Question Number 48067 by maxmathsup by imad last updated on 18/Nov/18
$${let}\:{y}>\mathrm{0}\:{give}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}^{{y}} }{{e}^{{x}} −\mathrm{1}}{dx}\:{at}\:{form}\:{of}\:{series}. \\ $$
Commented by maxmathsup by imad last updated on 19/Nov/18
$${let}\:{A}\left({y}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{x}^{{y}} }{{e}^{{x}} −\mathrm{1}}{dx}\:\:\Rightarrow{A}\left({y}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{x}} \:{x}^{{y}} }{\mathrm{1}−{e}^{−{x}} }{dx} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{x}} {x}^{{y}} \:\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{nx}} \right){dx}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left({n}+\mathrm{1}\right){x}} \:{x}^{{y}} {dx} \\ $$$$=_{\left({n}+\mathrm{1}\right){x}={t}} \:\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} \:\left(\frac{{t}}{{n}+\mathrm{1}}\right)^{{y}} \:\frac{{dt}}{{n}+\mathrm{1}} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{{y}+\mathrm{1}} }\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} \:{t}^{{y}} \:{dt}\:\:\:{but}\:\Gamma\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:{t}^{{x}−\mathrm{1}} {e}^{−{t}} {dt}\:\left({x}>\mathrm{0}\right)\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{t}^{{y}} {e}^{−{t}} {dt}\:=\Gamma\left({y}+\mathrm{1}\right)\:\:{and}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{{y}+\mathrm{1}} }\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{{y}+\mathrm{1}} }\:=\xi\left({y}+\mathrm{1}\right)\:\Rightarrow \\ $$$${A}\left({y}\right)=\xi\left({y}+\mathrm{1}\right)\Gamma\left({y}+\mathrm{1}\right)\:. \\ $$