How to prove that the inverse Laplace Transform of zero is zero itself?
$$\mathscr{L}^{-1}\{0\}=0$$
I know that the inverse Laplace Transform of a constant is Dirac's Delta. But I think that that applies only to positive constants.
Thanks
[Math] Inverse Laplace Transform of zero
laplace transform
Related Solutions
The Gamma function has poles at $z=0$ and at the negative integers. When $z=-k$, then the residue at that pole is $(-1)^k/k!$. Now, the inverse transform is
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, \Gamma(p+1+T s) e^{s (t-T \log{N})} $$
The poles of this Gamma function are at $p+1+T s = -k$, or $s=-(p+1+k)/T$ for $k \in \{0,-1,-2,\ldots \}$. Thus, the ILT is
$$e^{-(p+1)(t-T \log{N}))/T} H(t-T \log{N})\sum_{k=0}^{\infty} \frac{(-1)^k}{k!} e^{-k (t-T \log{N})/T} $$
where $H$ is the Heaviside function, or, summing the series,
$$e^{-(p+1)(t-T \log{N}))/T} \exp{\left [-e^{-(t-T \log{N})/N}\right ]} H(t-T \log{N}) $$
which, aside from the Heaviside factor, agrees with your expected result.
This problem is incorrectly stated. The Final Value Theorem is stated as follows (or check out a signals and systems text):
If $f$ is bounded on $(0,\infty)$ and $\lim\limits_{t\rightarrow\infty}f(t) < M$ for some $M$, then \begin{align*} \lim\limits_{t\rightarrow\infty}f(t) = \lim\limits_{s \searrow 0} sF(s) \end{align*} where $F(s)$ is the unilateral Laplace Transform of $f$.
With this, the final value (DC Gain) is $0$. The OP has this in their solution, since $\lim\limits_{t\rightarrow \infty} e^{-at} \rightarrow 0$, assuming $a>0$. Otherwise, the function is unbounded, and the final value theorem does not apply since it has a pole outside of the open left half plane.
Here's what I think the problem meant to say: Let $a>0$ and $Y(s) = G(s)U(s) = \frac{1}{s}\frac{A}{s+a}$. Using the FVT, find the DC gain, and the time domain response.
Then, applying FVT gives, \begin{align*} \lim\limits_{s \searrow 0} sY(s) = \lim\limits_{s \searrow 0} s\frac{A}{s+a}\frac{1}{s} = \lim\limits_{s \searrow 0} \frac{A}{s+a} = \frac{A}{a} \end{align*}
As the solution above suggests, partial fractions can be used to get the total response, or, if you have been introduced to some complex analysis, you can also use Cauchy's Residue Theorem.
Applying partial fractions should give $\frac{A/a}{s} + \frac{-A/a}{s+a}$, which when factored gives the final answer. Note that if you take $t\rightarrow \infty$, you find the final value (DC gain) as expected.
Best Answer
If $\mathcal{L}(f) = F$, then $\mathcal{L}^{-1}(F) =f$. $\mathcal{L}(0) = 0$ because $\mathcal{L}$ is a linear operator. Or you can actually compute $\mathcal{L}(0)$ using the definition.