$\mathcal{L} (\sin t) = \dfrac{1}{s^2 + 1}$
$\displaystyle \mathcal{L} \left( \dfrac{\sin t}{t} \right) = \int_s^\infty \dfrac{du}{u^2 + 1}$
$\displaystyle \mathcal{L} \left( \dfrac{\sin t}{t} \right) = \big[ \arctan u \big]_s^\infty$
$\displaystyle \mathcal{L} \left( \dfrac{\sin t}{t} \right) = \lim_{a \to \infty}\big[ \arctan u \big]_s^a$
$\displaystyle \mathcal{L} \left( \dfrac{\sin t}{t} \right) = \lim_{a \to \infty}\big[ \arctan a - \arctan s \big]$
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
