Question

Find the Laplace transform of f * g using Theorem 7.4.2. Do not evaluate the convolution integral before transforming.

$\mathbf{L}\left\{{\int }_0^t{e}^{-T}\mathrm{cosTdT}\right\}$

Short answer

Laplace transform of the function is$\text{L}\left\{{\int }_0^{\mathrm{t}}{\mathrm{e}}^{-\mathrm{T}}\mathrm{cosTdT}\right\}=\frac{\mathrm{s}+1}{\mathrm{s}[{\left(\mathrm{s}+1\right)}^2+1]}$

Step-by-step solution

Formulas and theorems

The Laplace transform of the product of a function f (t) with t can be found by differentiating the Laplace transform of g (t). According to theorem 7.4.2, convolution theorem.

If f (t) and g (t) are piecewise continuous on $[0,\infty )$and of exponential order, then$\mathrm{L}\left\{\mathrm{f}.\mathrm{g}\right\}=\mathrm{L}\left\{\mathrm{f}\left(\mathrm{t}\right)\right\}.\mathrm{L}\left\{\mathrm{g}\left(\mathrm{t}\right)\right\}=\mathrm{F}\left(\mathrm{s}\right).\mathrm{G}\left(\mathrm{s}\right)$

Now use these two formulas for find the transform:

$\begin{array}{c}L\left\{1\right\}=\frac{1}{s}\\ L\left\{e^{-T}\cos T\right\}=\frac{s+1}{{\left(s+1\right)}^2+1^2}\end{array}$

Application of theorem and formulas

By using theorem 7.4.2 transform is find as;

$\begin{array}{c}L\left\{{\int }_0^t{e}^{-T}\cos TdT\right\}=L\left\{{\int }_0^t{e}^{-T}\cos T.1dT\right\}\\ =L\left\{e^{-T}\cos T\right\}.L\left\{1\right\}\\ =\frac{s+1}{{\left(s+1\right)}^2+1}.\frac{1}{s}\\ =\frac{s+1}{s\left[{\left(s+1\right)}^2+1\right]}\end{array}$

Hence, Laplace transform of the function is $L\left\{{\int }_0^t{e}^{-T}\cos TdT\right\}=\frac{s+1}{s\left[{\left(s+1\right)}^2+1\right]}$