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6.41 Existence of Laplace transforms

Proposition.

Suppose that $f:(0,\infty)\rightarrow{\mathbb{R}}$ is a continuous function that satisfies

|f(x)|\leq Me^{ax}\qquad(x>0)

for some constants $a$ and $M$ . Then the Laplace transform integral converges for all $s>a$ and satisfies

|{\mathcal{L}}(f)(s)|\leq{{M}\over{s-a}}\qquad(s>a).

Proof. By the comparison test, $|\int_{0}^{\infty}f(x)e^{-sx}\,dx|\leq\int_{0}^{\infty}Me^{ax}e^{-sx}\,dx$ . Since this integral converges to $\frac{M}{s-a}$ by Example 1.60, the Laplace transform of $f$ converges too.