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