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Question Number 27345 by abdo imad last updated on 05/Jan/18

prove that  ∫_0 ^∞   (t^(x−1) /(e^t −1))dt  =ξ(x)Γ(x)  with ξ(x)= Σ_(n=1) ^∝  (1/n^x )   and Γ(x)=∫_0 ^∞  t^(x−1)  e^(−t) dt  ( x>1)

$${prove}\:{that}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{x}−\mathrm{1}} }{{e}^{{t}} −\mathrm{1}}{dt}\:\:=\xi\left({x}\right)\Gamma\left({x}\right) \\ $$ $${with}\:\xi\left({x}\right)=\:\sum_{{n}=\mathrm{1}} ^{\propto} \:\frac{\mathrm{1}}{{n}^{{x}} }\:\:\:{and}\:\Gamma\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:{t}^{{x}−\mathrm{1}} \:{e}^{−{t}} {dt} \\ $$ $$\left(\:{x}>\mathrm{1}\right) \\ $$

Commented byabdo imad last updated on 07/Jan/18

∫_0 ^∝   (t^(x−1) /(e^t  −1))dt = ∫_0 ^∝  ((e^(−t)  t^(x−1) )/(1−e^(−t) ))dt  = ∫_0 ^∝ (Σ_(n=0) ^∝  e^(−nt) )e^(−t) t^(x−1) dt  =Σ_(n=0) ^∝ (∫_0 ^∝  e^(−(n+1)t)  t^(x−1) dt)   the ch. (n+1)t=u give  ∫_0 ^∝   e^(−(n+1)t)  t^(x−1) dt= (1/(n+1)) ∫_0 ^∝  e^(−u) ((u/(n+1)))^(x−1) du  = (1/((n+1)^x )) ∫_0 ^∝  e^(−u)  u^(x−1)  du= ((Γ(x))/((n+1)^x ))  ⇒ ∫_0 ^∝   (t^(x−1) /(e^t −1))dt= Σ_(n=0) ^∝ (((Γ(x))/((n+1)^x )))=(Σ_(n=1) ^∝  (1/n^x ))Γ(x)  =ξ(x).Γ(x).

$$\int_{\mathrm{0}} ^{\propto} \:\:\frac{{t}^{{x}−\mathrm{1}} }{{e}^{{t}} \:−\mathrm{1}}{dt}\:=\:\int_{\mathrm{0}} ^{\propto} \:\frac{{e}^{−{t}} \:{t}^{{x}−\mathrm{1}} }{\mathrm{1}−{e}^{−{t}} }{dt} \\ $$ $$=\:\int_{\mathrm{0}} ^{\propto} \left(\sum_{{n}=\mathrm{0}} ^{\propto} \:{e}^{−{nt}} \right){e}^{−{t}} {t}^{{x}−\mathrm{1}} {dt} \\ $$ $$=\sum_{{n}=\mathrm{0}} ^{\propto} \left(\int_{\mathrm{0}} ^{\propto} \:{e}^{−\left({n}+\mathrm{1}\right){t}} \:{t}^{{x}−\mathrm{1}} {dt}\right)\:\:\:{the}\:{ch}.\:\left({n}+\mathrm{1}\right){t}={u}\:{give} \\ $$ $$\int_{\mathrm{0}} ^{\propto} \:\:{e}^{−\left({n}+\mathrm{1}\right){t}} \:{t}^{{x}−\mathrm{1}} {dt}=\:\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\propto} \:{e}^{−{u}} \left(\frac{{u}}{{n}+\mathrm{1}}\right)^{{x}−\mathrm{1}} {du} \\ $$ $$=\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{{x}} }\:\int_{\mathrm{0}} ^{\propto} \:{e}^{−{u}} \:{u}^{{x}−\mathrm{1}} \:{du}=\:\frac{\Gamma\left({x}\right)}{\left({n}+\mathrm{1}\right)^{{x}} } \\ $$ $$\Rightarrow\:\int_{\mathrm{0}} ^{\propto} \:\:\frac{{t}^{{x}−\mathrm{1}} }{{e}^{{t}} −\mathrm{1}}{dt}=\:\sum_{{n}=\mathrm{0}} ^{\propto} \left(\frac{\Gamma\left({x}\right)}{\left({n}+\mathrm{1}\right)^{{x}} }\right)=\left(\sum_{{n}=\mathrm{1}} ^{\propto} \:\frac{\mathrm{1}}{{n}^{{x}} }\right)\Gamma\left({x}\right) \\ $$ $$=\xi\left({x}\right).\Gamma\left({x}\right). \\ $$ $$ \\ $$

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