Use the convolution integral to find the inverse transforms of:$\frac{\mathbf{p}}{\left(p+a\right)\left(p^2+b^2\right)}$

The equation is $\frac{\mathrm{p}}{\left(\mathrm{p}+\mathrm{a}\right)\left({\mathrm{p}}^2+{\mathrm{b}}^2\right)}$.

The piecewise-continuous and exponentially-restricted real function f(t) is the inverse Laplace transform of a function $\mathbf{F}\left(s\right)$, and it has the property:

$$\begin{gathered} \mathbf{L}\left\{f\right\}\left(s\right)\mathbf{=}\mathbf{L}\left\{f\left(t\right)\right\}\left(s\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{F}\left(s\right) \end{gathered}$$

where $\mathbf{L}$is the Laplace transform.

As per Convolution theorem, if we have two functions, taking their convolution and then Laplace is the same as taking the Laplace first (of the two functions separately) and then multiplying the two Laplace Transforms.

Consider the equation.

$\frac{\mathrm{p}}{\left(\mathrm{p}+\mathrm{a}\right)\left({\mathrm{p}}^2-{\mathrm{b}}^2\right)}=\frac{1}{\left(\mathrm{p}+\mathrm{a}\right)}.\frac{\mathrm{p}}{\left({\mathrm{p}}^2-{\mathrm{b}}^2\right)}$

As per the convolution theorem.

$$\begin{gathered} {\mathrm{L}}^{-1}\left[\frac{1}{\left(\mathrm{p}+\mathrm{a}\right)}.\frac{\mathrm{p}}{\left({\mathrm{p}}^2-{\mathrm{b}}^2\right)}\right]={\int }_0^{\mathrm{t}}{\mathrm{g}}_1.{\mathrm{g}}_2\mathrm{dx} \\ ={\int }_0^{\mathrm{t}}{\mathrm{g}}_1\left(\mathrm{t}\right){\mathrm{g}}_2\left(\mathrm{t}-\mathrm{x}\right)\mathrm{d} \\ ={\int }_0^{\mathrm{t}}{\mathrm{e}}^{-\mathrm{ax}}\frac{\mathrm{coshb}\left(\mathrm{t}-\mathrm{x}\right)}{\mathrm{b}}\mathrm{dx} \\ ={\int }_0^{\mathrm{t}}{\mathrm{e}}^{-\mathrm{ax}}\mathrm{coshb}\left(\mathrm{t}-\mathrm{x}\right)\mathrm{dx} \end{gathered}$$

Solve further,

$$\begin{gathered} {\mathrm{L}}^{-1}\left[\frac{1}{\left(\mathrm{p}+\mathrm{a}\right)}.\frac{\mathrm{p}}{\left({\mathrm{p}}^2-{\mathrm{b}}^2\right)}\right]={\int }_0^{\mathrm{t}}{\mathrm{e}}^{-\mathrm{ax}}\left(\mathrm{coshbtcoshbx}-\mathrm{sinhbtsinhbx}\right)\mathrm{dx} \\ =\frac{\mathrm{sinhbt}}{\mathrm{b}}\frac{1}{\left({\mathrm{b}}^2-{\mathrm{a}}^2\right)}{\left[{\mathrm{be}}^{-\mathrm{ax}}\mathrm{sinhbx}+{\mathrm{ae}}^{-\mathrm{ax}}\mathrm{coshbx}\right]}_0^{\mathrm{t}} \\ -\frac{\mathrm{coshbt}}{\mathrm{b}}\frac{1}{\left({\mathrm{b}}^2-{\mathrm{a}}^2\right)}{\left[{\mathrm{be}}^{-\mathrm{ax}}\mathrm{sinhbx}+{\mathrm{ae}}^{-\mathrm{ax}}\mathrm{coshbx}\right]}_0^{\mathrm{t}} \\ =\frac{1}{\left({\mathrm{b}}^2-{\mathrm{a}}^2\right)}\left[{\mathrm{ae}}^{-\mathrm{at}}\left({\cosh }^2\mathrm{bt}-{\sinh }^2\mathrm{bt}\right)-\mathrm{acoshbt}+\mathrm{bsinhbt}\right] \\ =\frac{1}{\left({\mathrm{b}}^2-{\mathrm{a}}^2\right)}\left[{\mathrm{ae}}^{-\mathrm{at}}\left(1\right)-\mathrm{acoshbt}-\mathrm{bsinhbt}\right] \end{gathered}$$

Solve further,

${\mathrm{L}}^{-1}\left[\frac{1}{\left(\mathrm{p}+\mathrm{a}\right)}.\frac{\mathrm{p}}{\left({\mathrm{p}}^2-{\mathrm{b}}^2\right)}\right]$role="math" localid="1659258470122" $=\frac{1}{\left({\mathrm{a}}^2-{\mathrm{b}}^2\right)}$$(-{\mathrm{ae}}^{-\mathrm{at}}+\mathrm{acoshbt}-\mathrm{bsinhbt})$

Thus, the inverse transform of given equation is role="math" localid="1659258534291" $\frac{1}{\left({\mathrm{a}}^2-{\mathrm{b}}^2\right)}(-{\mathrm{ae}}^{-\mathrm{at}}+\mathrm{acoshbt}-\mathrm{bsinhbt})$.