First slide

Evaluation of definite integrals

Question

$I = \int_{0}^{π} [\cot x] d x$ where $[\cdot]$ denotes the greatest integer function, is equal to

Moderate

A

-1
B

$- π / 2$
C

$π / 2$
D

1

Solution

$\begin{array}{l} I = \int_{0}^{π} [\cot x] d x = \int_{0}^{π / 2} ([\cot x] + [\cot (π - x)]) d x \\ = \int_{0}^{π / 2} ([\cot x] + [- \cot x]) d x \end{array}$
Put $\cot x = t$ , so that
$\begin{array}{l} I = \int_{0}^{\infty} [t] + [- t] \frac{d t}{1 + t^{2}} \\ = lim_{n \to \infty} \sum_{k = 1}^{n} \int_{k - 1}^{k} ([t] + [- t]) \frac{d t}{1 + t^{2}} \end{array}$
But $[t] + [- t] = - 1$ for $k - 1 < t < k$ , there fore
$\begin{array}{l} \int_{k - 1}^{k} ([t] + [- t]) \frac{d t}{1 + t^{2}} \\ = \int_{k - 1}^{k} (- 1) \frac{d t}{1 + t^{2}} = [\tan^{- 1} k - \tan^{- 1} (k - 1)] \end{array}$
$∴ I = - lim_{n \to \infty} \sum_{k = 1}^{n} (\tan^{- 1} k - \tan^{- 1} (k - 1))$
$= - lim_{n \to \infty} [\tan^{- 1} n - \tan^{- 1} 0] = - \frac{π}{2} .$

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