0-1-ln-1-x-x-2-x-3-x-n-x-dx-

Question Number 130781 by EDWIN88 last updated on 28/Jan/21

\int_{0}^{1} \frac{\ln (1 + x + x^{2} + x^{3} + \dots + x^{n})}{x} d x ?

Answered by mathmax by abdo last updated on 29/Jan/21

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

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

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

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

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

= - \sum_{p = 1}^{\infty} \frac{1}{p (n + 1) p} = - \sum_{p = 1}^{\infty} \frac{1}{(n + 1) p^{2}} = - \frac{π^{2}}{6 (n + 1)} \Rightarrow

A_{n} = - \frac{1}{n + 1} . \frac{π^{2}}{6} + \frac{π^{2}}{6} = \frac{π^{2}}{6} (1 - \frac{1}{n + 1}) = \frac{n π^{2}}{6 (n + 1)}

Commented by mathmax by abdo last updated on 29/Jan/21

remark \lim_{n \to + \infty} A_{n} = \frac{π^{2}}{6}

Commented by Lordose last updated on 29/Jan/21

Very nice sir

Answered by mnjuly1970 last updated on 29/Jan/21

ϕ_{n} = \int_{0}^{1} \frac{1}{x} l n (1 - x^{n + 1}) + {l i}_{2} (1)

= - \sum_{m ⩾ 1} \frac{1}{m} \int_{0}^{1} \frac{x^{n m + m}}{x} d x + \frac{π^{2}}{6}

= - \sum_{m ⩾ 1} \frac{1}{m^{2} (n + 1)} + \frac{π^{2}}{6}

= - \frac{1}{n + 1} ζ (2) + \frac{π^{2}}{6}

ϕ_{n} = - \frac{π^{2}}{6 (n + 1)} + \frac{π^{2}}{6}

i f n \to \infty t h e n ϕ_{n} \to \frac{π^{2}}{6}

Answered by Dwaipayan Shikari last updated on 29/Jan/21

\int_{0}^{1} \frac{l o g (1 + x + x + x^{3} + . . + x^{n})}{x} d x

= \int_{0}^{1} \frac{l o g (1 - x^{n + 1})}{x} - \frac{l o g (1 - x)}{x} d x

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

= - \underset{k = 1}{\sum^{\infty}} \frac{1}{(n + 1) k^{2}} + \frac{π^{2}}{6} = \frac{π^{2}}{6} (\frac{n}{n + 1})