Question Number 40883 by prof Abdo imad last updated on 28/Jul/18
$${find}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{p}} }{{e}^{{t}} −\mathrm{1}}{dt}\:{with}\:{p}\in{N}^{\bigstar} \\ $$
Answered by maxmathsup by imad last updated on 31/Jul/18
![let A_p = ∫_0 ^∞ (t^p /(e^t −1))⇒A_p = ∫_0 ^∞ ((e^(−t) t^p )/(1−e^(−t) )) =∫_0 ^∞ e^(−t) t^p (Σ_(n=0) ^∞ e^(−nt) ) =Σ_(n=0) ^∞ ∫_0 ^∞ t^p e^(−(n+1)t) dt changement (n+1)t =x give A_p = Σ_(n=0) ^∞ ((x/(n+1)))^p e^(−x) (dx/(n+1)) = Σ_(n=0) ^∞ ∫_0 ^∞ ((x^p e^(−x) )/((n+1)^(p+1) ))dx =Σ_(n=0) ^∞ (1/((n+1)^(p+1) )) ∫_0 ^∞ x^p e^(−x) dx by parts w_p =∫_0 ^∞ x^p e^(−x) dx =[−x^p e^(−x) ]_0 ^(+∞) +∫_0 ^∞ px^(p−1) e^(−p) dx =pw_(p−1) ⇒w_p =p!w_0 and w_0 =∫_0 ^∞ e^(−x) dx =1 ⇒w_p =p! ⇒ A_p =p! Σ_(n=1) ^∞ (1/n^(p+1) ) =p! ξ(p+1) with ξ(x)=Σ_(n=1) ^∞ (1/n^x ) , x>1](Q41041.png)
$${let}\:{A}_{{p}} =\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{{p}} }{{e}^{{t}} −\mathrm{1}}\Rightarrow{A}_{{p}} =\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{t}} \:{t}^{{p}} }{\mathrm{1}−{e}^{−{t}} } \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{t}} {t}^{{p}} \left(\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{nt}} \right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\int_{\mathrm{0}} ^{\infty} \:\:{t}^{{p}} \:{e}^{−\left({n}+\mathrm{1}\right){t}} {dt}\:{changement}\:\left({n}+\mathrm{1}\right){t}\:={x} \\ $$$${give} \\ $$$${A}_{{p}} =\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\left(\frac{{x}}{{n}+\mathrm{1}}\right)^{{p}} \:{e}^{−{x}} \:\frac{{dx}}{{n}+\mathrm{1}}\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{p}} \:{e}^{−{x}} }{\left({n}+\mathrm{1}\right)^{{p}+\mathrm{1}} }{dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{{p}+\mathrm{1}} }\:\int_{\mathrm{0}} ^{\infty} \:\:{x}^{{p}} \:{e}^{−{x}} {dx}\:\:{by}\:{parts} \\ $$$${w}_{{p}} =\int_{\mathrm{0}} ^{\infty} \:\:{x}^{{p}} \:{e}^{−{x}} {dx}\:=\left[−{x}^{{p}} \:{e}^{−{x}} \right]_{\mathrm{0}} ^{+\infty} \:\:+\int_{\mathrm{0}} ^{\infty} \:{px}^{{p}−\mathrm{1}} \:{e}^{−{p}} {dx} \\ $$$$={pw}_{{p}−\mathrm{1}} \:\:\:\Rightarrow{w}_{{p}} ={p}!{w}_{\mathrm{0}} \:\:\:\:{and}\:{w}_{\mathrm{0}} =\int_{\mathrm{0}} ^{\infty} \:{e}^{−{x}} {dx}\:=\mathrm{1}\:\Rightarrow{w}_{{p}} ={p}!\:\Rightarrow \\ $$$${A}_{{p}} ={p}!\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{{p}+\mathrm{1}} }\:={p}!\:\xi\left({p}+\mathrm{1}\right)\:\:\:{with}\:\xi\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{{x}} }\:,\:\:\:{x}>\mathrm{1} \\ $$$$ \\ $$