Question

Find the Laplace transform of f * g . Do not evaluate the convolution integral before transforming where$\mathbf{L}\left\{e^{2t}\mathrm{sint}\right\}$.

Short answer

The Laplace transformation is$L\{e^{-t}*e^t\cos t\}=\frac{s-1}{\left(s+1\right)\left[{\left(s-1\right)}^2+1\right]}$

Step-by-step solution

Perform the Laplace transformation:

$\begin{array}{l}L\left\{t^{-t}*e^t\cos t\right\}=L\left\{e^{-t}\right\}L\left\{e^t\cos t\right\}\\ L\left\{e^t\cos t\right\}=L{\left.\left\{\cos t\right\}\right|}_{s\to s-1}\\ ={\left.\frac{s}{s^2+1}\right|}_{s\to s-1}\\ =\left.\frac{s-1}{{\left(s-1\right)}^2+1}\right|\\ L\left\{e^{-t}\right\}=\frac{1}{s+1}\end{array}$

Using the formula to solve further: 

As we clearly see that

$\mathrm{L}\{{\mathrm{e}}^{-\mathrm{t}}*{\mathrm{e}}^{\mathrm{t}}\mathrm{cost}\}=\mathrm{L}\left\{{\mathrm{e}}^{\mathrm{t}}\right\}\mathrm{L}\{{\mathrm{e}}^{\mathrm{t}}\mathrm{cost}\}$

The Laplace transformation is $$\begin{gathered} L\{e^{-t}*e^{t}Cost\}=\frac{s-1}{\left(s+1\right)\left[{\left(s-1\right)}^2+1\right]} \\ =\frac{s-1}{\left(s+1\right)\left[{\left(s-1\right)}^2+1\right]} \end{gathered}$$