Question

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

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

Short answer

Laplace transform of the given function is$L\left\{{\int }_0^t{e}^{T}dT\right\}=\frac{1}{s^2\left(s-1\right)}$

Step-by-step solution

Statement of theorem

According to convolution 7.4.2 theorem:

If $f\left(t\right)$and $g\left(t\right)$are piecewise continuous on$[0,\mathrm{\infty })$ and of exponential order, then

$L\{f.g\}=L\left\{f\left(t\right)\right\}.L\left\{g\left(t\right)\right\}=F\left(s\right).G\left(s\right)$.

Application of theorem

By using theorem 7.4.2 given equation is written as:

$\begin{array}{c}L\left\{{\int }_0^t{e}^{T}dT\right\}=L\left\{1.e^t\right\}\\ L\left\{{\int }_0^{t}T{e}^{t-T}dT\right\}=L\left\{1\right\}.L\left\{e^t\right\}\\ =\frac{0!}{s^{0+1}}.\frac{1}{s-1}\\ =\frac{1}{s}.\frac{1}{s-1}\end{array}$

Further solve the above expression,

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

Laplace transform of the given function is $L\left\{{\int }_0^t{e}^{T}dT\right\}=\frac{1}{s^2\left(s-1\right)}$