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

∫ (x/(e^x − 1))dx, for x > 0

$$\int\:\frac{\mathrm{x}}{\mathrm{e}^{\mathrm{x}} \:−\:\mathrm{1}}\mathrm{dx},\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{for}\:\:\mathrm{x}\:>\:\mathrm{0} \\ $$

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 .

$$\int\:\:\frac{{x}}{{e}^{{x}} −\mathrm{1}}{dx}\:=\int\:\:\frac{{x}\:{e}^{−{x}} }{\mathrm{1}−{e}^{−{x}} }{dx}\:=\int\:\:\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{nx}} \right){xe}^{−{x}} \:{dx} \\ $$ $$=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\int\:\:{x}\:{e}^{−\left({n}+\mathrm{1}\right){x}} \:{dx}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:{A}_{{n}} \\ $$ $${A}_{{n}} =\int\:\:{x}\:{e}^{−\left({n}+\mathrm{1}\right){x}} {dx}\:=_{\left({n}+\mathrm{1}\right){x}\:={t}} \:\:\:\:\:\int\:\:\frac{{t}}{{n}+\mathrm{1}}\:{e}^{−{t}} \:\:\frac{{dt}}{{n}+\mathrm{1}} \\ $$ $$=\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\int\:\:{t}\:{e}^{−{t}} \:{dt}\:\:\:\:{and}\:{by}\:{parts}\: \\ $$ $$\int\:{t}\:{e}^{−{t}} \:{dt}\:=−{te}^{−{t}} \:+\int\:{e}^{−{t}} \:{dt}\:\:=−{t}\:{e}^{−{t}} \:−{e}^{−{t}} \:=−\left({t}+\mathrm{1}\right){e}^{−{t}} \:\Rightarrow \\ $$ $${A}_{{n}} =−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\left\{\:\left({t}+\mathrm{1}\right){e}^{−{t}} \:+{c}\right\}\:\Rightarrow\sum_{{n}=\mathrm{0}} ^{\infty} \:{A}_{{n}} =−\left({t}+\mathrm{1}\right){e}^{−{t}} \:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:−{c}\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$ $$=−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\left({t}+\mathrm{1}\right){e}^{−{t}} \:−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}{c}\:\:. \\ $$

Commented byTawa1 last updated on 21/Jun/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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

you are most welcome.

$${you}\:{are}\:{most}\:{welcome}. \\ $$

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