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Question Number 48067 by maxmathsup by imad last updated on 18/Nov/18

let y>0 give ∫_0 ^∞ (x^y /(e^x −1))dx at form of series.

$${let}\:{y}>\mathrm{0}\:{give}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}^{{y}} }{{e}^{{x}} −\mathrm{1}}{dx}\:{at}\:{form}\:{of}\:{series}. \\ $$

Commented bymaxmathsup by imad last updated on 19/Nov/18

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

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

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