First slide

Evaluation of definite integrals

Question

If $\int_{0}^{\infty} \frac{\sin x}{x} d x = \frac{π}{2}$ then
Statement-1: $\int_{0}^{\infty} \frac{\sin a x \cos b x}{x} d x = π / 2 (a > b > 0)$
Statement-2: $lim_{x \to 0} \frac{\sin a x \cos b x}{x} = a$

Moderate

A

STATEMENT-1 is True, STATEMENT-2 is True;
STATEMENT-2 is a correct explanation for STATEMENT-1
B

STATEMENT-1 is True, STATEMENT-2 is True;
STATEMENT-2 is NOT a correct explanation for STATEMENT-1
C

STATEMENT-1 is True, STATEMENT-2 is False
D

STATEMENT-1 is False, STATEMENT-2 is True

Solution

$\sin a x \cos b x = \frac{1}{2} [\sin (a + b) x + \sin (a - b) x]$
$\begin{array}{l} \int_{0}^{\infty} \frac{\sin a x \cos b x}{x} d x \\ = \frac{1}{2} [\int_{0}^{\infty} \frac{\sin (a + b) x}{x} d x + \int_{0}^{\infty} \frac{\sin (a - b) x}{x} d x] \\ = \frac{1}{2} [\int_{0}^{\infty} \frac{\sin t}{t} d t + \int_{0}^{\infty} \frac{\sin u}{u} d u] \\ = π / 2 \end{array}$

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