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Find the value of integral $$\int^{2\pi}_{-\frac{\pi}{2}}\lfloor \cot^{-1}(x)\rfloor dx$$

$\lfloor x \rfloor = x-\{x\}$ and $0\leq \{x\}<1$

for $0\leq x<\cot(1),\lfloor \cot^{-1}(x)\rfloor = 0$ and for $\cot (1)\leq x<2\pi,\lfloor \cot^{-1}(x)\rfloor = 0$

but i want go further for negative interval of $x,$ could some help with me this.

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$$\int^{2\pi}_{-\frac{\pi}{2}}\lfloor \cot^{-1}(x)\rfloor dx\\= \int^{2\pi}_{0}\lfloor \cot^{-1}(x)\rfloor dx+\int^{0}_{-\frac{\pi}{2}}\lfloor \cot^{-1}(x)\rfloor dx=$$ $$\int^{0.642}_{0}(+1) dx+\int^{0.642}_{2\pi}(0)+\int^{0}_{-\frac{\pi}{2}}\lfloor \cot^{-1}(x)\rfloor dx=$$ $$0.642(+1)+0+\int^{-0.458}_{-\frac{\pi}{2}}\lfloor \cot^{-1}(x)\rfloor dx+\int^{0}_{-0.458}\lfloor \cot^{-1}(x)\rfloor dx=\\ 0.642(+1)+0+\int^{-0.458}_{-\frac{\pi}{2}}(+2) dx+\int^{0}_{-0.458}(+1)dx=\\0.642(+1)+0+(+2)\times(-0.458-(-\frac{\pi}{2}))+(+1)\times(0-(-0.458)$$

remark$$\cot^{-1}(0)=\frac{\pi}{2}=1.57.. \\\cot^{-1}(-0.458)=2\\ \cot^{-1}(-\frac{\pi}{2})=2.575\\ 0 <x<-0.458 \to \lfloor \cot^{-1}(x)\rfloor=1\\ -0.458 <x<-\frac{\pi}{2} \to \lfloor \cot^{-1}(x)\rfloor=2$$

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  • $\begingroup$ Should it not $\displaystyle \int^{2\pi}_{0}\lfloor \cot^{-1}(x) \rfloor dx = \int^{\cot 1}_{0}1dx+\int^{2\pi}_{\cot 1}0 dx = \cot 1$ $\endgroup$
    – DXT
    Oct 14, 2016 at 7:23
  • $\begingroup$ yes . you are right ,I' correct it $\endgroup$
    – Khosrotash
    Oct 14, 2016 at 7:24

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