Question Number 96514 by Farruxjano last updated on 02/Jun/20
![∫_0 ^1 (1/(1+[(1/x)]))dx=?](https://www.tinkutara.com/question/Q96514.png)
$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{1}}{\mathrm{1}+\left[\frac{\mathrm{1}}{\boldsymbol{{x}}}\right]}\boldsymbol{{dx}}=? \\ $$
Commented by Farruxjano last updated on 02/Jun/20
![I don′t know english perfect, but i can say [x]−the whole part of x. x=[x]+{x}](https://www.tinkutara.com/question/Q96517.png)
$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{english}\:\mathrm{perfect},\:\mathrm{but}\:\mathrm{i}\:\mathrm{can}\:\mathrm{say} \\ $$$$\left[\mathrm{x}\right]−\mathrm{the}\:\mathrm{whole}\:\mathrm{part}\:\mathrm{of}\:\mathrm{x}.\:\:\:\mathrm{x}=\left[\mathrm{x}\right]+\left\{\mathrm{x}\right\} \\ $$
Commented by prakash jain last updated on 02/Jun/20
![[(1/x)]=⌊(1/x)⌋ floor function?](https://www.tinkutara.com/question/Q96516.png)
$$\left[\frac{\mathrm{1}}{{x}}\right]=\lfloor\frac{\mathrm{1}}{{x}}\rfloor\:\mathrm{floor}\:\mathrm{function}? \\ $$
Commented by prakash jain last updated on 02/Jun/20
![∫_0 ^1 (1/(1+⌊(1/x)⌋))= x∈((1/2),1] ⇒⌊(1/x)⌋=1 area=(1/(1+⌊(1/x)⌋))×(1−(1/2)) x∈((1/3),(1/2)] ⇒⌊(1/x)⌋=2 area=(1/3)×((1/2)−(1/3)) x∈((1/(n+1)),(1/n)]⇒⌊(1/x)⌋=n area=(1/(n+1))×((1/n)−(1/(n+1)))=(1/(n+1)) =(1/(n(n+1)^2 )) I=Σ_(n=1) ^∞ (1/(n(n+1)^2 ))=2−(π^2 /6)](https://www.tinkutara.com/question/Q96518.png)
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+\lfloor\frac{\mathrm{1}}{{x}}\rfloor}= \\ $$$${x}\in\left(\frac{\mathrm{1}}{\mathrm{2}},\mathrm{1}\right]\:\Rightarrow\lfloor\frac{\mathrm{1}}{{x}}\rfloor=\mathrm{1}\:\:\mathrm{area}=\frac{\mathrm{1}}{\mathrm{1}+\lfloor\frac{\mathrm{1}}{{x}}\rfloor}×\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${x}\in\left(\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{2}}\right]\:\Rightarrow\lfloor\frac{\mathrm{1}}{{x}}\rfloor=\mathrm{2}\:\:\mathrm{area}=\frac{\mathrm{1}}{\mathrm{3}}×\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$${x}\in\left(\frac{\mathrm{1}}{{n}+\mathrm{1}},\frac{\mathrm{1}}{{n}}\right]\Rightarrow\lfloor\frac{\mathrm{1}}{{x}}\rfloor={n} \\ $$$$\:\:\:\:\:\:\mathrm{area}=\frac{\mathrm{1}}{{n}+\mathrm{1}}×\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)=\frac{\mathrm{1}}{{n}+\mathrm{1}} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${I}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{2}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$
Answered by abdomathmax last updated on 02/Jun/20
![I =∫_0 ^1 (dx/(1+[(1/x)])) changement (1/x)=t give I =−∫_1 ^(+∞) (1/(1+[t]))(−(dt/t^2 )) =∫_1 ^(+∞) (dt/(t^2 (1+[t]))) =Σ_(n=1) ^∞ ∫_n ^(n+1) (1/(n+1))×(dt/t^2 ) =Σ_(n=1) ^∞ (1/(n+1))[−(1/t)]_n ^(n+1) =Σ_(n=1) ^∞ (1/(n+1))((1/n)−(1/(n+1))) =Σ_(n=1) ^∞ (1/(n(n+1)))−Σ_(n=1) ^∞ (1/((n+1)^2 )) Σ_(n=1) ^∞ (1/(n(n+1))) =lim_(n→+∞) Σ_(k=1) ^n ((1/(k(k+1)))) =lim_(n→+∞) Σ_(k=1) ^n ((1/k)−(1/(k+1))) =lim_(n→+∞) (1−(1/(n+1)))=1 Σ_(n=1) ^∞ (1/((n+1)^2 )) =Σ_(n=2) ^∞ (1/n^2 ) =(π^2 /6)−1 ⇒ I =1−((π^2 /6)−1) =2−(π^2 /6)](https://www.tinkutara.com/question/Q96519.png)
$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dx}}{\mathrm{1}+\left[\frac{\mathrm{1}}{\mathrm{x}}\right]}\:\mathrm{changement}\:\frac{\mathrm{1}}{\mathrm{x}}=\mathrm{t}\:\mathrm{give} \\ $$$$\mathrm{I}\:=−\int_{\mathrm{1}} ^{+\infty} \:\frac{\mathrm{1}}{\mathrm{1}+\left[\mathrm{t}\right]}\left(−\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} }\right)\:=\int_{\mathrm{1}} ^{+\infty} \:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} \left(\mathrm{1}+\left[\mathrm{t}\right]\right)} \\ $$$$=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\int_{\mathrm{n}} ^{\mathrm{n}+\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}×\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} }\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\left[−\frac{\mathrm{1}}{\mathrm{t}}\right]_{\mathrm{n}} ^{\mathrm{n}+\mathrm{1}} \\ $$$$=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\left(\frac{\mathrm{1}}{\mathrm{n}}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\right) \\ $$$$=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}−\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}\:=\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \left(\frac{\mathrm{1}}{\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)}\right) \\ $$$$=\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \left(\frac{\mathrm{1}}{\mathrm{k}}−\frac{\mathrm{1}}{\mathrm{k}+\mathrm{1}}\right) \\ $$$$=\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\right)=\mathrm{1} \\ $$$$\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} }\:=\sum_{\mathrm{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{I}\:=\mathrm{1}−\left(\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\mathrm{1}\right)\:=\mathrm{2}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$