Using a series expansion under the integral, show that $$\int _ { 0 } ^ { + \infty } \ln \left( 1 - e ^ { - u } \right) \mathrm { d } u = - \frac { \pi ^ { 2 } } { 6 }$$
Using a series expansion under the integral, show that
$$\int _ { 0 } ^ { + \infty } \ln \left( 1 - e ^ { - u } \right) \mathrm { d } u = - \frac { \pi ^ { 2 } } { 6 }$$