Question

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

$\mathbf{L}\left\{{\int }_0^t\mathrm{TsinTdT}\right\}$

Short answer

Laplace transform of the function is$\mathrm{L}\left\{{\int }_0^{\mathrm{t}}{\mathrm{e}}^{-\mathrm{T}}\mathrm{cosTdT}\right\}=\frac{\mathrm{s}+1}{\mathrm{s}\left[{\left(\mathrm{s}+1\right)}^2+1\right]}$

Step-by-step solution

Formulas and theorems

The Laplace transform of the product of a function f (t) with t can be found by differentiating the Laplace transform of f (t). According to theorem 7.4.2, convolution theorem.

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

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

Now use the formulas for find the solution:

$\begin{array}{c}L\left\{t^{n}f\left(t\right)\right\}={\left(-1\right)}^n\frac{d^n}{d{s}^n}F\left(s\right)\\ L\left\{1\right\}=\frac{1}{s}\end{array}$

Applications of theorems and formulas

By using theorem 7.4.2 Find the transform of given equation:

$\begin{array}{c}L\left\{{\int }_0^{t}T\sin TdT\right\}=L\left\{{\int }_0^{t}T\sin T.1dT\right\}\\ =L\left\{T.\sin T\right\}.L\left\{1\right\}\end{array}$

Laplace transform of the function is $\mathrm{L}\left\{{\int }_0^{\mathrm{t}}{\mathrm{e}}^{-\mathrm{T}}\mathrm{cosTdT}\right\}=\frac{\mathrm{s}+1}{\mathrm{s}\left[{\left(\mathrm{s}+1\right)}^2+1\right]}$

Since it knows from theorem 7.4.1 $L\left\{t^{n}f\left(t\right)\right\}={\left(-1\right)}^n\frac{d^n}{d{s}^n}F\left(s\right)$, $f\left(t\right)=\cos T,F\left(s\right)=\frac{1}{s^2+1}$

where n=1.

Thus, it is written as:

$\begin{array}{c}L\left\{T\cos T\right\}={\left(-1\right)}^1\frac{d}{ds}\left(\frac{1}{s^2+1}\right)\\ =-\frac{d}{ds}\left(\frac{1}{s^2+1}\right)\\ =-\left\{-\frac{2s}{{\left(s^2+1\right)}^2}\right\}\\ L\left\{T\cos T\right\}=\frac{2s}{{\left(s^2+1\right)}^2}\end{array}$

Substituting in equation $L\left\{T\cos T\right\}=\frac{2s}{{\left(s^2+1\right)}^2}$, $L\left\{t^{n}f\left(t\right)\right\}={\left(-1\right)}^n\frac{d^n}{d{s}^n}F\left(s\right)$it gives

$$\begin{gathered} L\left\{{\int }_0^{t}T\sin TdT\right\}=\frac{2s}{{\left(s^2+1\right)}^2}.\frac{1}{s} \\ L\left\{{\int }_0^{t}T\sin TdT\right\}=\frac{2}{{\left(s^2+1\right)}^2} \end{gathered}$$

Hence, Laplace transform of the function is$L\left\{{\int }_0^t{e}^{-T}\cos TdT\right\}=\frac{s+1}{s\left[{\left(s+1\right)}^2+1\right]}$