Question Number 40890 by abdo.msup.com last updated on 28/Jul/18
![1)calculate ∫_(1/(n+1)) ^(1/n) [(1/t)−[(1/t)]]dt 2)prove that ∫_0 ^1 [(1/t)−[(1/t)]]dt=1−γ γ is constant number of euler](Q40890.png)
$$\left.\mathrm{1}\right){calculate}\:\int_{\frac{\mathrm{1}}{{n}+\mathrm{1}}} ^{\frac{\mathrm{1}}{{n}}} \left[\frac{\mathrm{1}}{{t}}−\left[\frac{\mathrm{1}}{{t}}\right]\right]{dt} \\ $$$$\left.\mathrm{2}\right){prove}\:{that}\:\int_{\mathrm{0}} ^{\mathrm{1}} \left[\frac{\mathrm{1}}{{t}}−\left[\frac{\mathrm{1}}{{t}}\right]\right]{dt}=\mathrm{1}−\gamma \\ $$$$\gamma\:{is}\:{constant}\:{number}\:{of}\:{euler} \\ $$
Commented by maxmathsup by imad last updated on 01/Aug/18
![1) let A_n = ∫_(1/(n+1)) ^(1/n) {(1/t)−[(1/t)]}dt changement (1/t)=x give A_n = −∫_n ^(n+1) {x−[x]}(−(dx/x^2 )) = ∫_n ^(n+1) {(1/x)−(([x])/x^2 )}dx =∫_n ^(n+1) (dx/x) −∫_n ^(n+1) (n/x^2 )dx =[ln(x)]_n ^(n+1) +n [ (1/x)]_n ^(n+1) =ln(n+1)−ln(n) +n{(1/(n+1)) −(1/n)} =ln(n+1)−ln(n) +(n/(n+1)) −1](Q41069.png)
$$\left.\mathrm{1}\right)\:{let}\:{A}_{{n}} =\:\int_{\frac{\mathrm{1}}{{n}+\mathrm{1}}} ^{\frac{\mathrm{1}}{{n}}} \left\{\frac{\mathrm{1}}{{t}}−\left[\frac{\mathrm{1}}{{t}}\right]\right\}{dt}\:\:{changement}\:\frac{\mathrm{1}}{{t}}={x}\:{give} \\ $$$${A}_{{n}} =\:−\int_{{n}} ^{{n}+\mathrm{1}} \left\{{x}−\left[{x}\right]\right\}\left(−\frac{{dx}}{{x}^{\mathrm{2}} }\right)\:=\:\int_{{n}} ^{{n}+\mathrm{1}} \left\{\frac{\mathrm{1}}{{x}}−\frac{\left[{x}\right]}{{x}^{\mathrm{2}} }\right\}{dx} \\ $$$$=\int_{{n}} ^{{n}+\mathrm{1}} \:\frac{{dx}}{{x}}\:−\int_{{n}} ^{{n}+\mathrm{1}} \:\:\frac{{n}}{{x}^{\mathrm{2}} }{dx}\:=\left[{ln}\left({x}\right)\right]_{{n}} ^{{n}+\mathrm{1}} \:+{n}\:\left[\:\frac{\mathrm{1}}{{x}}\right]_{{n}} ^{{n}+\mathrm{1}} \\ $$$$={ln}\left({n}+\mathrm{1}\right)−{ln}\left({n}\right)\:+{n}\left\{\frac{\mathrm{1}}{{n}+\mathrm{1}}\:−\frac{\mathrm{1}}{{n}}\right\}\:={ln}\left({n}+\mathrm{1}\right)−{ln}\left({n}\right)\:+\frac{{n}}{{n}+\mathrm{1}}\:−\mathrm{1} \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 01/Aug/18
![2) let I = ∫_0 ^1 { (1/t)−[(1/t)]}dt changement (1/t)=x give I = −∫_1 ^(+∞) {x−[x]}(−(dx/x^2 )) = ∫_1 ^(+∞) ((x−[x])/x^2 )dx =Σ_(n=1) ^∞ ∫_n ^(n+1) ((x−[x])/x^2 )dx =Σ_(n=1) ^∞ A_n but Σ_(n=1) ^∞ A_n =lim_(n→+∞) Σ_(k=1) ^n A_k Σ_(k=1) ^n A_k =Σ_(k=1) ^n {ln(k+1)−ln(k)} −Σ_(k=1) ^n (1/(k+1)) =ln(n+1)−Σ_(k=2) ^(n+1) (1/k) =ln(n+1)−(H_(n+1) −1) =1−(H_(n+1) −ln(n+1)) but H_(n+1) =ln(n+1) +γ +o((1/n))⇒H_(n+1) −ln(n+1)=γ +o((1/n))⇒ Σ_(k=1) ^n A_k =1−γ +o((1/n)) ⇒lim_(n→+∞) Σ_(k=1) ^n A_k =1−γ =I .](Q41070.png)
$$\left.\mathrm{2}\right)\:{let}\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left\{\:\frac{\mathrm{1}}{{t}}−\left[\frac{\mathrm{1}}{{t}}\right]\right\}{dt}\:{changement}\:\frac{\mathrm{1}}{{t}}={x}\:{give} \\ $$$${I}\:\:=\:\:−\int_{\mathrm{1}} ^{+\infty} \left\{{x}−\left[{x}\right]\right\}\left(−\frac{{dx}}{{x}^{\mathrm{2}} }\right)\:=\:\int_{\mathrm{1}} ^{+\infty} \:\frac{{x}−\left[{x}\right]}{{x}^{\mathrm{2}} }{dx} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\int_{{n}} ^{{n}+\mathrm{1}} \:\:\frac{{x}−\left[{x}\right]}{{x}^{\mathrm{2}} }{dx}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:{A}_{{n}} \:\:{but} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:{A}_{{n}} ={lim}_{{n}\rightarrow+\infty} \:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:{A}_{{k}} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:{A}_{{k}} =\sum_{{k}=\mathrm{1}} ^{{n}} \left\{{ln}\left({k}+\mathrm{1}\right)−{ln}\left({k}\right)\right\}\:−\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{1}} \\ $$$$={ln}\left({n}+\mathrm{1}\right)−\sum_{{k}=\mathrm{2}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{1}}{{k}}\:={ln}\left({n}+\mathrm{1}\right)−\left({H}_{{n}+\mathrm{1}} −\mathrm{1}\right) \\ $$$$=\mathrm{1}−\left({H}_{{n}+\mathrm{1}} −{ln}\left({n}+\mathrm{1}\right)\right)\:{but} \\ $$$${H}_{{n}+\mathrm{1}} ={ln}\left({n}+\mathrm{1}\right)\:+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\Rightarrow{H}_{{n}+\mathrm{1}} −{ln}\left({n}+\mathrm{1}\right)=\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\Rightarrow \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:{A}_{{k}} =\mathrm{1}−\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:\sum_{{k}=\mathrm{1}} ^{{n}} \:{A}_{{k}} =\mathrm{1}−\gamma\:\:={I}\:. \\ $$$$ \\ $$