calcilate-0-1-ln-1-x-2-x-2-dx-

Question Number 32477 by prof Abdo imad last updated on 25/Mar/18

c a l c i l a t e \int_{0}^{1} \frac{l n (1 - x^{2})}{x^{2}} d x

Commented by prof Abdo imad last updated on 31/Mar/18

f o r ∣ t ∣< 1 \frac{1}{1 - t} = \sum_{n = 0}^{\infty} t^{n} \Rightarrow - l n ∣ 1 - t ∣= \sum_{n = 0}^{\infty} \frac{t^{n + 1}}{n + 1}

= \sum_{n = 1}^{\infty} \frac{t^{n}}{n} \Rightarrow \sum_{n = 1}^{\infty} \frac{t^{n}}{n} = - l n (1 - t) a n d

l n (1 - x^{2}) = - \sum_{n = 1}^{\infty} \frac{x^{2 n}}{n} \Rightarrow \frac{l n (1 - x^{2})}{x^{2}} = - \sum_{n = 1}^{\infty} \frac{x^{2 n - 2}}{n}

\Rightarrow \int_{0}^{1} \frac{l n (1 - x^{2})}{x^{2}} d x = - \sum_{n = 1}^{\infty} \frac{1}{n (2 n - 1)}

l e t p u t S_{n} = \sum_{k = 1}^{n} \frac{1}{k (2 k - 1)} w e h a v e

\frac{1}{2} S_{n} = \sum_{k = 1}^{n} \frac{1}{2 k (2 k - 1)} = \sum_{k = 1}^{n} (\frac{1}{2 k - 1} - \frac{1}{2 k})

= \sum_{k = 1}^{n} \frac{1}{2 k - 1} - \frac{1}{2} H_{n} b u t

\sum_{k = 1}^{n} \frac{1}{2 k - 1} = 1 + \frac{1}{3} + \dots . + \frac{1}{2 n - 1}

= 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{2 n} - \frac{1}{2} - \frac{1}{4} - \dots . - \frac{1}{2 n}

= H_{2 n} - \frac{1}{2} H_{n}

\frac{1}{2} S_{n} = H_{2 n} - H_{n} = l n (2 n) + γ - l n (n) - γ + o (1)

= l n (\frac{2 n}{n}) + γ + o (1) \to l n (2) \Rightarrow {l i m}_{n \to \infty} S_{n} = 2 l n (2)

\Rightarrow \int_{0}^{1} \frac{l n (1 - x^{2})}{x^{2}} = - 2 l n (2) .