What is the correct answer for $\mathcal{L}^{-1}\left(\frac{p^2}{(p-3)^2}\right)?$

Question

What is the correct answer for $$\mathcal{L}^{-1}\left(\frac{p^2}{(p-3)^2}\right)?$$ Using partial fraction technique I got the answer as: $\delta(t)+e^{3t}9t+6e^{3t}$ and uing shifting thorem for inverse Laplace I gotthe answer as: $e^{3t}(\delta(t)+6+9t)$.

Answer 1

The frequency shift property of Laplace Transform states that: $$ \bbox[5px,border:1.1px solid black] { F(p+a)\iff e^{-at}\cdot f(t) } $$ It's implicit in this property that you must apply the inverse Laplace transform to $F(p)$, and then multiply the result by $e^{-at}$.

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We can use it in the case you've presented, look:

$$ F(p-3)=\frac{p^2}{(p-3)^2} $$ If $p'=p-3$, then $p=p'+3$. So, we have: $$ F(p'+3-3)=\frac{(p'+3)^2}{(p'+3-3)^2}\iff F(p')=\frac{(p'+3)^2}{(p')^2} \implies F(p)=\frac{(p+3)^2}{p^2} $$ Applying the inverse Laplace transform to $F(p)$, we have: $$ \begin{alignat}{1} \mathscr{L}^{-1}\left[F(p)\right]&=\mathscr{L}^{-1}\left[\frac{(p+3)^2}{p^2}\right] \\&=\mathscr{L}^{-1}\left[\frac{p^2+6p+9}{p^2}\right] \\&=\mathscr{L}^{-1}\left[1+\frac 6 p + \frac{9}{p^2}\right] \\&=\delta(t)+6+9t=f(t) \end{alignat} $$

Using the frequency shift property: $$ \bbox[5px,border:1.1px solid black] { \mathscr{L}^{-1}[F(p-3)]=e^{-(-3)t}\cdot f(t)=e^{3t}\cdot[\delta(t)+6+9t] } $$ So, both answers are correct

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Having that in mind, we need to prove that $$e^{3t}\delta(t)=\delta(t)$$ First, it's necessary to make some clarifications:

A "delta function" is not really a function, it's what is called a "generalized function", or distribution. So, when you see people write $\delta(0) = \infty$, this is just a "formality", and a very misleading one at that. The only correct way to think about a delta function is under an integral sign with another ("normal") function $f(x)$:
$$ \int \delta(x)f(x)dx $$ This expression makes perfect sense, and as long as $f(x)$ is "nice enough" (e.g. continuous at 0), it will evaluate to $f(0)$.
                                                                                                        _{Source: this answer, given by the user icurays1}

So, all we need is this definition: $$ \bbox[9px,border:1.1px solid black] { \int f(t)\cdot \delta(t)\,\,dt \triangleq f(0) } $$ The proof: $$ \begin{align} \delta(t)=1\cdot\delta(t) &\implies \int 1\cdot \delta(t)\,\,dt=1 \\ e^{3t}\cdot\delta(t)&\implies \int e^{3t}\cdot\delta(t)\,\,dt=e^{3\cdot\color{red}{0}}=1 \end{align} $$ So, we can conclude that: $$ \\ \int\delta(t)\,\,dt=\int e^{3t}\cdot\delta(t)\,\,dt \implies\delta(t)=e^{3t}\delta(t) $$

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One important note:

You should avoid interpreting the Dirac Delta Function this way: $$ \delta(x) = \left\{\begin{array}{cc} \infty & x = 0 \\ 0 & x\neq 0 \end{array}\right. $$

I'll show one example that illustrates why:

What is $\mathscr{L}\left[\,\delta(t)\,\right]$? And $\mathscr{L}\left[\,\delta(3t)\,\right]$?

Using the above interpretation, we have: $$ \begin{alignat}{0} \delta(t) = \left\{\begin{array}{cc} \infty & t = 0 \\ 0 & t\neq 0 \end{array}\right. &&\delta(3t) = \left\{\begin{array}{cc} \infty & 3t = 0 \\ 0 & 3t\neq 0 \end{array}\right. = \left\{\begin{array}{cc} \infty & t = 0 \\ 0 & t\neq 0 \end{array}\right. =\delta(t) \end{alignat} $$ So, since we have the same "function"... $\mathscr{L}\left[\,\delta(t)\,\right]$ must be equal to $\mathscr{L}\left[\,\delta(3t)\,\right]$, right? WRONG! $$ \begin{alignat}{1} \mathscr{L}\left[\,\delta(t)\,\right]&=1 \\ \mathscr{L}\left[\,\delta(3t)\,\right]&=\frac 1 3 \end{alignat} $$ But Laplace Transform Uniqueness Theorem states that:

$$\mathscr{L}[\,f_1\,]\neq\mathscr{L}[\,f_2\,] \iff f_1 \neq f_2$$

(And this theorem is valid for Dirac Delta 'function').

Then: $$ \mathscr{L}\left[\,\delta(t)\,\right] \neq \mathscr{L}[\,\delta(3t)\,] \iff \delta(t) \neq \delta(3t) $$

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Note:

One could argue: "But Dirac Delta 'function' isn't continuous. You have omitted the part of Uniqueness theorem that says that $f_1$ and $f_2$ must be continuous on $[0,∞)$.Then, the theorem is not valid for this case."

My counter argument: Yes, Delta 'function' isn't continuous. But there is a theorem stronger than this one which says that Uniqueness also applies to the Dirac Delta Function. See this excerpt, extracted from Laplace Transform Lecture Notes - Ante Mimica (Suggested by Calvin Khor here)

Answer 2

Both are correct as $\delta(t)=e^{3t}\delta(t)$. See the definitions of delta function and delta sequence for more details.

Answer 3

Both are correct because $\delta(t)=e^{3t}\delta(t).$

$$\delta(t)=\{_{0,\;\;x\neq0}^{+\infty,\;x=0}$$ and $$\int_{-\infty}^{+\infty}\delta(t)dt=1$$ Likewise $$e^{3t}\delta(t)=\{_{0,\;\;x\neq0}^{+\infty,\;x=0}$$ and $$\int_{-\infty}^{+\infty}e^{3t}\delta(t)dt=1$$ since $$\int_{-\infty}^{+\infty}f(t)\delta(t)dt=f(0)$$ Therefore $$\delta(t)=e^{3t}\delta(t)$$