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Question Number 140531 by Willson last updated on 09/May/21

Prove that {\int_{0}}^{x} \frac{t}{e^{t} - 1} dt = \underset{n = 1}{\sum^{+ \infty}} \frac{{(1 - e^{- x})}^{n}}{n^{2}}

Answered by mathmax by abdo last updated on 09/May/21

\int_{0}^{x} \frac{t}{e^{t} - 1} dt = \int_{0}^{x} \frac{{te}^{- t}}{1 - e^{- t}} dt = \int_{0}^{x} {te}^{- t} \sum_{n . 0}^{\infty} e^{- nt} dt

= \sum_{n = 0}^{\infty} \int_{0}^{x} t e^{- (n + 1) t} dt =_{(n + 1) t = y} \sum_{n = 0}^{\infty} \int_{0}^{(n + 1) x} \frac{y}{n + 1} e^{- y} \frac{dy}{n + 1}

= \sum_{n = 0}^{\infty} \frac{1}{{(n + 1)}^{2}} \int_{0}^{(n + 1) x} {ye}^{- y} dy and

\int_{0}^{(n + 1) x} y e^{- y} dy = {[- y e^{- y}]}_{0}^{(n + 1) x} + \int_{0}^{(n + 1) x} e^{- y} dy

= - (n + 1) x e^{- (n + 1) x} + {[- e^{- y}]}_{0}^{(n + 1) x}

= - (n + 1) {xe}^{- (n + 1) x} + 1 - e^{- (n + 1) x}

= - ((n + 1) x + 1) e^{- (n + 1) x} + 1 \Rightarrow

\int_{0}^{x} \frac{t}{e^{t} - 1} dt = \sum_{n = 0}^{\infty} \frac{1}{{(n + 1)}^{2}} (1 - ((n + 1) x) e^{- (n + 1) x} + 1)

= \sum_{n = 1}^{\infty} \frac{1}{n^{2}} (1 - nx) e^{- nx} + \sum_{n = 1}^{\infty} \frac{1}{n^{2}}

= \frac{π^{2}}{6} + \sum_{n = 1}^{\infty} \frac{(1 - nx) e^{- nx}}{n^{2}} \dots .