Question-156617

Question Number 156617 by mnjuly1970 last updated on 13/Oct/21

Answered by mindispower last updated on 14/Oct/21

e^{- x} = t

= \int_{0}^{1} \frac{l n (1 + t) (1 - t)}{1 + t} . \frac{d t}{t}

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

= - {l n}^{2} (2) - {L i}_{2} (- 1) = \frac{π^{2}}{12} - {l n}^{2} (2) = \frac{\frac{π^{2}}{12} - {l n}^{2} (2)}{\frac{π^{2}}{12} - \frac{{l n}^{2} (2)}{2}} . {L i}_{2} (\frac{1}{2})

{L i}_{2} (\frac{1}{2}) = \frac{π^{2}}{12} - \frac{{l n}^{2} (2)}{2}