Question

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

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

Short answer

Laplace transform of the function is$\mathrm{L}\left\{{\int }_0^{\mathrm{t}}\mathrm{cosTdT}\right\}=\frac{1}{{\mathrm{s}}^2+1}$

Step-by-step solution

Theorems and formulas

The Laplace transform of the product of a function$f\left(t\right)$ with t can be found by differentiating the Laplace transform of $f\left(t\right)$.Accordingto theorem 7.4.2, convolution theorem.

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

Now use these two formulae for find the transformation.

Application of formulas and theorems

By using theorem 7.4.2 transform is find as:

$\begin{array}{c}L\left\{{\int }_0^t\cos TdT\right\}=L\left\{1\right\}.L\left\{\cos T\right\}\\ =\frac{1}{s}.\frac{s}{s^2+1}\\ =\frac{1}{s^2+1}\end{array}$

Hence, Laplace transform of the function is$L\left\{{\int }_0^t\cos TdT\right\}=\frac{1}{s^2+1}$