Question Number 35675 by abdo imad last updated on 21/May/18
$${calculate}\:\:\int_{\mathrm{1}} ^{\mathrm{3}} \:\:\:\frac{{x}}{{e}^{{x}} \:−\mathrm{1}}{dx}\:.. \\ $$
Commented by prof Abdo imad last updated on 25/May/18
![I = ∫_1 ^3 ((x e^(−x) )/(1−e^(−x) ))dx =∫_1 ^3 ( Σ_(n=0) ^∞ e^(−nx) )x e^(−x) dx = Σ_(n=0) ^∞ ∫_1 ^3 x e^(−(n+1)x) dx = Σ_(n=0) ^∞ A_n with A_n = ∫_1 ^3 x.e^(−(n+1)x) dx .changement (n+1)x=t give A_n =∫_(n+1) ^(3(n+1)) (t/(n+1)) e^(−t) (dt/(n+1)) = (1/((n+1)^2 )) ∫_(n+1) ^(3(n+1)) t^ e^(−t) dt by parts ∫_(n+1) ^(3(n+1)) t e^(−t) dt = [ −t e^(−t) ]_(n+1) ^(3(n+1)) +∫_(n+1) ^(3(n+1)) e^(−t) dt = (n+1)e^(−(n+1)) −3(n+1) e^(−3(n+1)) +[ −e^(−t) ]_(n+1) ^(3(n+1)) =(n+1) e^(−(n+1)) −3(n+1) e^(−3(n+1)) +e^(−(n+1)) −e^(−3(n+1)) A_n = (e^(−(n+1)) /(n+1)) −3(e^(−3(n+1)) /(n+1)) + (e^(−(n+1)) /((n+1)^2 )) −(e^(−3(n+1)) /((n+1)^2 )) Σ_(n=0) ^∞ A_n = Σ_(n=0) ^∞ (e^(−(n+1)) /(n+1)) −3 Σ_(n=0) ^∞ (e^(−3(n+1)) /(n+1)) + Σ_(n=0) ^∞ (e^(−(n+1)) /((n+1)^2 )) −Σ_(n=0) ^∞ (e^(−3(n+1)) /((n+1)^2 )) = Σ_(n=1) ^∞ (e^(−n) /n) −3 Σ_(n=1) ^∞ (e^(−3n) /n) +Σ_(n=1) ^∞ (e^(−n) /n^2 ) −Σ_(n=1) ^∞ (e^(−3n) /n^2 ) ....be continued....](https://www.tinkutara.com/question/Q35858.png)
$${I}\:\:=\:\int_{\mathrm{1}} ^{\mathrm{3}} \:\:\frac{{x}\:{e}^{−{x}} }{\mathrm{1}−{e}^{−{x}} }{dx}\:=\int_{\mathrm{1}} ^{\mathrm{3}} \left(\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{nx}} \right){x}\:{e}^{−{x}} \:{dx} \\ $$$$=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\:\int_{\mathrm{1}} ^{\mathrm{3}} \:\:{x}\:{e}^{−\left({n}+\mathrm{1}\right){x}} \:{dx}\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{A}_{{n}} \:\:{with} \\ $$$${A}_{{n}} \:=\:\int_{\mathrm{1}} ^{\mathrm{3}} \:\:{x}.{e}^{−\left({n}+\mathrm{1}\right){x}} {dx}\:\:.{changement}\:\left({n}+\mathrm{1}\right){x}={t} \\ $$$${give}\:{A}_{{n}} \:=\int_{{n}+\mathrm{1}} ^{\mathrm{3}\left({n}+\mathrm{1}\right)} \:\:\frac{{t}}{{n}+\mathrm{1}}\:{e}^{−{t}} \:\:\:\frac{{dt}}{{n}+\mathrm{1}} \\ $$$$=\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\int_{{n}+\mathrm{1}} ^{\mathrm{3}\left({n}+\mathrm{1}\right)} \:\:{t}^{} \:{e}^{−{t}} \:{dt}\:\:{by}\:{parts} \\ $$$$\int_{{n}+\mathrm{1}} ^{\mathrm{3}\left({n}+\mathrm{1}\right)} \:\:{t}\:{e}^{−{t}} \:{dt}\:=\:\left[\:−{t}\:{e}^{−{t}} \right]_{{n}+\mathrm{1}} ^{\mathrm{3}\left({n}+\mathrm{1}\right)} \:+\int_{{n}+\mathrm{1}} ^{\mathrm{3}\left({n}+\mathrm{1}\right)} \:\:{e}^{−{t}} {dt} \\ $$$$=\:\left({n}+\mathrm{1}\right){e}^{−\left({n}+\mathrm{1}\right)} \:−\mathrm{3}\left({n}+\mathrm{1}\right)\:{e}^{−\mathrm{3}\left({n}+\mathrm{1}\right)} \: \\ $$$$+\left[\:−{e}^{−{t}} \right]_{{n}+\mathrm{1}} ^{\mathrm{3}\left({n}+\mathrm{1}\right)} \\ $$$$=\left({n}+\mathrm{1}\right)\:{e}^{−\left({n}+\mathrm{1}\right)} \:−\mathrm{3}\left({n}+\mathrm{1}\right)\:{e}^{−\mathrm{3}\left({n}+\mathrm{1}\right)} \:+{e}^{−\left({n}+\mathrm{1}\right)} \:−{e}^{−\mathrm{3}\left({n}+\mathrm{1}\right)} \\ $$$${A}_{{n}} \:\:=\:\frac{{e}^{−\left({n}+\mathrm{1}\right)} }{{n}+\mathrm{1}}\:−\mathrm{3}\frac{{e}^{−\mathrm{3}\left({n}+\mathrm{1}\right)} }{{n}+\mathrm{1}}\:\:+\:\frac{{e}^{−\left({n}+\mathrm{1}\right)} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:−\frac{{e}^{−\mathrm{3}\left({n}+\mathrm{1}\right)} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:{A}_{{n}} \:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{−\left({n}+\mathrm{1}\right)} }{{n}+\mathrm{1}}\:−\mathrm{3}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{e}^{−\mathrm{3}\left({n}+\mathrm{1}\right)} }{{n}+\mathrm{1}} \\ $$$$+\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{−\left({n}+\mathrm{1}\right)} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\:−\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{−\mathrm{3}\left({n}+\mathrm{1}\right)} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{e}^{−{n}} }{{n}}\:\:−\mathrm{3}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{e}^{−\mathrm{3}{n}} }{{n}}\:\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{e}^{−{n}} }{{n}^{\mathrm{2}} } \\ $$$$−\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{{e}^{−\mathrm{3}{n}} }{{n}^{\mathrm{2}} }\:\:\:\:….{be}\:{continued}…. \\ $$