Let f(x)f(x) and g(x)g(x) be functions whose Laplace transforms exist for all s>αs>\alpha, and let cc be a constant. Then ℒ(f+g)=ℒ(f)+ℒ(g){\mathcal{L}}(f+g)={\mathcal{L}}(f)+{\mathcal{L}}(g) and ℒ(cf)=cℒ(f){\mathcal{L}}(cf)=c{\mathcal{L}}(f).
Proof. The first statement follows from observing that
and taking limits as R→∞R\rightarrow\infty. By the same argument, the second statement comes down to the fact that