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Question Number 166437 by mnjuly1970 last updated on 20/Feb/22
calculate Ω=∫_0 ^( 1) (( ln(1−x ).ln(1+ x ))/x^( 2) ) dx=? −−−−−−−
calculateΩ=∫01ln(1−x).ln(1+x)x2dx=?−−−−−−−
Answered by qaz last updated on 21/Feb/22
∫_0 ^1 ((ln(1−x)ln(1+x))/x^2 )dx =∫_0 ^1 ((1/x)−1)(((ln(1+x))/(x−1))+((ln(1−x))/(1+x)))dx =−∫_0 ^1 ((ln(1+x))/x)dx+∫_0 ^1 ((ln(1−x))/x)dx−2∫_0 ^1 ((ln(1−x))/(1+x))dx =−(1/(12))π^2 +∫_0 ^1 ((ln(1−x))/x)dx−2∫_1 ^2 ((ln(2−x))/x)dx =−(1/(12))π^2 +∫_0 ^1 ((ln(1−x))/x)dx−2∫_(1/2) ^1 ((ln2+ln(1−x))/x)dx =−(1/(12))π^2 −2ln^2 2+Li_2 (1)−2Li_2 ((1/2)) =−(1/(12))π^2 −3ln^2 2
∫01ln(1−x)ln(1+x)x2dx=∫01(1x−1)(ln(1+x)x−1+ln(1−x)1+x)dx=−∫01ln(1+x)xdx+∫01ln(1−x)xdx−2∫01ln(1−x)1+xdx=−112π2+∫01ln(1−x)xdx−2∫12ln(2−x)xdx=−112π2+∫01ln(1−x)xdx−2∫1/21ln2+ln(1−x)xdx=−112π2−2ln22+Li2(1)−2Li2(12)=−112π2−3ln22

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