ln-1-e-u-du-

Question Number 93553 by Ar Brandon last updated on 13/May/20

\int l n (1 + e^{u}) du

Commented by Tony Lin last updated on 13/May/20

l e t - e^{u} = z d z = z d u

\int \frac{l n (1 - z)}{z} d z

= - {L i}_{2} (z) + c

= - {L i}_{2} (- e^{u}) + c

{L i}_{2} (z) = \underset{k = 1}{\sum^{\infty}} \frac{z^{k}}{k^{2}}

Commented by Ar Brandon last updated on 13/May/20

Okay thanks ��