Question

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

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

Short answer

Laplace transform of the given function is,

$L\left\{t{\int }_0^{t}T{e}^{-T}dT\right\}=\frac{3s+1}{s^2{\left(s+1\right)}^2}$

Step-by-step solution

State the convolution theorem

The Laplace transform of the product of a function f (t)with t can be found by differentiating the Laplace transform of f (t).Accordingto convolution theorem, If f (t) and g (t) are piecewise continuous on and of exponential order, then $\mathrm{L}\left\{\mathrm{f}.\mathrm{g}\right\}=\mathrm{L}\left\{\mathrm{f}\left(\mathrm{t}\right)\right\}.\mathrm{L}\left\{\mathrm{g}\left(\mathrm{t}\right)\right\}=\mathrm{F}\left(\mathrm{s}\right).\mathrm{G}\left(\mathrm{s}\right)$. We use following formulas

Take the Laplace

By using formulas stated in above step, we can write

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

By using convolution theorem, we can write.

Simplify further as,

Substitute $forL\left\{{\int }_0^{t}T{e}^{-T}dT\right\}inL\left\{t{\int }_0^{t}T{e}^{-T}dT\right\}=\left(-1\right)\frac{d}{ds}L\left\{{\int }_0^{t}T{e}^{-T}dT\right\}$

$\begin{array}{c}L\left\{t{\int }_0^{t}T{e}^{-T}dT\right\}={\left(-1\right)}^1\frac{d}{ds}\left\{\frac{1}{s{\left(s+1\right)}^2}\right\}\\ =-\frac{d}{ds}\left\{\frac{1}{s{\left(s+1\right)}^2}\right\}\\ =-\left\{\frac{s{\left(s+1\right)}^2.\left(0\right)-1.{\left(s+1\right)}^2+2s\left(s+1\right)}{{\left(s{\left(s+1\right)}^2\right)}^2}\right\}\end{array}$

Simplify further as,

$\begin{array}{c}L\left\{t{\int }_0^{t}T{e}^{-T}dT\right\}=\frac{\left(s+1\right)\left(2s+s+1\right)}{s^2{\left(s+1\right)}^2}\\ =\frac{3s+1}{s^2{\left(s+1\right)}^2}\end{array}$

Thus, Laplace of given expression is $=\frac{3s+1}{s^2{\left(s+1\right)}^2}$