76 Laplace transform of derivative

$$=|f(0)|+\frac{M{e}^{\beta x}}{\beta }-\frac{M}{\beta },$$

so $f$ satisfies $(E)$. Now for $s>\beta$, we integrate by parts to get

$${\int }_0^R{e}^{-sx}{f}^{\prime }(x)𝑑x={\left[e^{-sx}f(x)\right]}_0^R+s{\int }_0^R{e}^{-sx}f(x)𝑑x$$

$$=e^{-sR}f(R)-f(0)+s{\int }_0^R{e}^{-sx}f(x)𝑑x$$

so we let $R\to \mathrm{\infty }$ to get

$${\int }_0^{\mathrm{\infty }}{e}^{-sx}{f}^{\prime }(x)𝑑x=-f(0)+s{\int }_0^{\mathrm{\infty }}{e}^{-sx}f(x)𝑑x.$$