Solution:

Consider

$\int_0^{\infty{}}e^{-st}\frac{{sin}^2t}{t}dt= L\ \left[\frac{{sin}^2t}{t}\right] \\[2ex] L\ \left[{sin}^2t\right] = L\ \left[\frac{1-\cos{2t}}{2}\right] \\[2ex] \frac{1}{2}\left[L\left(1\right)-L(cos2t)\right] \\[2ex] \frac{1}{2}\left[\frac{1}{s}\ -\frac{s}{{s^2+2}^2}\ \right] $

$If \ L [f(t)]=\phi (s), \\[2ex] L[\dfrac{1}{\ t}f(t)] = \int_s^{\infty{}}\phi (s)ds \\[2ex] \therefore{} L\ \left[\dfrac{{sin}^2t}{t}\right] = \displaystyle \int_s^{\infty{}}\dfrac{1}{2}\left[\dfrac{1}{s}\ -\dfrac{s}{s^2+4}\ \right]ds \\[2ex] =\dfrac{1}{2}{\left[\log{s-}\ \dfrac{1}{2}log(s^2+4)\right]}_s^{\infty{}} \\[2ex] =\dfrac{1}{4}{\left[2log{s}-\ log(s^2+4)\right]}_s^{\infty{}} \\[2ex] =\dfrac{1}{4}{\left[\ log\dfrac{s^2}{s^2+4}\right]}_s^{\infty{}}\\ -\dfrac{1}{4}\ log\dfrac{s^2}{s^2+4}$

$= \dfrac{1}{4}\ log\frac{s^2+4}{s^2}$

$\therefore\displaystyle \int_0^{\infty{}}e^{-st}\frac{{sin}^2t}{t}dt= \frac{1}{4}\ log\frac{s^2+4}{s^2}$

Putting s = 1,

$\displaystyle \int_0^{\infty{}}e^{-t}\frac{{sin}^2t}{t}dt = \frac{1}{4}\ log\frac{1^2+4}{1^2}= \frac{1}{4}log\ 5$