Question Number 33342 by prof Abdo imad last updated on 14/Apr/18
$${let}\:{I}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}}{{x}}\left(\mathrm{1}−\left(\mathrm{1}−\frac{{x}}{{n}}\right)^{{n}} \right){dx}\:{and} \\ $$$${J}_{{n}} \:=\:\int_{\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{x}}\left(\mathrm{1}−\frac{{x}}{{n}}\right)^{{n}} {dx}\:\:,{n}\:{integr}\:{not}\:\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:{lim}_{} \:{I}_{{n}} \:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}−{e}^{−{x}} }{{x}}{dx} \\ $$$${lim}\:{J}_{{n}} \:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{e}^{−\frac{\mathrm{1}}{{x}}} }{{x}}\:{dx}\:\left({n}\rightarrow\infty\right) \\ $$$$\left.\mathrm{2}\right)\:{prove}\:{that}\:\:\forall{n}\in\:{N}^{\bigstar} \:\:\:{I}_{{n}} \:−{J}_{{n}} \:=\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:−{ln}\left({n}\right) \\ $$$$\left.\mathrm{3}\right)\:{prove}\:{that}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{1}−{e}^{−{x}} \:−{e}^{−\frac{\mathrm{1}}{{x}}} }{{x}}\:=\gamma \\ $$
Commented by prof Abdo imad last updated on 24/Apr/18
![I_n = ∫_R (1/x)(1−(1−(x/n))^n )χ_(]0^ 1]) (x)dx let put f_n (x)= (1/x)(1−(1−(x/n))^n )χ_(]0,1]) (x) f_n (x)^(c.s) → ((1−e^(−x) )/x) χ_(]0,1]) (x) (n→+∞) and f_n (x) ≤ ((1−e^(−x) )/x) ∀ x∈]0,1] conv.dominee ⇒ ∫_R f_n (x)dx _(n→+∞) → ∫_0 ^1 ((1−e^(−x) )/x)dx ⇒ lim_(n→+∞) I_n = ∫_0 ^1 ((1−e^(−x) )/x) dx .](Q33797.png)
$${I}_{{n}} \:=\:\int_{{R}} \:\frac{\mathrm{1}}{{x}}\left(\mathrm{1}−\left(\mathrm{1}−\frac{{x}}{{n}}\right)^{{n}} \right)\chi_{\left.\right]\left.\bar {\mathrm{0}1}\right]} \left({x}\right){dx}\:{let}\:{put} \\ $$$${f}_{{n}} \left({x}\right)=\:\frac{\mathrm{1}}{{x}}\left(\mathrm{1}−\left(\mathrm{1}−\frac{{x}}{{n}}\right)^{{n}} \right)\chi_{\left.\right]\left.\mathrm{0},\mathrm{1}\right]} \left({x}\right) \\ $$$${f}_{{n}} \left({x}\right)\:^{{c}.{s}} \rightarrow\:\frac{\mathrm{1}−{e}^{−{x}} }{{x}}\:\chi_{\left.\right]\left.\mathrm{0},\mathrm{1}\right]} \left({x}\right)\:\left({n}\rightarrow+\infty\right)\:{and} \\ $$$$\left.{f}_{{n}} \left.\left({x}\right)\:\leqslant\:\frac{\mathrm{1}−{e}^{−{x}} }{{x}}\:\forall\:{x}\in\right]\mathrm{0},\mathrm{1}\right]\:{conv}.{dominee}\:\Rightarrow \\ $$$$\int_{{R}} {f}_{{n}} \left({x}\right){dx}\:_{{n}\rightarrow+\infty} \rightarrow\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}−{e}^{−{x}} }{{x}}{dx}\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:{I}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}−{e}^{−{x}} }{{x}}\:{dx}\:. \\ $$
Commented by prof Abdo imad last updated on 24/Apr/18
