Question

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

$\mathbf{L}\left\{{\int }_0^t{\mathrm{Te}}^{t-T}\mathrm{dT}\right\}$

Short answer

Laplace transform of the given function is

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

Step-by-step solution

State convolution theorem

Accordingto convolution theorem, If f (t) and g (t) are piecewise continuous on$\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)}$ and of exponential order, then $\mathbf{L}\left\{f.g\right\}\mathbf{=}\mathbf{L}\left\{f\left(t\right)\right\}\mathbf{.}\mathbf{L}\left\{g\left(t\right)\right\}\mathbf{=}\mathbf{F}\left(s\right)\mathbf{.}\mathbf{G}\left(s\right)$.

Take the Laplace

By using convolution theorem, we can write.

$\begin{array}{l}L\left\{{\int }_0^{t}T{e}^{t-T}dT\right\}=L\left\{t.e^t\right\}\\ L\left\{{\int }_0^{t}T{e}^{t-T}dT\right\}=L\left\{t\right\}.L\left\{e^t\right\}\end{array}$

Take the Laplace on right side as follows and simplify.

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

Thus Laplace of given expression is $\frac{1}{s^2\left(s-1\right)}$