dilogarithm-integral-calculate-0-1-li-2-1-x-2-dx-

Question Number 135055 by mnjuly1970 last updated on 09/Mar/21

\dots . d i l o g a r i t h m i n t e g r a l \dots .

c a l c u l a t e :::

ϕ = \int_{0}^{1} {l i}_{2} (1 - x^{2}) d x = ?

Answered by Ñï= last updated on 10/Mar/21

\int_{0}^{1} {L i}_{2} (1 - x^{2}) d x = {x L i}_{2} (1 - x^{2}) ∣_{0}^{1} - \int_{0}^{1} 4 x^{2} \cdot \frac{l n x}{1 - x^{2}} d x

= 4 \int_{0}^{1} (1 - \frac{1}{1 - x^{2}}) l n x d x = 4 \int_{0}^{1} l n x d x - 4 \int_{0}^{1} \frac{l n x}{1 - x^{2}} d x

= 4 (x l n x - x) ∣_{0}^{1} - 4 [\frac{1}{2} l n \frac{1 + x}{1 - x} l n x ∣_{0}^{1} - \frac{1}{2} \int_{0}^{1} \frac{l n (1 + x) - l n (1 - x)}{x} d x]

= - 4 + 2 \int_{0}^{1} \frac{l n (1 + x) - l n (1 - x)}{x} d x

= - 4 + 2 [- {L i}_{2} (- 1) + {L i}_{2} (1)]

= - 4 + 3 {L i}_{2} (1)

= \frac{π^{2}}{2} - 4

Commented by mnjuly1970 last updated on 10/Mar/21

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