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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 by abdo 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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