Question Number 140827 by mathsuji last updated on 13/May/21
![Find 1. ∫(dx/([x]^2 )) , x≥1 2. ∫(([x]^λ )/x^(λ+1) ) , x≥1 Where [∗] denote the integer part](https://www.tinkutara.com/question/Q140827.png)
$${Find} \\ $$$$\mathrm{1}.\:\int\frac{{dx}}{\left[{x}\right]^{\mathrm{2}} }\:,\:{x}\geqslant\mathrm{1} \\ $$$$\mathrm{2}.\:\int\frac{\left[{x}\right]^{\lambda} }{{x}^{\lambda+\mathrm{1}} }\:,\:{x}\geqslant\mathrm{1} \\ $$$${Where}\:\left[\ast\right]\:{denote}\:{the}\:{integer}\:{part} \\ $$
Answered by mathmax by abdo last updated on 13/May/21
![if youmean ∫_1 ^∞ (dx/([x]^2 )) let I =∫_1 ^∞ (dx/([x]^2 )) ⇒ I =Σ_(n=1) ^∞ ∫_n ^(n+1) (dx/n^2 ) =Σ_(n=1) ^∞ (1/n^2 )=(π^2 /6) if I =∫_1 ^x (dt/([t]^2 )) ⇒ I =Σ_(n=1) ^([x]−1) ∫_n ^(n+1) (dt/n^2 ) =Σ_(n=1) ^([x]−1w) (1/n^2 )](https://www.tinkutara.com/question/Q140851.png)
$$\mathrm{if}\:\mathrm{youmean}\:\int_{\mathrm{1}} ^{\infty} \:\frac{\mathrm{dx}}{\left[\mathrm{x}\right]^{\mathrm{2}} }\:\:\mathrm{let}\:\mathrm{I}\:=\int_{\mathrm{1}} ^{\infty} \:\frac{\mathrm{dx}}{\left[\mathrm{x}\right]^{\mathrm{2}} }\:\Rightarrow \\ $$$$\mathrm{I}\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\int_{\mathrm{n}} ^{\mathrm{n}+\mathrm{1}} \:\frac{\mathrm{dx}}{\mathrm{n}^{\mathrm{2}} }\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\mathrm{if}\:\mathrm{I}\:=\int_{\mathrm{1}} ^{\mathrm{x}} \:\frac{\mathrm{dt}}{\left[\mathrm{t}\right]^{\mathrm{2}} }\:\Rightarrow\:\mathrm{I}\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\left[\mathrm{x}\right]−\mathrm{1}} \:\int_{\mathrm{n}} ^{\mathrm{n}+\mathrm{1}} \:\frac{\mathrm{dt}}{\mathrm{n}^{\mathrm{2}} }\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\left[\mathrm{x}\right]−\mathrm{1w}} \:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} } \\ $$
Commented by mathsuji last updated on 16/May/21
$${tankyou}\:{Sir}… \\ $$