_{n = 1}^{\infty} \frac{1}{n} \int_{0}^{1} x^{n - 1} dx = - \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = - \frac{π^{2}}{6}

also \log (1 - x^{3}) = - \sum_{n = 1}^{\infty} \frac{x^{3 n}}{n} \Rightarrow \frac{\log (1 - x^{3})}{x} = - \sum_{n = 1}^{\infty} \frac{x^{3 n - 1}}{n}

\Rightarrow \int_{0}^{1} \frac{\log (1 - x^{3})}{x} dx = - \sum_{n = 1}^{\infty} \frac{1}{n} \times \frac{1}{3 n} = - \frac{1}{3} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = - \frac{1}{3} \times \frac{π^{2}}{6}

= - \frac{π^{2}}{18} \Rightarrow Φ = - \frac{π^{2}}{18} + \frac{π^{2}}{6} = - \frac{π^{2}}{18} + \frac{3 π^{2}}{18} = \frac{2 π^{2}}{18} = \frac{π^{2}}{9}