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Find the Laplace transform of $\frac{d}{dt}\bigg(\frac{1 - cos2t}{t}\bigg)$
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$L[1 - cos2t] = L[1] - L[cos2t]$ $L[1 - cos2t] = \frac{1}{s} - \frac{s}{s^2 + 2^2}$ $\therefore L\bigg[\frac{1 - cos2t}{t}\bigg] = \int\limits_{s}^{\infty}\bigg[\frac{1}{s} - \frac{s}{s^2 + 4}\bigg] ds$ By division property $\therefore L\bigg[\frac{1 - cos2t}{t}\bigg] = \int\limits_{s}^{\infty}\bigg[\frac{1}{s} - \frac{2s}{2(s^2 + 4)}\bigg]ds$ By integration $\int\frac{f'(x)}{f(x)} = logf(x)$ $$ = \bigg[logs - \frac{1}{2}log(s^2 + 4)\bigg]_{s}^{\infty}..........(1)$$

$\therefore$ Taking $\frac{1}{2}$ common from equation (1) we get $$ = \frac{1}{2}\big[2logs - log(s^2 + 4)\big]_{s}^{\infty}$$ $$ = \frac{1}{2}\bigg[log\bigg[\frac{s^2}{s^2 + 4}\bigg]\bigg]_{s}^{\infty}$$ $$ = \frac{1}{2}log\bigg[\frac{1}{1 + \frac{4}{s^2}}\bigg]_{s}^{\infty}$$

Now substituting value of s = $\infty$ we get $$\frac{1}{2}log\bigg[\frac{1}{1 + 0}\bigg] - \frac{1}{2}log\bigg[\frac{1}{1 + 4/s^2}\bigg]$$ $$ = 0 - \frac{1}{2}log\bigg[\frac{s^2}{s^2 + 4}\bigg]$$ $$ = \frac{1}{2}log\bigg[\frac{s^2 + 4}{s^2}\bigg]$$

$\therefore$ Now applying the differentiation property $f'(t) = sL(t) -f(0)$

$f(0) = \bigg[\frac{1 - cos0}{0}\bigg] = \frac{0}{0}$ Not defined

Taking derivative on both numerator and denominator according to L hospital rule we get

$f(0) = \frac{1 - (-sin2t) * 2}{1}$ substituting t = 0

$f(0) = \frac{0}{1} = 0 $

$\therefore L\bigg[\frac{d}{dt}\bigg(\frac{1 - cos2t}{t}\bigg)\bigg] = \frac{s}{2}log\frac{s^2 + 4}{s^2}$

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