$${changement}\:{x}=\frac{\mathrm{1}}{{t}\:}\:{give}\: \\ $$$${J}_{{n}} =\:−\int_{\mathrm{1}} ^{\frac{\mathrm{1}}{{n}}} \:\:{t}\left(\mathrm{1}−\frac{\mathrm{1}}{{nt}}\right)^{{n}} \:\frac{{dt}}{{t}^{\mathrm{2}} }\:=\:\int_{\frac{\mathrm{1}}{{n}}} ^{\mathrm{1}} \:\frac{\mathrm{1}}{{t}}\left(\mathrm{1}−\frac{\mathrm{1}}{{nt}}\right)^{{n}} {dt}\:\Rightarrow \\ $$$${J}_{{n}} \:\rightarrow\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{e}^{−\frac{\mathrm{1}}{{t}}} }{{t}}\:{dt}\:\left({n}\rightarrow+\infty\right)\:\:{so} \\ $$$${lim}_{{n}\rightarrow+\infty} \:{J}_{{n}} \:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{e}^{−\frac{\mathrm{1}}{{t}}} }{{t}}\:{dt}\:. \\ $$
Commented by math khazana by abdo last updated on 25/Apr/18
$${changement}\:\frac{{x}}{{n}}={t}\:{give} \\ $$$${J}_{{n}} \:=\:\int_{\frac{\mathrm{1}}{{n}}} ^{\mathrm{1}} \:\frac{\mathrm{1}}{{nt}}\left(\mathrm{1}−{t}\right)^{{n}} \:{ndt}\:=\:\int_{\frac{\mathrm{1}}{{n}}} ^{\mathrm{1}} \:\:\frac{\left(\mathrm{1}−{t}\right)^{{n}} }{{t}}{dt}\:\Rightarrow \\ $$$${I}_{{n}} \:−{J}_{{n}} \:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}}{{x}}\left(\:\mathrm{1}−\left(\mathrm{1}−\frac{{x}}{{n}}\right)^{{n}} \right){dx}\:+\int_{\mathrm{1}} ^{\frac{\mathrm{1}}{{n}}} \:\:\frac{\left(\mathrm{1}−{x}\right)^{{n}} }{{x}}{dx} \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{{n}}} \:\:\:\frac{\mathrm{1}}{{x}}\left(\:\mathrm{1}−\left(\mathrm{1}−\frac{{x}}{{n}}\right)^{{n}} \:+\left(\mathrm{1}−{x}\right)^{{n}} \right){dx} \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{{n}}} \:\frac{\mathrm{1}}{{x}}\left(\:\mathrm{1}+\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left(−\mathrm{1}\right)^{{k}} \:{x}^{{k}\:} \:−\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left(−\mathrm{1}\right)^{{k}} \:\frac{{x}^{{k}} }{{n}^{{k}} }\right){dx} \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{{n}}} \:\:\frac{\mathrm{1}}{{x}}\left(\:\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left(−\mathrm{1}\right)^{{k}} {x}^{{k}} \:−\sum_{{k}=\mathrm{1}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left(−\mathrm{1}\right)^{{k}} \:\frac{{x}^{{k}} }{{n}^{{k}} }\right){dx} \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{{n}}} \:\:\frac{\mathrm{1}}{{x}}\left(\:\:\:\mathrm{1}+\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left(−\mathrm{1}\right)^{{k}} \:\left(\mathrm{1}+\:\frac{\mathrm{1}}{{n}^{{k}} }\right){x}^{{k}} \right){dx}...{be} \\ $$$${continued}\:... \\ $$
Commented by math khazana by abdo last updated on 25/Apr/18
$$\left.\mathrm{3}\right)\:{we}\:{have}\:\:{I}_{{n}} \:−{J}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:−{ln}\left({n}\right)\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \left({I}_{{n}} \:−{J}_{{n}} \right)={lim}_{{n}\rightarrow+\infty} \left(\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:−{ln}\left({n}\right)\right)=\gamma \\ $$$$\Rightarrow\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}−\:{e}^{−{x}} }{{x}}{dx}\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{e}^{−\frac{\mathrm{1}}{{x}}} }{{x}}{dx}\:=\gamma\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{1}−{e}^{−{x}} \:−{e}^{−\frac{\mathrm{1}}{{x}}} }{{x}}\:=\gamma\:\:. \\ $$