"Evaluating the integral,∫0∞ln⁡(1−e−x)dx" solved and explained

Adolfo Hebert

Answered question

2022-04-05

Evaluating the integral,

\int_{0}^{\infty} \ln (1 - e^{- x}) d x

armejantm925

Beginner2022-04-06Added 20 answers

One route to evaluating the integral is

- \int_{0}^{\infty} \ln (1 - e^{- x}) d x = \int_{0}^{\infty} (e^{- x} + \frac{e^{- 2 x}}{2} + \frac{e^{- 3 x}}{3} + \dots) d x

= \int_{0}^{\infty} e^{- x} d x + \frac{1}{2} \int_{0}^{\infty} e^{- 2 x} d x + \frac{1}{3} \int_{0}^{\infty} e^{- 3 x} d x + \dots

= 1 + \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{3} \cdot \frac{1}{3} + \frac{1}{4} \cdot \frac{1}{4} \dots

= ζ (2) = \frac{π^{2}}{6}

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