${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}(t-1)e^{-st}dt=-\frac{(t-1)e^{-st}}{s}-\frac{e^{-st}}{s^2}+C$

This allows $usto$ evaluate the improper integral and find $L\{f(t)\},$ such that $s>0.$ The result $is$ a function $ofs.$

$L\{f(t)\}=2{\displaystyle \underset{1}{\overset{}{}}}(t-1)e^{-st}dt$

$=2(-\frac{(t-1)e^{-st}}{s}-\frac{e^{-st}}{s^2}){|}_1^{}$