_0^ [ x],dx, where [.] denotes the greatest integer function, is equal to:

Question

$\displaystyle\int_0^\pi [\cot x]\,dx$, where $[.]$ denotes the greatest integer function, is equal to:

Accepted Answer

Let $$I = \int_0^\pi  {\left[ {\cot x} \right]dx\,\,\,\,\,\,...\left( 1 \right)} $$

$$ = \int_0^\pi  {\left[ {\cot \left( {\pi  - x} \right)} \right]} dx$$

$$ = \int_0^\pi  {\left[ { - \cot x} \right]dx\,\,\,\,\,\,...\left( 2 \right)} $$

Adding two values of $I$ in $$e{q^n}s\left( 1 \right)\,\,\& \,\,\left( 2 \right),$$

We get $$2I = \int_0^\pi  {\left( {\left[ {\cot x} \right] + \left[ { - \cot x} \right]} \right)dx} $$

$$ = \int_0^\pi  {\left( { - 1} \right)dx} $$

$$\left[ {} \right.$$ As $$\left[ x \right] + \left[ { - x} \right] =  - 1,\,\,if\,\,$$ $$x \notin z$$

and $$\left[ x \right] + \left[ { - x} \right] = 0,\,\,if\,\,x \in z$$ $$\left. {} \right]$$

$$ = \left[ { - x} \right]_0^\pi  =  - \pi  \Rightarrow I =  - {\pi  \over 2}$$

Solutions · 1