Correct Answer - Option 2 : π
\(\begin{array}{l} \mathop \smallint \limits_0^\infty \frac{1}{{1 + {x^2}}}dx = \left[ {{{\tan }^{ - 1}}x} \right]_0^\infty \\ \mathop \smallint \limits_0^\infty \frac{1}{{1 + {x^2}}}dx= {\tan ^{ - 1}}\infty - {\tan ^{ - 1}}0 = \frac{\pi }{2}\\ we know that\\L\left( {\sin x} \right) = \frac{1}{{{s^2} + 1}}\\ L\left( {\frac{{sinx}}{x}} \right) = \mathop \smallint \limits_s^\infty \frac{1}{{{s^2} + 1}}dx \end{array}\)
(Using “division by x”)
\(= \left[ {{{\tan }^{ - 1}}s} \right]_s^\infty \)
= tan-1 ∞ - tan-1 (s) = cot-1(s)
\(\Rightarrow \mathop \smallint \limits_0^\infty {e^{ - sx}}\frac{{sinx}}{x}dx = {\cot ^{ - 1}}\left( s \right)\)
(Using definition of Laplace transform)
Put s = 0, we get
\(\begin{array}{l} \mathop \smallint \limits_0^\infty \frac{{sinx}}{x}dx = {\cot ^{ - 1}}\left( 0 \right) = \frac{\pi }{2}\\ \mathop \smallint \limits_0^\infty\frac{1}{{1 + {x^2}}}dx + \mathop \smallint \limits_0^\infty \frac{{sinx}}{x}dx = \frac{\pi }{2}+\frac{\pi }{2}=\pi \end{array}\)
