∫ (x/(e^x − 1))dx, for x


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

$\int \frac{x}{e^{x} - 1} dx, for x > 0$
Commented bymathmax by abdo last updated on 21/Jun/19
$∫ (x/(e^x −1))dx =∫ ((x e^(−x) )/(1−e^(−x) ))dx =∫ (Σ_(n=0) ^∞ e^(−nx) )xe^(−x) dx = Σ_(n=0) ^∞ ∫ x e^(−(n+1)x) dx =Σ_(n=0) ^∞ A_n A_n =∫ x e^(−(n+1)x) dx =_((n+1)x =t) ∫ (t/(n+1)) e^(−t) (dt/(n+1)) =(1/((n+1)^2 )) ∫ t e^(−t) dt and by parts ∫ t e^(−t) dt =−te^(−t) +∫ e^(−t) dt =−t e^(−t) −e^(−t) =−(t+1)e^(−t) ⇒ A_n =−(1/((n+1)^2 )){ (t+1)e^(−t) +c} ⇒Σ_(n=0) ^∞ A_n =−(t+1)e^(−t) Σ_(n=0) ^∞ (1/((n+1)^2 )) −cΣ_(n=0) ^∞ (1/((n+1)^2 )) =−(π^2 /6)(t+1)e^(−t) −(π^2 /6)c .$
$\int \frac{x}{e^{x} - 1} d x = \int \frac{x e^{- x}}{1 - e^{- x}} d x = \int (\sum_{n = 0}^{\infty} e^{- n x}) {x e}^{- x} d x$ $= \sum_{n = 0}^{\infty} \int x e^{- (n + 1) x} d x = \sum_{n = 0}^{\infty} A_{n}$ $A_{n} = \int x e^{- (n + 1) x} d x =_{(n + 1) x = t} \int \frac{t}{n + 1} e^{- t} \frac{d t}{n + 1}$ $= \frac{1}{{(n + 1)}^{2}} \int t e^{- t} d t a n d b y p a r t s$ $\int t e^{- t} d t = - {t e}^{- t} + \int e^{- t} d t = - t e^{- t} - e^{- t} = - (t + 1) e^{- t} \Rightarrow$ $A_{n} = - \frac{1}{{(n + 1)}^{2}} {(t + 1) e^{- t} + c} \Rightarrow \sum_{n = 0}^{\infty} A_{n} = - (t + 1) e^{- t} \sum_{n = 0}^{\infty} \frac{1}{{(n + 1)}^{2}} - c \sum_{n = 0}^{\infty} \frac{1}{{(n + 1)}^{2}}$ $= - \frac{π^{2}}{6} (t + 1) e^{- t} - \frac{π^{2}}{6} c .$
Commented byTawa1 last updated on 21/Jun/19

$God bless you sir$
Commented bymathmax by abdo last updated on 22/Jun/19

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