Question

Find the Laplace transform of.$\mathit{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\underset{0}{\overset{t}{\int }}{e}^{v}sin(t-v)dv$

Short answer

The Laplace transformation of${\int }_0^t{e}^v\sin (t-v)dv$is$\frac{1}{(s^2+1)(s-1)}$.

Step-by-step solution

Given that,

$f\left(t\right)={\int }_0^t{e}^v\sin (t-v)dv$

Applying the Convolution Theorem

$$\begin{gathered} g\star h={\int }_0^{t}g(t-v)h\left(v\right)dv \\ g\left(t\right)=\sin t \\ g\left(s\right)=\frac{1}{s^2+1} \\ h\left(t\right)=e^t \\ g\left(s\right)=\frac{1}{s-1} \end{gathered}$$

Step 3: Simplify

$\begin{array}{c}\mathcal{L}[g\star h]=g\left(s\right)\cdot h\left(s\right)\\ =\frac{1}{s^2+1}\cdot \frac{1}{s-1}\\ =\frac{1}{(s^2+1)(s-1)}\end{array}$

Hence,

$f\left(s\right)=\frac{1}{(s^2+1)(s-1)}$