Question Number 66446 by ~ À ® @ 237 ~ last updated on 15/Aug/19
$$\:\:{Find}\:\:\:\:\int_{\mathrm{1}} ^{\infty} \:\left(\frac{\mathrm{1}}{{E}\left({x}\right)}\:−\frac{\mathrm{1}}{{x}}\right){dx} \\ $$
Commented by mathmax by abdo last updated on 15/Aug/19
![let A =∫_1 ^(+∞) ((1/([x]))−(1/x))dx ⇒A =Σ_(n=1) ^∞ ∫_n ^(n+1) ((1/n)−(1/x))dx =Σ_(n=1) ^∞ ( (1/n)−∫_n ^(n+1) (dx/x)) =Σ_(n=1) ^∞ ((1/n)−ln(n+1)+ln(n)) let S_n =Σ_(k=1) ^n ((1/k)−ln(k+1)+ln(k)) we have A=lim_(n→+∞) S_n but S_(n ) =H_n +Σ_(k=1) ^n {ln(k)−ln(k+1)} =H_n +(ln(1)−ln(2)+ln(2)−ln(3)+...ln(n)−ln(n+1) =H_n −ln(n+1) =H_n −ln(n)−ln(1+(1/n)) but we know lim_(n→+∞) H_n −ln(n) =γ ⇒ A=lim_(n→+∞) S_n =γ (constant of Euler)](https://www.tinkutara.com/question/Q66460.png)
$${let}\:{A}\:=\int_{\mathrm{1}} ^{+\infty} \left(\frac{\mathrm{1}}{\left[{x}\right]}−\frac{\mathrm{1}}{{x}}\right){dx}\:\Rightarrow{A}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\int_{{n}} ^{{n}+\mathrm{1}} \left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{x}}\right){dx} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \left(\:\frac{\mathrm{1}}{{n}}−\int_{{n}} ^{{n}+\mathrm{1}} \:\frac{{dx}}{{x}}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(\frac{\mathrm{1}}{{n}}−{ln}\left({n}+\mathrm{1}\right)+{ln}\left({n}\right)\right) \\ $$$${let}\:{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \left(\frac{\mathrm{1}}{{k}}−{ln}\left({k}+\mathrm{1}\right)+{ln}\left({k}\right)\right)\:{we}\:{have}\:{A}={lim}_{{n}\rightarrow+\infty} {S}_{{n}} \\ $$$${but}\:\:{S}_{{n}\:} \:={H}_{{n}} \:+\sum_{{k}=\mathrm{1}} ^{{n}} \left\{{ln}\left({k}\right)−{ln}\left({k}+\mathrm{1}\right)\right\} \\ $$$$={H}_{{n}} +\left({ln}\left(\mathrm{1}\right)−{ln}\left(\mathrm{2}\right)+{ln}\left(\mathrm{2}\right)−{ln}\left(\mathrm{3}\right)+…{ln}\left({n}\right)−{ln}\left({n}+\mathrm{1}\right)\right. \\ $$$$={H}_{{n}} −{ln}\left({n}+\mathrm{1}\right)\:={H}_{{n}} −{ln}\left({n}\right)−{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\:{but}\:{we}\:{know} \\ $$$${lim}_{{n}\rightarrow+\infty} \:{H}_{{n}} −{ln}\left({n}\right)\:=\gamma\:\:\Rightarrow\:{A}={lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =\gamma \\ $$$$\left({constant}\:{of}\:{Euler}\right) \\ $$