Question

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

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

Short answer

Laplace transform of the function is$L\left\{t{\int }_0^t\sin TdT\right\}=\frac{3s^2+1}{{\left(s^3+s\right)}^2}$

Step-by-step solution

State the convolution theorem

The Laplace transform of the product of a function f (t)with t can be found by differentiating the Laplace transform of f (t).Accordingto convolution theorem, If f (t) and g (t) are piecewise continuous on $[0,\infty )$and of exponential order, then localid="1668595985835" $\mathbf{L}\left\{f.g\right\}\mathbf{=}\mathbf{L}\left\{f\left(t\right)\right\}\mathbf{.}\mathbf{L}\left\{g\left(t\right)\right\}\mathbf{=}\mathbf{F}\left(s\right)\mathbf{.}\mathbf{G}\left(s\right)$. We use following formulas

role="math" localid="1668596026232"

Take the Laplace

By using formulas stated in above step, we can write

$L\left\{t{\int }_0^t\sin TdT\right\}=\left(-1\right)\frac{d}{ds}\left\{\frac{L\left\{\sin t\right\}}{s}\right\}$

Take the Laplace on right side as follows and simplify.

$\begin{array}{c}L\left\{t{\int }_0^t\sin TdT\right\}=\left(-1\right)\frac{d}{ds}\left\{\frac{1}{s\left(s^2+1\right)}\right\}\\ =-\frac{d}{ds}\left\{\frac{1}{\left(s^3+s\right)}\right\}\\ =-\left\{-\frac{-3s^2-1}{{\left(s^3+s\right)}^2}\right\}\\ =\frac{3s^2+1}{{\left(s^3+s\right)}^2}\end{array}$

Thus, Laplace of given expression is$L\left\{t{\int }_0^t\sin TdT\right\}=\frac{3s^2+1}{{\left(s^3+s\right)}^2}$