Using the definition of the Laplace transform
Formula for the definition of the Laplace transform
To solve a Laplace transform in the form ???L\left\{f(t)\right\}??? using the definition of the Laplace transform, you’ll need to solve the equation
???F(s)=\int^\infty_0{e}^{-st}f(t)\ dt???
where ???F(s)??? is the Laplace transform of the function ???f(t)???, ???s??? is a constant, and ???f(t)??? is the given function.
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Applying the definition of the Laplace transform in order to transform a differential equation
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Laplace transform for an exponential function
Example
Use the definition to find the Laplace transform of the function.
???f(t)=e^{-2t}???
Plugging the given function into the definition of the Laplace transform gives
???F(s)=\int^\infty_0{e}^{-st}f(t)\ dt???
???F(s)=\int^\infty_0{e}^{-st}e^{-2t}\ dt???
???F(s)=\int^\infty_0{e}^{-st-2t}\ dt???
???F(s)=\int^\infty_0{e}^{(-s-2)t}\ dt???
To Find the Laplace transform of a function using the definition of the Laplace transform, we’ll need to solve an integral equation.
Since ???s??? is a constant, we can integrate and say
???F(s)=\frac{1}{-s-2}e^{(-s-2)t}\Big|^\infty_0???
???F(s)=\lim_{b\to\infty}\frac{1}{-s-2}e^{(-s-2)t}\Big|^b_0???
Evaluate over the interval.
???F(s)=\lim_{b\to\infty}\frac{1}{-s-2}e^{(-s-2)b}-\frac{1}{-s-2}e^{(-s-2)0}???
???F(s)=\lim_{b\to\infty}\frac{1}{-s-2}e^{-(sb+2b)}-\frac{1}{-s-2}e^0???
???F(s)=\frac{1}{-s-2}e^{-\infty}-\frac{1}{-s-2}e^0???
Since ???e^{-\infty}=0???,
???F(s)=\frac{1}{-s-2}(0)-\frac{1}{-s-2}(1)???
???F(s)=-\frac{1}{-s-2}???
???F(s)=\frac{1}{s+2}???
This is the Laplace transform of ???f(t)=e^{-2t}???.
