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Question Number 33119 by abdo imad last updated on 10/Apr/18
find ∫_0 ^∞ (t^n /(e^t −1)) dt by using ξ(x) for n integr ξ(x)=Σ_(n=1) ^∞ (1/n^x ) with x>1 .
$${find}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{{n}} }{{e}^{{t}} \:−\mathrm{1}}\:{dt}\:{by}\:{using}\:\xi\left({x}\right)\:{for}\:{n}\:{integr} \\ $$$$\xi\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{{x}} }\:\:\:{with}\:{x}>\mathrm{1}\:. \\ $$
Commented by prof Abdo imad last updated on 15/Apr/18
let put A_n = ∫_0 ^∞ (t^n /(e^t −1))dt A_n = ∫_0 ^∞ ((e^(−t) t^n )/(1−e^(−t) ))dt =∫_0 ^∞ t^n e^(−t) ( Σ_(p=0) ^∞ e^(−pt) )dt = Σ_(p=0) ^∞ ∫_0 ^∞ t^n e^(−(p+1)t) dt thech (p+1)t =u give A_n = Σ_(p=0) ^∞ ∫_0 ^∞ (u^n /((p+1)^n )) e^(−u) (du/((p+1))) = Σ_(p=0) ^∞ (1/((p+1)^(n+1) )) ∫_0 ^∞ u^n e^(−u) du but we know that Γ(x) = ∫_0 ^∞ t^(x−1) e^(−t) dt for x>0 ⇒ ∫_0 ^∞ u^n e^(−u) du = Γ(n+1) and Σ_(p=0) ^∞ (1/((p+1)^(n+1) )) =Σ_(p=1) ^∞ (1/p^(n+1) ) =ξ(n+1) ⇒ A_n = Γ(n+1).ξ(n+1) .
$${let}\:{put}\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{n}} }{{e}^{{t}} \:−\mathrm{1}}{dt}\: \\ $$$${A}_{{n}} =\:\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{t}} \:{t}^{{n}} }{\mathrm{1}−{e}^{−{t}} }{dt}\:\:=\int_{\mathrm{0}} ^{\infty} \:{t}^{{n}} {e}^{−{t}} \:\left(\:\sum_{{p}=\mathrm{0}} ^{\infty} \:{e}^{−{pt}} \right){dt} \\ $$$$=\:\sum_{{p}=\mathrm{0}} ^{\infty} \:\:\int_{\mathrm{0}} ^{\infty} \:{t}^{{n}} \:\:{e}^{−\left({p}+\mathrm{1}\right){t}} \:{dt}\:\:{thech}\:\left({p}+\mathrm{1}\right){t}\:={u}\:{give} \\ $$$${A}_{{n}} \:=\:\sum_{{p}=\mathrm{0}} ^{\infty} \:\:\int_{\mathrm{0}} ^{\infty} \:\frac{{u}^{{n}} }{\left({p}+\mathrm{1}\right)^{{n}} }\:{e}^{−{u}} \:\frac{{du}}{\left({p}+\mathrm{1}\right)} \\ $$$$=\:\sum_{{p}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left({p}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\:\int_{\mathrm{0}} ^{\infty} \:{u}^{{n}} \:{e}^{−{u}} \:{du}\:\:{but}\:{we}\:{know}\:{that} \\ $$$$\Gamma\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:{t}^{{x}−\mathrm{1}} \:{e}^{−{t}} {dt}\:\:{for}\:{x}>\mathrm{0}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{u}^{{n}} \:{e}^{−{u}} \:{du}\:=\:\Gamma\left({n}+\mathrm{1}\right)\:\:{and}\:\sum_{{p}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left({p}+\mathrm{1}\right)^{{n}+\mathrm{1}} } \\ $$$$=\sum_{{p}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{p}^{{n}+\mathrm{1}} }\:=\xi\left({n}+\mathrm{1}\right)\:\Rightarrow \\ $$$${A}_{{n}} =\:\Gamma\left({n}+\mathrm{1}\right).\xi\left({n}+\mathrm{1}\right)\:. \\ $$

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