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# 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), $I=\frac{4}{\pi }\times \frac{\pi }{2}\Rightarrow 2$

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