Question

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

$\mathbf{L}\left\{{\int }_0^t\mathrm{sinT}.\cos \left(t-T\right)\mathrm{dT}\right\}$

Short answer

Laplace transform of the function is

$\mathrm{L}\left\{{\int }_0^{\mathrm{t}}\mathrm{sinTcos}\left(\mathrm{t}-\mathrm{T}\right)\mathrm{dT}\right\}=\frac{\mathrm{s}}{{\mathrm{s}}^2{\left({\mathrm{s}}^2+1\right)}^2}$

Step-by-step solution

State convolution theorem

Accordingto convolution theorem, If f (t) and g (t) are piecewise continuous on$\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)}$$\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)$

Take the Laplace

$\mathrm{L}\left\{\mathrm{coskt}\right\}=\frac{\mathrm{s}}{{\mathrm{s}}^2+{\mathrm{k}}^2}$By using convolution theorem, we can write.

$\begin{array}{l}\mathrm{L}\left\{{\int }_0^{\mathrm{t}}\mathrm{sinT}.\cos \left(\mathrm{t}-\mathrm{T}\right)\mathrm{dT}\right\}=\mathrm{L}\left\{\mathrm{sint}.\mathrm{cost}\right\}\\ \mathrm{L}\left\{{\int }_0^{\mathrm{t}}\mathrm{sinT}.\cos \left(\mathrm{t}-\mathrm{T}\right)\mathrm{dT}\right\}=\mathrm{L}\left\{\mathrm{sint}\right\}.\mathrm{L}\left\{\mathrm{cost}\right\}\end{array}$

Take the Laplace on right side using $L\left\{\sin kt\right\}=\frac{k}{s^2+k^2}$, and simplify as follows:

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

Thus Laplace of given expression is $\begin{array}{c}\frac{s}{s^2{\left(s^2+1\right)}^2}\end{array}$