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1.60 Laplace transforms

Example.

Let $f(x)=e^{ax}.$ Then the Laplace transform is defined for $s>a$ with

F(s)={{1}\over{s-a}}.

Solution. If $a\neq s$ then we have

\int_{0}^{R}e^{ax}e^{-sx}dx={\left[{{e^{(a-s)x}}\over{(a-s)}}\right]_{0}^{R}}% \,{=\frac{e^{(a-s)R}-1}{a-s}.}

If $a<s$ then $e^{(a-s)R}\rightarrow{0}$ as $R\rightarrow\infty$ , which establishes the value of $F(s)$ for $s>a$ . If $a>s$ then $e^{(a-s)R}\rightarrow{\infty}$ as $R\rightarrow\infty$ , so the integral doesn’t converge. Finally, if $a=s$ then $\int_{0}^{R}e^{(a-s)x}\;dx={\int_{0}^{R}1\,dx}\,{=R}$ doesn’t converge either.