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expanding example
The unilateral Laplace transform is defined by
The unilateral Laplace transform is defined by
:<math>\mathcal{L} \left\{f(t)\right\}=\int_0^{\infty} e^{-st} f(t) \,dt </math>
:<math>\mathcal{L} \left\{f(t)\right\}=\int_0^{\infty} e^{-st} f(t) \,dt </math>
Given the integral converges. A necessary condition for this integral to converge is
\lim_{b\to\infty} e^{-sb}f(b)=0
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:<math>\int_0^{\infty}e^{-st}y'\,dt + \int_0^{\infty}e^{-st}y'\,dt </math>
:<math>\int_0^{\infty}e^{-st}y'\,dt + \int_0^{\infty}e^{-st}y'\,dt </math>
:<math>=e^{-st}y + s\int_0^{\infty}e^{-st}y\,dt + \int_0^{\infty}e^{-st}y\,dt </math>
:<math>=e^{-st}y + s\int_0^{\infty}e^{-st}y\,dt + \int_0^{\infty}e^{-st}y\,dt </math>
:<math>=e^{-st}y + (s+1) \int_0^{\infty}e^{-st}y\,dt </math>
:<math>=\lim_{b\to\infty}e^{-sb}y(b)-e^{-st}y(0) + (s+1) \int_0^{\infty}e^{-st}y\,dt </math>
For the transformation to converge,
:<math>\lim_{b\to\infty}e^{-sb}y(b)=0</math>
Therefore, substituting the initial condition y(0)=0 the left side becomes
:<math>(s+1) \int_0^{\infty}e^{-st}y\,dt </math>
Equating the two sides of the equation:
:<math>(s+1) \int_0^{\infty}e^{-st}y\,dt =\frac{1}{a-s}</math>
:<math>\int_0^{\infty}e^{-st}y\,dt =\frac{1}{(s+1)(a-s)}</math>
If <math>a \ne 1</math>, we can use partial fractions to changet the right side into
:<math>\int_0^{\infty}e^{-st}y\,dt =\frac{1}{a+1}\left(\frac{1}{a-s}-\frac{1}{s+1}\right)</math>
and the solution is obtained by noticing:
:<math>\int_0^{\infty}e^{-st+at} \,dt=\frac{1}{a-s}</math>
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