0-1-ln-1-x-x-dx-

Question Number 169873 by sciencestudent last updated on 11/May/22

\underset{0}{\int^{1}} \frac{l n (1 + x)}{x} d x = ?

Answered by ArielVyny last updated on 11/May/22

\sum_{n ⩾ 0} {(- 1)}^{n} x^{n} = \frac{1}{1 + x}

\sum_{n ⩾ 0} {(- 1)}^{n} \frac{x^{n + 1}}{n + 1} = l n (1 + x)

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

\sum_{n ⩾ 0} {(- 1)}^{n} \frac{1}{{(n + 1)}^{2}} = \int_{0}^{1} \frac{l n (1 + x)}{x} = \frac{π^{2}}{12}

Answered by Mathspace last updated on 11/May/22

\frac{d}{d x} l n (1 + x) = \frac{1}{1 + x} = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{n}

w i t h ∣ x ∣< 1 \Rightarrow l n (1 + x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} x^{n + 1} + c (c = 0)

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

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

= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} \int_{0}^{1} x^{n - 1} d x

= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n^{2}} = - δ (2)

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

= (2^{1 - x} - 1) ξ (x) \Rightarrow

δ (2) = (\frac{1}{2} - 1) ξ (2) = - \frac{1}{2} \frac{π^{2}}{6}

= - \frac{π^{2}}{12} \Rightarrow \int_{0}^{1} \frac{l n (1 + x)}{x} d x = \frac{π^{2}}{12}