Question Number 87301 by M±th+et£s last updated on 03/Apr/20
Commented by mathmax by abdo last updated on 04/Apr/20
![let I =∫_1 ^(+∞) (dx/(2[x]^2 +[x])) ⇒ I =Σ_(n=1) ^∞ ∫_n ^(n+1) (dx/(2n^2 +n)) =Σ_(n=1) ^∞ (1/(n(2n+1))) =lim_(n→+∞) Σ_(k=1) ^n (1/(k(2k+1))) =lim_(n→+∞) S_n (1/2)S_n =Σ_(k=1) ^n ((1/(2k))−(1/(2k+1))) =(1/2)Σ_(k=1) ^n (1/k)−Σ_(k=1) ^n (1/(2k+1)) Σ_(k=1) ^n (1/k) =H_n Σ_(k=1) ^n (1/(2k+1)) =(1/3)+(1/5)+....+(1/(2n+1)) =1+(1/2)+(1/3)+(1/4)+...+(1/(2n))+(1/(2n+1)) −1−(1/2)−(1/4)−....−(1/(2n)) =H_(2n+1) −1−(1/2)H_n ⇒ (1/2)S_n =(1/2)H_n −H_(2n+1) +(1/2)H_n +1 =(ln(n)+γ +o((1/n)))−ln(2n+1)−γ +o((1/(2n+1))) +1 =ln((n/(2n+1)))+1 ⇒1−ln(2) ⇒S_n →2−2ln(2) (n→+∞) we have ln((e^2 /4)) =2ln(e)−2ln(2) =2−2ln(2) I =lim S_n =2−2ln(2)=ln((e^2 /4)) .](Q87323.png)
$${let}\:\:{I}\:=\int_{\mathrm{1}} ^{+\infty} \:\frac{{dx}}{\mathrm{2}\left[{x}\right]^{\mathrm{2}} \:+\left[{x}\right]}\:\Rightarrow\:{I}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\int_{{n}} ^{{n}+\mathrm{1}} \:\:\frac{{dx}}{\mathrm{2}{n}^{\mathrm{2}} \:+{n}}\: \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}\left(\mathrm{2}{n}+\mathrm{1}\right)}\:={lim}_{{n}\rightarrow+\infty} \:\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}\left(\mathrm{2}{k}+\mathrm{1}\right)}\:={lim}_{{n}\rightarrow+\infty} \:\:{S}_{{n}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \left(\frac{\mathrm{1}}{\mathrm{2}{k}}−\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}−\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:={H}_{{n}} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}+....+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+...+\frac{\mathrm{1}}{\mathrm{2}{n}}+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}} \\ $$$$−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}−....−\frac{\mathrm{1}}{\mathrm{2}{n}}\:={H}_{\mathrm{2}{n}+\mathrm{1}} −\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{H}_{{n}} \:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{S}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}{H}_{{n}} −{H}_{\mathrm{2}{n}+\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{2}}{H}_{{n}} +\mathrm{1} \\ $$$$=\left({ln}\left({n}\right)+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\right)−{ln}\left(\mathrm{2}{n}+\mathrm{1}\right)−\gamma\:+{o}\left(\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right)\:+\mathrm{1} \\ $$$$={ln}\left(\frac{{n}}{\mathrm{2}{n}+\mathrm{1}}\right)+\mathrm{1}\:\Rightarrow\mathrm{1}−{ln}\left(\mathrm{2}\right)\:\Rightarrow{S}_{{n}} \rightarrow\mathrm{2}−\mathrm{2}{ln}\left(\mathrm{2}\right)\:\left({n}\rightarrow+\infty\right) \\ $$$${we}\:{have}\:{ln}\left(\frac{{e}^{\mathrm{2}} }{\mathrm{4}}\right)\:=\mathrm{2}{ln}\left({e}\right)−\mathrm{2}{ln}\left(\mathrm{2}\right)\:=\mathrm{2}−\mathrm{2}{ln}\left(\mathrm{2}\right) \\ $$$${I}\:={lim}\:{S}_{{n}} =\mathrm{2}−\mathrm{2}{ln}\left(\mathrm{2}\right)={ln}\left(\frac{{e}^{\mathrm{2}} }{\mathrm{4}}\right)\:. \\ $$
Commented by M±th+et£s last updated on 04/Apr/20
$${god}\:{bless}\:{you}\:{sir}.\:{great}\:{solution} \\ $$
Commented by abdomathmax last updated on 04/Apr/20
$${you}\:{are}\:{welcome}. \\ $$