# GATE 2016 EE – SET 2 Question 29

Using Laplace Transform, consider, $$2\int_{-\infty }^{\infty }\frac{\sin 2\pi t}{\pi t}dt=I$$

Since $$\frac{\sin 2\pi t}{\pi t}$$ is an even function, the integral now becomes,

$$2\int_{-\infty }^{\infty }\frac{\sin 2\pi t}{\pi t}dt\Rightarrow 2\left [ 2\int_{0}^{\infty }\frac{\sin 2\pi t}{\pi t} \right ]dt\Rightarrow 4\int_{0}^{\infty }\frac{\sin 2\pi t}{\pi t}dt\Rightarrow \frac{4}{\pi }\int_{0}^{\infty }\frac{\sin 2\pi t}{t}dt$$ ________ (1)

Laplace transform of $$\frac{\sin 2\pi t}{t}=\int_{s}^{\infty }L[\sin 2\pi t]ds$$ $$\because L[\frac{f(t)}{t}]=\int_{s}^{\infty }L[f(t)]ds$$

$$\int_{s}^{\infty }L[\sin 2\pi t]ds\Rightarrow \int_{s}^{\infty }\frac{2\pi }{s^{2}+(2\pi )^{2}}ds$$

$$\because L(\sin at)=\frac{a}{s^{2}+a^{2}}$$ also,  $$\because \int \frac{a}{x^{2}+a^{2}}dx=\tan ^{-1}(\frac{x}{a})$$

$$\Rightarrow\tan ^{-1}\left [ \frac{s}{2\pi } \right ]_{s}^{\infty }\Rightarrow \tan ^{-1}\infty -\tan ^{-1}\left ( \frac{s}{2\pi } \right )$$

$$\Rightarrow \frac{\pi }{2}-\tan ^{-1}\left ( \frac{s}{2\pi } \right )\Rightarrow \cot ^{-1}\left ( \frac{s}{2\pi } \right )$$

We now have, $$L\left [ \frac{\sin 2\pi t}{t} \right ]=\cot ^{-1}\left ( \frac{s}{2\pi } \right )$$

Taking s = 0, $$\int_{0}^{\infty }\frac{\sin 2\pi t}{t}dt=\cot ^{-1}(0)=\frac{\pi }{2}$$ _______(2)

From (1) and (2), $$I=\frac{4}{\pi }\times \frac{\pi }{2}\Rightarrow 2$$