Let $ y(t)=\int_{-\infty}^{\infty}2x(t)dt $
Then $ h(t) =2u(t) $
And $ H(s) = \int_{-\infty}^{\infty}h(t)e^{-st}dt $
$ =\int_{-\infty}^{\infty}2u(t)e^{-st}dt $
$ =\int_{0}^{\infty}2e^{-st}dt $
$ =(\frac{-2}{s}e^{-st})|_{0}^{\infty} $
$ =\frac{2}{s} $
