I=∫0π [cot⁡x]dx where  [·] denotes the greatest integer function, is equal to

I=∫0π [cot⁡x]dx=∫0π/2 ([cot⁡x]+[cot⁡(π−x)])dx=∫0π/2 ([cot⁡x]+[−cot⁡x])dxPut cot⁡x=t, so thatI=∫0∞ [t]+[−t]dt1+t2=limn→∞ ∑k=1n ∫k−1k ([t]+[−t])dt1+t2But [t]+[−t]=−1 for k−1