Collin's questions via email about Inverse Laplace Transforms

[Prove It]

Evaluate $\displaystyle \begin{align*} \mathcal{L}^{-1}\,\left\{ \frac{5\,s^2 + 20\,s + 26}{\left( s + 3 \right) ^3} \right\} \end{align*}$

As the denominator is a function of s + 3, it suggests a shift had to have been utilised. As such, we also need the numerator to be a function of s + 3...

Let $\displaystyle \begin{align*} u = s + 3 \end{align*}$, then $\displaystyle \begin{align*} s = u-3 \end{align*}$ and thus

$\displaystyle \begin{align*} 5\,s^2 + 20\,s + 26 &= 5\,\left( u - 3 \right) ^2 + 20\,\left( u - 3 \right) + 26 \\ &= 5\,\left( u^2 - 6\,u + 9 \right) + 20\,u - 60 + 26 \\ &= 5\,u^2 - 30\,u + 45 + 20\,u - 34 \\ &= 5\,u^2 - 10\,u + 11 \\ &= 5\,\left( s + 3 \right) ^2 - 10\,\left( s + 3 \right) + 11 \end{align*}$

and thus

$\displaystyle \begin{align*} \mathcal{L}^{-1}\,\left\{ \frac{5\,s^2 + 20\,s + 26 }{\left( s + 3 \right) ^3} \right\} &= \mathcal{L}^{-1}\,\left\{ \frac{5\,\left( s + 3 \right) ^2 - 10\,\left( s + 3 \right) + 11}{\left( s + 3 \right) ^3} \right\} \\ &= \mathrm{e}^{-3\,t}\,\mathcal{L}^{-1}\,\left\{ \frac{5\,s^2 - 10\,s + 11}{s^3} \right\} \\ &= \mathrm{e}^{-3\,t}\,\mathcal{L}^{-1}\,\left\{ \frac{5}{s} - \frac{10}{s^2} + \frac{11}{s^3} \right\} \\ &= \mathrm{e}^{-3\,t}\,\left( 5\,\mathcal{L}^{-1}\,\left\{ \frac{0!}{s^{0 + 1}} \right\} - 10\,\mathcal{L}^{-1}\,\left\{ \frac{1!}{s^{1 + 1}} \right\} + \frac{11}{2}\,\mathcal{L}^{-1}\,\left\{ \frac{2!}{s^{2 + 1}} \right\} \right) \\ &= \mathrm{e}^{-3\,t}\,\left( 5\,t^0 - 10\,t^1 + \frac{11}{2}\,t^2 \right) \\ &= \mathrm{e}^{-3\,t} \,\left( 5 - 10\,t + \frac{11}{2}\,t^2 \right) \end{align*}$

Evaluate $\displaystyle \begin{align*} \mathcal{L}^{-1}\,\left\{ 7\log{ \left( \frac{7 + s}{s} \right) } \right\} \end{align*}$

As logarithms have a very simple integral - becoming a rational function, I would make use of this rule:

$\displaystyle \begin{align*} \mathcal{L}\,\left\{ \frac{f\left( t \right) }{t} \right\} = \int_s^{\infty}{ F\left( u \right) \,\mathrm{d}u } \end{align*}$

Now we should note that

$\displaystyle \begin{align*} \int{\left( \frac{1}{7 + u} - \frac{1}{u} \right) \,\mathrm{d}u } &= \log{ \left| 7 + u \right| } - \log{ \left| u \right| } + C \\ &= \log{ \left| \frac{7 + u}{u} \right| } + C \end{align*}$

and thus (since $\displaystyle \begin{align*} s > 0 \end{align*}$)

$\displaystyle \begin{align*} \int_s^{\infty}{ \left( \frac{1}{7 + u} - \frac{1}{u} \right) \,\mathrm{d}u } &= \lim_{b \to \infty} \int_s^b{ \left( \frac{1}{7 + u} - \frac{1}{u} \right) \,\mathrm{d}u } \\ &= \lim_{b \to \infty} \left[ \log{ \left( \frac{7 + u}{u} \right) } \right] _s^b \\ &= \lim_{b \to \infty} \left[ \log{\left( \frac{7 + b}{b} \right) } \right] - \log{ \left( \frac{7 + s}{s} \right) } \\ &= \lim_{b \to \infty} \left[ \log{ \left( \frac{7}{b} + 1 \right) } \right] - \log{ \left( \frac{7 + s}{s} \right) } \\ &= \log{ \left( 0 + 1 \right) } - \log{ \left( \frac{7 + s}{s} \right) } \\ &= 0 -\log{ \left( \frac{7 + s}{s} \right) } \\ &= -\log{ \left( \frac{7 + s}{s} \right) } \end{align*}$

So this suggests that $\displaystyle \begin{align*} F\left( s \right) = - \left( \frac{1}{7 + s} - \frac{1}{s} \right) = \frac{1}{s} - \frac{1}{7 + s} \end{align*}$, therefore

$\displaystyle \begin{align*} f\left( t \right) &= \mathcal{L}^{-1}\,\left\{ \frac{1}{s} - \frac{1}{7 + s} \right\} \\ &= \mathcal{L}^{-1}\,\left\{ \frac{0!}{s^{0 + 1}} \right\} - \mathrm{e}^{-7\,t}\,\mathcal{ L }^{-1}\,\left\{ \frac{0!}{s^{0 + 1}} \right\} \\ &= t^0 - \mathrm{e}^{-7\,t}\,t^0 \\ &= 1 - \mathrm{e}^{-7\,t} \end{align*}$

so finally $\displaystyle \begin{align*} \mathcal{L}^{-1}\,\left\{ 7 \log{ \left( \frac{7 + s}{s} \right) } \right\} &= 7\,\left( \frac{1 - \mathrm{e}^{-7\,t}}{t} \right) \end{align*}$.

 

[benorin]

This is correct. Ty been wanting to study integral transforms recently.