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Question Number 140827 by mathsuji last updated on 13/May/21

$$Find$$$$1.\int \frac{dx}{{\left[x\right]}^2},x⩾1$$$$2.\int \frac{{\left[x\right]}^{\lambda }}{x^{\lambda +1}},x⩾1$$$$Where[\ast ]denotetheintegerpart$$

Answered by mathmax by abdo last updated on 13/May/21

$$\mathrm{if}\mathrm{youmean}{\int }_1^{\mathrm{\infty }}\frac{\mathrm{dx}}{{\left[\mathrm{x}\right]}^2}\mathrm{let}\mathrm{I}={\int }_1^{\mathrm{\infty }}\frac{\mathrm{dx}}{{\left[\mathrm{x}\right]}^2}\Rightarrow$$$$\mathrm{I}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\int }_{\mathrm{n}}^{\mathrm{n}+1}\frac{\mathrm{dx}}{{\mathrm{n}}^2}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{1}{{\mathrm{n}}^2}=\frac{{\pi }^2}{6}$$$$\mathrm{if}\mathrm{I}={\int }_1^{\mathrm{x}}\frac{\mathrm{dt}}{{\left[\mathrm{t}\right]}^2}\Rightarrow \mathrm{I}=\sum _{\mathrm{n}=1}^{\left[\mathrm{x}\right]-1}{\int }_{\mathrm{n}}^{\mathrm{n}+1}\frac{\mathrm{dt}}{{\mathrm{n}}^2}=\sum _{\mathrm{n}=1}^{\left[\mathrm{x}\right]-1\mathrm{w}}\frac{1}{{\mathrm{n}}^2}$$

Commented by mathsuji last updated on 16/May/21

$$tankyouSir...$$

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