MHB Evaluating Definite Integrals with Floor Function

[juantheron]

Evaluation of $$\displaystyle \int_{0}^{\pi}\lfloor \cot x \rfloor dx$$ and $$\displaystyle \int_{0}^{\pi}\lfloor \cos x \rfloor dx\;,$$ where $$\lfloor x \rfloor $$ denote Floor function of $$x$$

 

[chisigma]

Evaluation of $$\displaystyle \int_{0}^{\pi}\lfloor \cot x \rfloor dx$$ and $$\displaystyle \int_{0}^{\pi}\lfloor \cos x \rfloor dx\;,$$ where $$\lfloor x \rfloor $$ denote Floor function of $$x$$

[sp]By treating the integral as 'improper' You obtain a telescopic sum...

$\displaystyle \int_{0}^{\pi} \lfloor \cot x \rfloor dx = \int_{- \frac{\pi}{2}}^{\frac{\pi}{2}} \lfloor \tan x \rfloor dx = \lim_{n \rightarrow \infty} \tan^{-1} 0 - \tan^{-1} 1 + \tan^{-1} 1 - \tan^{-1} 2 + ... + \tan^{-1} (n-1) - \tan^{-1} n = - \frac{\pi}{2}$ [/sp]

Kind regards

$\chi$ $\sigma$