Question

In Problems 3 - 24 fill in the blanks or answer true or false.

24.

$\mathit{L}\left\{\underset{0}{\overset{t}{\int }}{e}^{aT}f\left(T\right)dT\right\}\mathbf{=}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}$ whereas

$\mathit{L}\left\{e^{at}\underset{0}{\overset{t}{\int }}f\left(T\right)dT\right\}\mathbf{=}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}$

Short answer

The solution is

$\begin{array}{rcl}L\left\{\underset{0}{\overset{t}{\int }}{e}^{aT}f\left(T\right)dT\right\}& =& F\left(s-a\right).\frac{1}{s}\\ L\left\{e^{at}\underset{0}{\overset{t}{\int }}f\left(T\right)dT\right\}& =& F\left(s-a\right).\frac{1}{s-a}\\ & & \end{array}$

Step-by-step solution

given information

The given value is:

$L\left\{\underset{0}{\overset{t}{\int }}{e}^{aT}f\left(T\right)dT\right\}=\underset{¯}{}$ whereas

$L\left\{e^{at}\underset{0}{\overset{t}{\int }}f\left(T\right)dT\right\}=\underset{¯}{}$

find L{∫0teaTfTdT}

$L\left\{\underset{0}{\overset{t}{\int }}{e}^{aT}f\left(T\right)dT\right\}=L\left\{e^{at}f\left(t\right)*1\right\}$

$=L\left\{e^{at}f\left(t\right)\right\}L\left\{1\right\}$ $\left(L\left\{f*g\right\}=L\left\{f\right\}L\left\{g\right\}\right)$

$L\left\{\underset{0}{\overset{t}{\int }}{e}^{aT}f\left(T\right)dT\right\}=L\left\{e^{at}f\left(t\right)\right\}.\frac{1}{s}$ $\left(L\left\{1\right\}=\frac{1}{s}\right)$

$=F\left(s-a\right).\frac{1}{s}$ $\left(L\left\{e^{at}f\left(t\right)\right\}=F\left(s-a\right)\right)$

Whereas

$L\left\{e^{at}\underset{0}{\overset{t}{\int }}f\left(T\right)dT\right\}={\left.L\left\{\underset{0}{\overset{t}{\int }}f\left(T\right)dT\right\}\right|}_{s\to s-a}$ $\left(\left\{e^{at}f\left(t\right)\right\}={\left.Lf\left(t\right)\right|}_{s\to s-a}\right)$

$L\left\{\underset{0}{\overset{t}{\int }}f\left(T\right)dT\right\}=L\left\{f\left(t\right)*1\right\}$

$=L\left\{f\left(t\right)\right\}L\left\{1\right\}$ $\left(L\left\{f*g\right\}=L\left\{f\right\}L\left\{g\right\}\right)$

$=L\left\{f\left(t\right)\right\}.\frac{1}{s}$ $\left(L\left\{1\right\}=\frac{1}{s}\right)$

Let $L\left\{f\left(t\right)\right\}=F\left(s\right)$, then

$$\begin{gathered} L\left\{\underset{0}{\overset{t}{\int }}f\left(T\right)dT\right\}=F\left(s\right).\frac{1}{s}\left(1\right) \\ L\left\{e^{at}\underset{0}{\overset{t}{\int }}f\left(T\right)dT\right\}={\left.L\left\{\underset{0}{\overset{t}{\int }}f\left(T\right)dT\right\}\right|}_{s\to s-a} \end{gathered}$$

$$\begin{gathered} \stackrel{1}{=}{\left.\left[F\left(s\right).\frac{1}{s}\right]\right|}_{s\to s-a} \\ =F\left(s-a\right).\frac{1}{s-a} \end{gathered}$$

Final answer

The solution is

$$\begin{gathered} L\left\{\underset{0}{\overset{t}{\int }}{e}^{aT}f\left(T\right)dT\right\}=F\left(s-a\right).\frac{1}{s} \\ L\left\{e^{at}\underset{0}{\overset{t}{\int }}f\left(T\right)dT\right\}=F\left(s-a\right).\frac{1}{s-a} \end{gathered}$$