Question Number 87969 by M±th+et£s last updated on 07/Apr/20
Answered by mind is power last updated on 07/Apr/20
![=Σ_(k≥1) ∫_k ^((2k+1)/2) ((√(x−⌊x]))/([2x]^2 ))dx_(=S) +Σ_(k≥1) ∫_((2k+1)/2) ^(k+1) ((√(x−[x]))/([x]^2 ))dx_(=T) S=Σ_(k≥1) ∫_k ^((2k+1)/2) ((√(x−k))/(4k^2 ))dx=Σ_(k≥1) (1/(6k^2 ))[((1/2))^(3/2) ]=Σ_(k≥1) (1/(12k^2 (√2)))=(π^2 /(72(√2))) T=Σ_(k≥1) ∫_((2k+1)/2) ^(k+1) ((√(x−[x]))/([2x]^2 ))dx=Σ_(k≥1) ∫_((2k+1)/2) ^(k+1) ((√(x−[k]))/((2k+1)^2 ))dx =Σ_(k≥1) (2/(3(2k+1)^2 ))[1−((1/2))^(3/2) ]=−Σ_(k≥1) (1/(3(√2)(2k+1)^2 ))+Σ_(k≥1) (2/(3(2k+1)^2 )) =Σ_(k≥0) (1/((2k+1)^2 ))=(3/4)ζ(2)=(π^2 /8) T=(−(1/(3(√2)))+(2/3))(π^2 /8)−(2/3)(1−(1/(2(√2)))) S+T=(π^2 /(72(√2)))−(π^2 /(24(√2)))+((2π^2 )/(24))−(2/3)+(1/(3(√2))) =((−(√2)π^2 +6π^2 )/(72))+((12(√2)−48)/(72))=((π^2 (6−(√2))+12((√2)−4))/(72))](https://www.tinkutara.com/question/Q87981.png)
$$=\underset{{k}\geqslant\mathrm{1}} {\sum}\int_{{k}} ^{\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{2}}} \frac{\sqrt{\left.{x}−\lfloor{x}\right]}}{\left[\mathrm{2}{x}\right]^{\mathrm{2}} }{d}\underset{={S}} {{x}}+\underset{{k}\geqslant\mathrm{1}} {\sum}\int_{\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{2}}} ^{{k}+\mathrm{1}} \frac{\sqrt{{x}−\left[{x}\right]}}{\left[{x}\right]^{\mathrm{2}} }{d}\underset{={T}} {{x}} \\ $$$${S}=\underset{{k}\geqslant\mathrm{1}} {\sum}\int_{{k}} ^{\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{2}}} \frac{\sqrt{{x}−{k}}}{\mathrm{4}{k}^{\mathrm{2}} }{dx}=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{6}{k}^{\mathrm{2}} }\left[\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{12}{k}^{\mathrm{2}} \sqrt{\mathrm{2}}}=\frac{\pi^{\mathrm{2}} }{\mathrm{72}\sqrt{\mathrm{2}}} \\ $$$${T}=\underset{{k}\geqslant\mathrm{1}} {\sum}\int_{\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{2}}} ^{{k}+\mathrm{1}} \frac{\sqrt{{x}−\left[{x}\right]}}{\left[\mathrm{2}{x}\right]^{\mathrm{2}} }{dx}=\underset{{k}\geqslant\mathrm{1}} {\sum}\int_{\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{2}}} ^{{k}+\mathrm{1}} \frac{\sqrt{{x}−\left[{k}\right]}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{2}}{\mathrm{3}\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }\left[\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]=−\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{2}}\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }+\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{2}}{\mathrm{3}\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{3}}{\mathrm{4}}\zeta\left(\mathrm{2}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$${T}=\left(−\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{2}}}+\frac{\mathrm{2}}{\mathrm{3}}\right)\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\right) \\ $$$${S}+{T}=\frac{\pi^{\mathrm{2}} }{\mathrm{72}\sqrt{\mathrm{2}}}−\frac{\pi^{\mathrm{2}} }{\mathrm{24}\sqrt{\mathrm{2}}}+\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{24}}−\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{2}}} \\ $$$$=\frac{−\sqrt{\mathrm{2}}\pi^{\mathrm{2}} +\mathrm{6}\pi^{\mathrm{2}} }{\mathrm{72}}+\frac{\mathrm{12}\sqrt{\mathrm{2}}−\mathrm{48}}{\mathrm{72}}=\frac{\pi^{\mathrm{2}} \left(\mathrm{6}−\sqrt{\mathrm{2}}\right)+\mathrm{12}\left(\sqrt{\mathrm{2}}−\mathrm{4}\right)}{\mathrm{72}} \\ $$
Commented by M±th+et£s last updated on 07/Apr/20
$${nice}\:{solution}\:{thanx}\:{sir} \\ $$