$I={\int }_0^{\pi } [\cot x]dx$ where $[·]$denotes the greatest integer function, is equal to

$\begin{array}{l}I={\int }_0^{\pi } [\cot x]dx={\int }_0^{\pi /2} \left(\right[\cot x]+[\cot (\pi -x)\left]\right)dx\\ ={\int }_0^{\pi /2} \left(\right[\cot x]+[-\cot x\left]\right)dx\end{array}$

Put $\cot x=t$, so that

$\begin{array}{l}I={\int }_0^{\mathrm{\infty }} \left[t\right]+[-t]\frac{dt}{1+t^2}\\ =\underset{n\to \mathrm{\infty }}{lim} \sum _{k=1}^n {\int }_{k-1}^k \left(\right[t]+[-t\left]\right)\frac{dt}{1+t^2}\end{array}$

But $\left[t\right]+[-t]=-1$for $k-1<t<k$, there fore

$\begin{array}{l}{\int }_{k-1}^k \left(\right[t]+[-t\left]\right)\frac{dt}{1+t^2}\\ ={\int }_{k-1}^k (-1)\frac{dt}{1+t^2}=\left[{\tan }^{-1}k-{\tan }^{-1}(k-1)\right]\end{array}$

$\therefore I=-\underset{n\to \mathrm{\infty }}{lim} \sum _{k=1}^n \left({\tan }^{-1}k-{\tan }^{-1}(k-1)\right)$

$=-\underset{n\to \mathrm{\infty }}{lim} \left[{\tan }^{-1}n-{\tan }^{-1}0\right]=-\frac{\pi }{2}.$