Question Number 26571 by abdo imad last updated on 26/Dec/17
$${let}\:{give}\:\:{I}\left({x}\right)=\:\:\int_{\mathrm{1}} ^{\propto} \:\frac{{t}−{E}\left({t}\right)}{{t}^{{x}+\mathrm{1}} }{dt}\:\:\:{prove}\:{that} \\ $$$$\xi\left({x}\right)=\:\frac{{x}}{{x}−\mathrm{1}}\:−{xI}\left({x}\right)\:{then}\:{chow}\:{that}\:\left({x}−\mathrm{1}\right)_{{x}−\mathrm{1}^{+\:{ew}} } \xi\left({x}\right)−−>\mathrm{1} \\ $$$${we}\:{remind}\:\:\xi\left({x}\right)\:=\:\sum_{{n}\geqslant\mathrm{1}} \:\frac{\mathrm{1}}{{n}^{{x}} }\:\:\:{and}\:\:{x}>\mathrm{1} \\ $$
Commented by abdo imad last updated on 02/Jan/18
![we have I(x)= ∫_1 ^∝ (dt/t^x ) − ∫_1 ^∝ ((E(t))/t^(x+1) )dt but ∫_1 ^∝ (dt/t^x ) = ∫_1 ^∝ t^(−x) dt = [(1/(1−x)) t^(1−x) ]_(t=1) ^(t−>∝) = (1/(x−1)) ∫_1 ^∝ ((E(t))/t^(x+1) )dt= lim_(n−>∝) A_n with = A_n = Σ_(k=1) ^(n−1) ∫_k ^(k+1) (k/t^(x+1) )dt =Σ_(k=1) ^(n−1) k ∫_k ^(k+1) t^(−x−1) dt = Σ_(k=1) ^(n−1) k[ −(1/x) t^(−x) ]_k ^(k+1) = Σ_(k=1) ^(k=n−1) (k/x)( (1/k^x ) − (1/((k+1)^x ))) x A_n = Σ_(k=1) ^(k=n−1) (1/k^(x−1) ) − Σ_(k=1) ^(n−1) ((k+1−1)/((k+1)^x )) = Σ_(k=1) ^(k=n−1) (1/k^(x−1) ) − Σ_(k=1) ^(n−1) (1/((k+1)^(x−1) )) +Σ_(k=1) ^(n−1) (1/((k+1)^x )) = Σ_(k=1) ^(n−1) (1/k^(x−1) ) − Σ_(k=2) ^n (1/k^(x−1) ) + Σ_(k=2) ^n (1/k^x ) =1− (1/n^(x−1) ) + Σ_(k=2) ^n (1/k^x ) ⇒lim_(n−>∝) = ((ξ(x))/x) I(x)= (1/(x−1)) −((ξ(x))/x) ⇒xI(x) = (x/(x−1)) −ξ(x) ⇒ξ(x)= (x/(x−1)) −xI(x) ⇒ (x−1)ξ(x) =x −x(x−1) I(x) ⇒lim_(x−>1^+ ) (x−1)ξ(x)= 1 .](https://www.tinkutara.com/question/Q27175.png)
$${we}\:{have}\:{I}\left({x}\right)=\:\int_{\mathrm{1}} ^{\propto} \frac{{dt}}{{t}^{{x}} }\:−\:\int_{\mathrm{1}} ^{\propto} \:\frac{{E}\left({t}\right)}{{t}^{{x}+\mathrm{1}} }{dt}\:\:{but} \\ $$$$\int_{\mathrm{1}} ^{\propto} \frac{{dt}}{{t}^{{x}} }\:=\:\int_{\mathrm{1}} ^{\propto} \:{t}^{−{x}} {dt}\:=\:\:\:\left[\frac{\mathrm{1}}{\mathrm{1}−{x}}\:{t}^{\mathrm{1}−{x}} \right]_{{t}=\mathrm{1}} ^{{t}−>\propto} =\:\:\frac{\mathrm{1}}{{x}−\mathrm{1}} \\ $$$$\int_{\mathrm{1}} ^{\propto} \:\frac{{E}\left({t}\right)}{{t}^{{x}+\mathrm{1}} }{dt}=\:{lim}_{{n}−>\propto} \:\:{A}_{{n}} \:\:{with}\:= \\ $$$${A}_{{n}} \:\:=\:\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\int_{{k}} ^{{k}+\mathrm{1}} \:\frac{{k}}{{t}^{{x}+\mathrm{1}} }{dt}\:=\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} {k}\:\int_{{k}} ^{{k}+\mathrm{1}} {t}^{−{x}−\mathrm{1}} {dt} \\ $$$$=\:\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} {k}\left[\:−\frac{\mathrm{1}}{{x}}\:{t}^{−{x}} \:\:\right]_{{k}} ^{{k}+\mathrm{1}} \:\:=\:\sum_{{k}=\mathrm{1}} ^{{k}={n}−\mathrm{1}} \frac{{k}}{{x}}\left(\:\frac{\mathrm{1}}{{k}^{{x}} }\:−\:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{{x}} }\right) \\ $$$${x}\:{A}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{k}={n}−\mathrm{1}} \:\frac{\mathrm{1}}{{k}^{{x}−\mathrm{1}} }\:−\:\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\:\frac{{k}+\mathrm{1}−\mathrm{1}}{\left({k}+\mathrm{1}\right)^{{x}} } \\ $$$$=\:\sum_{{k}=\mathrm{1}} ^{{k}={n}−\mathrm{1}} \frac{\mathrm{1}}{{k}^{{x}−\mathrm{1}} }\:\:−\:\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{{x}−\mathrm{1}} }\:+\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{{x}} } \\ $$$$=\:\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\frac{\mathrm{1}}{{k}^{{x}−\mathrm{1}} }\:−\:\sum_{{k}=\mathrm{2}} ^{{n}} \:\:\:\frac{\mathrm{1}}{{k}^{{x}−\mathrm{1}} }\:+\:\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{{x}} } \\ $$$$=\mathrm{1}−\:\frac{\mathrm{1}}{{n}^{{x}−\mathrm{1}} }\:+\:\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{{x}} }\:\Rightarrow{lim}_{{n}−>\propto} \:\:=\:\frac{\xi\left({x}\right)}{{x}} \\ $$$${I}\left({x}\right)=\:\frac{\mathrm{1}}{{x}−\mathrm{1}}\:−\frac{\xi\left({x}\right)}{{x}}\:\:\Rightarrow{xI}\left({x}\right)\:=\:\frac{{x}}{{x}−\mathrm{1}}\:−\xi\left({x}\right) \\ $$$$\Rightarrow\xi\left({x}\right)=\:\frac{{x}}{{x}−\mathrm{1}}\:−{xI}\left({x}\right) \\ $$$$\Rightarrow\:\:\left({x}−\mathrm{1}\right)\xi\left({x}\right)\:\:={x}\:−{x}\left({x}−\mathrm{1}\right)\:{I}\left({x}\right) \\ $$$$\Rightarrow{lim}_{{x}−>\mathrm{1}^{+} } \:\:\left({x}−\mathrm{1}\right)\xi\left({x}\right)=\:\mathrm{1}\:. \\ $$$$ \\ $$