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1.60 Laplace transforms

Example.

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

F⁢(s)=1s-a.F(s)={{1}\over{s-a}}.

Solution. If a≠sa\neq s then we have

∫0Rea⁢x⁢e-s⁢x⁢d⁢x=[e(a-s)⁢x(a-s)]0R=e(a-s)⁢R-1a-s.\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<sa<s then e(a-s)⁢R→0e^{(a-s)R}\rightarrow{0} as R→∞R\rightarrow\infty, which establishes the value of F⁢(s)F(s) for s>as>a. If a>sa>s then e(a-s)⁢R→∞e^{(a-s)R}\rightarrow{\infty} as R→∞R\rightarrow\infty, so the integral doesn’t converge. Finally, if a=sa=s then ∫0Re(a-s)⁢x⁢d⁢x=∫0R1⁢d⁢x=R\int_{0}^{R}e^{(a-s)x}\;dx={\int_{0}^{R}1\,dx}\,{=R} doesn’t converge either.