Problem 01 | Division by t

$\mathcal{L} (\sin t) = \dfrac{1}{s^2 + 1}$
 

$\displaystyle \mathcal{L} \left( \dfrac{\sin t}{t} \right) = \int_s^\infty \dfrac{du}{u^2 + 1}$

$\displaystyle \mathcal{L} \left( \dfrac{\sin t}{t} \right) = \big[ \arctan u \big]_s^\infty$

$\displaystyle \mathcal{L} \left( \dfrac{\sin t}{t} \right) = \lim_{a \to \infty}\big[ \arctan u \big]_s^a$

$\displaystyle \mathcal{L} \left( \dfrac{\sin t}{t} \right) = \lim_{a \to \infty}\big[ \arctan a - \arctan s \big]$
 

Let
$z = \arctan a - \arctan s$

$x = \arctan a, \, \tan x = a$

$y = \arctan s, \, \tan y = s$
 

Hence,
$z = x - y$

$\tan z = \tan (x - y)$

$\tan z = \dfrac{\tan x - \tan y}{1 + \tan x \tan y}$

$\tan z = \dfrac{a - s}{1 + as}$

$z = \arctan \left( \dfrac{a - s}{1 + as} \right)$

$\arctan a - \arctan s = \arctan \left( \dfrac{a - s}{1 + as} \right)$
 

Thus,
$\displaystyle \mathcal{L} \left( \dfrac{\sin t}{t} \right) = \lim_{a \to \infty} \left[ \arctan \left( \dfrac{a - s}{1 + as} \right) \right]$

$\displaystyle \mathcal{L} \left( \dfrac{\sin t}{t} \right) = \lim_{a \to \infty} \left[ \arctan \left( \dfrac{\dfrac{a}{a} - \dfrac{s}{a}}{\dfrac{1}{a} + \dfrac{as}{a}} \right) \right]$

$\displaystyle \mathcal{L} \left( \dfrac{\sin t}{t} \right) = \lim_{a \to \infty} \left[ \arctan \left( \dfrac{1 - \dfrac{s}{a}}{\dfrac{1}{a} + s} \right) \right]$

$\mathcal{L} \left( \dfrac{\sin t}{t} \right) = \arctan \left( \dfrac{1 - \dfrac{s}{\infty}}{\dfrac{1}{\infty} + s} \right)$

$\mathcal{L} \left( \dfrac{\sin t}{t} \right) = \arctan \left( \dfrac{1 - 0}{0 + s} \right)$

$\mathcal{L} \left( \dfrac{\sin t}{t} \right) = \arctan \left( \dfrac{1}{s} \right)$           answer