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6.40 Basic properties of Laplace transforms

Theorem.

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 (f+g)=(f)+(g){\mathcal{L}}(f+g)={\mathcal{L}}(f)+{\mathcal{L}}(g) and (cf)=c(f){\mathcal{L}}(cf)=c{\mathcal{L}}(f).

Proof. The first statement follows from observing that

0Re-sx(f(x)+g(x))dx=0Re-sxf(x)dx+0Re-sxg(x)dx\int_{0}^{R}e^{-sx}(f(x)+g(x))\,dx=\int_{0}^{R}e^{-sx}f(x)\,dx+\int_{0}^{R}e^{% -sx}g(x)\,dx

and taking limits as RR\rightarrow\infty. By the same argument, the second statement comes down to the fact that

0Re-sx.cf(x)dx=c0Re-sxf(x)dx.\int_{0}^{R}e^{-sx}.cf(x)\,dx=c\int_{0}^{R}e^{-sx}f(x)\,dx.