Question-53483

Question Number 53483 by dwdkswd last updated on 22/Jan/19

Commented by maxmathsup by imad last updated on 22/Jan/19

let A_s =∫_0 ^∞ (x^s /(e^x −1)) dx ⇒A_s =∫_0 ^∞ ((e^(−x) x^s )/(1−e^(−x) )) dx =∫_0 ^∞ e^(−x) x^s (Σ_(n=0) ^∞ e^(−nx) )dx =Σ_(n=0) ^∞ ∫_0 ^∞ x^s e^(−(n+1)x) dx =_((n+1)x =t) Σ_(n=0) ^∞ ∫_0 ^∞ (t^s /((n+1)^s )) e^(−t) (dt/(n+1)) =Σ_(n=0) ^∞ (1/((n+1)^(s+1) )) ∫_0 ^∞ t^s e^(−t) dt but we know Γ(x)=∫_0 ^∞ t^(x−1) e^(−t) dt (x>0)⇒ ∫_0 ^∞ t^s e^(−t) dt =Γ(s+1) ⇒A_s =ξ(s+1)Γ(s+1) .

$${let}\:{A}_{{s}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{s}} }{{e}^{{x}} −\mathrm{1}}\:{dx}\:\Rightarrow{A}_{{s}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{x}} {x}^{{s}} }{\mathrm{1}−{e}^{−{x}} }\:{dx}\:=\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{x}} {x}^{{s}} \left(\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{nx}} \right){dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\int_{\mathrm{0}} ^{\infty} \:{x}^{{s}} {e}^{−\left({n}+\mathrm{1}\right){x}} \:{dx}\:=_{\left({n}+\mathrm{1}\right){x}\:={t}} \:\:\sum_{{n}=\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{{s}} }{\left({n}+\mathrm{1}\right)^{{s}} }\:{e}^{−{t}} \:\frac{{dt}}{{n}+\mathrm{1}} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{{s}+\mathrm{1}} }\:\int_{\mathrm{0}} ^{\infty} \:\:{t}^{{s}} \:{e}^{−{t}} {dt}\:\:\:{but}\:{we}\:{know}\:\Gamma\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:{t}^{{x}−\mathrm{1}} \:{e}^{−{t}} {dt}\:\left({x}>\mathrm{0}\right)\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:{t}^{{s}} {e}^{−{t}} {dt}\:=\Gamma\left({s}+\mathrm{1}\right)\:\Rightarrow{A}_{{s}} =\xi\left({s}+\mathrm{1}\right)\Gamma\left({s}+\mathrm{1}\right)\:. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 22/Jan/19