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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

$$=\sum _{k⩾1}{\int }_k^{\frac{2k+1}{2}}\frac{\sqrt{x-\lfloor x]}}{{\left[2x\right]}^2}d\underset{=S}{x}+\sum _{k⩾1}{\int }_{\frac{2k+1}{2}}^{k+1}\frac{\sqrt{x-\left[x\right]}}{{\left[x\right]}^2}d\underset{=T}{x}$$$$S=\sum _{k⩾1}{\int }_k^{\frac{2k+1}{2}}\frac{\sqrt{x-k}}{4k^2}dx=\sum _{k⩾1}\frac{1}{6k^2}\left[{\left(\frac{1}{2}\right)}^{\frac{3}{2}}\right]=\sum _{k⩾1}\frac{1}{12k^2\sqrt{2}}=\frac{{\pi }^2}{72\sqrt{2}}$$$$T=\sum _{k⩾1}{\int }_{\frac{2k+1}{2}}^{k+1}\frac{\sqrt{x-\left[x\right]}}{{\left[2x\right]}^2}dx=\sum _{k⩾1}{\int }_{\frac{2k+1}{2}}^{k+1}\frac{\sqrt{x-\left[k\right]}}{{(2k+1)}^2}dx$$$$=\sum _{k⩾1}\frac{2}{3{(2k+1)}^2}[1-{\left(\frac{1}{2}\right)}^{\frac{3}{2}}]=-\sum _{k⩾1}\frac{1}{3\sqrt{2}{(2k+1)}^2}+\sum _{k⩾1}\frac{2}{3{(2k+1)}^2}$$$$=\sum _{k⩾0}\frac{1}{{(2k+1)}^2}=\frac{3}{4}\zeta \left(2\right)=\frac{{\pi }^2}{8}$$$$T=(-\frac{1}{3\sqrt{2}}+\frac{2}{3})\frac{{\pi }^2}{8}-\frac{2}{3}(1-\frac{1}{2\sqrt{2}})$$$$S+T=\frac{{\pi }^2}{72\sqrt{2}}-\frac{{\pi }^2}{24\sqrt{2}}+\frac{2{\pi }^2}{24}-\frac{2}{3}+\frac{1}{3\sqrt{2}}$$$$=\frac{-\sqrt{2}{\pi }^2+6{\pi }^2}{72}+\frac{12\sqrt{2}-48}{72}=\frac{{\pi }^2(6-\sqrt{2})+12(\sqrt{2}-4)}{72}$$

Commented by M±th+et£s last updated on 07/Apr/20

$$nicesolutionthanxsir$$

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