x-e-x-1-dx-for-x-gt-0-

Question Number 62453 by Tawa1 last updated on 21/Jun/19

\int \frac{x}{e^{x} - 1} dx, for x > 0

Commented by mathmax by abdo last updated on 21/Jun/19

\int \frac{x}{e^{x} - 1} d x = \int \frac{x e^{- x}}{1 - e^{- x}} d x = \int (\sum_{n = 0}^{\infty} e^{- n x}) {x e}^{- x} d x

= \sum_{n = 0}^{\infty} \int x e^{- (n + 1) x} d x = \sum_{n = 0}^{\infty} A_{n}

A_{n} = \int x e^{- (n + 1) x} d x =_{(n + 1) x = t} \int \frac{t}{n + 1} e^{- t} \frac{d t}{n + 1}

= \frac{1}{{(n + 1)}^{2}} \int t e^{- t} d t a n d b y p a r t s

\int t e^{- t} d t = - {t e}^{- t} + \int e^{- t} d t = - t e^{- t} - e^{- t} = - (t + 1) e^{- t} \Rightarrow

A_{n} = - \frac{1}{{(n + 1)}^{2}} {(t + 1) e^{- t} + c} \Rightarrow \sum_{n = 0}^{\infty} A_{n} = - (t + 1) e^{- t} \sum_{n = 0}^{\infty} \frac{1}{{(n + 1)}^{2}} - c \sum_{n = 0}^{\infty} \frac{1}{{(n + 1)}^{2}}

= - \frac{π^{2}}{6} (t + 1) e^{- t} - \frac{π^{2}}{6} c .

Commented by Tawa1 last updated on 21/Jun/19

God bless you sir

Commented by mathmax by abdo last updated on 22/Jun/19

y o u a r e m o s t w e l c o m e .