I'm trying to prove the propositions which are in the comments of Willie Wong's answer. Let $S = \{ (-2)^k : k \in \mathbb{N} \}, 0\lt \epsilon \lt \frac14$ and $$\chi(t) = \begin{cases} 1 & \exists s\in S, \quad |t-s| < \epsilon \\ 0 & \text{otherwise} \end{cases}$$ Also the Lapace transform of $x(t)$ is defined as the $$X(s)=\mathcal{L}\{x(t)\}(s) = \int_{-\infty}^{+\infty} x(t)\exp(-st)dt, \ \ s \in \mathbb{C}$$We want to show that the following propositions hold:
- $\mathcal{L}\{\chi(t)\}(s)$ diverges for all $s \in \mathbb{C}$.
- $\mathcal{L}\{\exp(-\eta|t|)\chi(t)\}(s)$ converges only for $\Re(s)\in (-\eta , \eta)$ where $\eta \gt 0$ is a positive number.
- $\mathcal{L}\{\exp(-\eta|t| + 5\eta t)\chi(t)\}(s)$ converges only for $\Re(s)\in (4\eta , 6\eta)$ where $\eta \gt 0$ is a positive number.
My try:
Proposition 1: Note that we can write the integral as $$\mathcal{L}\{\chi(t)\}(s) = \sum_{k=1}^{\infty}\int_{(-2)^k-\epsilon}^{(-2)^k+\epsilon} \exp(-st)dt = \sum_{k=1}^{\infty}\left(\frac{\exp(-st)}{-s}\Bigr|_{t = (-2)^k-\epsilon}^{t = (-2)^k+\epsilon}\right) = \left( \frac{\exp(-s\epsilon) - \exp(s\epsilon)}{-s}\right) \times \sum_{k=1}^{\infty}\exp(-s(-2)^k)$$ It can be shown that the series $\sum_{k=1}^{\infty}\exp(-s(-2)^k)$ diverges for all $s \in \mathbb{C}$. So $\mathcal{L}\{\chi(t)\}(s)$ diverges for all $s \in \mathbb{C}, s \not = 0$. In the case $s = 0$, the integral clearly diverges: $$\mathcal{L}\{\chi(t)\}(0) = \sum_{k=1}^{\infty}\int_{(-2)^k-\epsilon}^{(-2)^k+\epsilon} dt = \sum_{k=1}^{\infty} 2\epsilon$$ Proposition 2: We have the inequality$$|\exp(-\eta|t|)\chi(t)\exp(-st)| = |\chi(t)\exp(-\eta |t|-st)| = |\chi(t)\exp(-\eta |t|-(\Re(s)+i\Im(s))t)| = \\ |\chi(t)||\exp(-\eta |t| - \Re(s)t)| \le \exp(-\eta |t| - \Re(s)t)$$ If $\Re(s)\in (-\eta , \eta)$ holds then the integral of RHS of inequality is convergent: $$\int_{-\infty}^{\infty} \exp(-\eta |t| - \Re(s)t)dt = \int_{-\infty}^{0}\exp(\underbrace{(\eta - \Re(s))}_{\gt 0}t)dt + \int_{0}^{+\infty}\exp(\underbrace{(-\eta - \Re(s))}_{\lt 0}t)dt$$ This shows that $\exp(-\eta|t|)\chi(t)\exp(-st)$ is absolutely integrable and so $\exp(-\eta|t|)\chi(t)\exp(-st)$ is integrable. In order to show that the Laplace transform is divergent in the other cases, I think we should explicitly compute the integral as in the previous proposition: $$\mathcal{L}\{\exp(-\eta|t|)\chi(t)\}(s) = \sum_{k=1}^{\infty}\int_{(-2)^k-\epsilon}^{(-2)^k+\epsilon} \exp(-st-\eta|t|)dt = \sum_{k=1}^{\infty}\left(\frac{\exp((\eta-s)t)}{\eta-s}\Bigr|_{t = (-2)^{2k-1}-\epsilon}^{t = (-2)^{2k-1}+\epsilon}\right) + \sum_{k=1}^{\infty}\left(\frac{\exp((-\eta-s)t)}{-\eta-s}\Bigr|_{t = (-2)^{2k}-\epsilon}^{t = (-2)^{2k}+\epsilon}\right)= \left( \frac{\exp((\eta - s)\epsilon) - \exp(-(\eta - s)\epsilon)}{\eta - s}\right) \times \sum_{k=1}^{\infty}\exp((\eta - s)(-2)^{2k-1}) + \left( \frac{\exp((-\eta - s)\epsilon) - \exp(-(-\eta - s)\epsilon)}{-\eta - s}\right) \times \sum_{k=1}^{\infty}\exp((-\eta - s)(-2)^{2k})$$ In the case $\Re(s) \ge \eta$, the first series is divergent since we have: $$\lim_{k \to \infty} |\exp((\eta - s)(-2)^{2k-1})| = \lim_{k \to \infty} \exp((\eta-\Re(s))(-2)^{2k-1}) = \begin{cases} 1 & \Re(s) = \eta \\ \infty &\Re(s) \gt \eta\end{cases}$$ In a similar manner, the second series diverges when $\Re(s) \le -\eta$ and we conclude that the Laplace transform converges only for $\Re(s)\in (-\eta , \eta)$.
Proposition 3: Now the hard work is done. We can apply the same steps as in the proposition 2. It's clear that the inequality $$|\exp(-\eta|t| + 5\eta t)\chi(t)\exp(-st)| = |\chi(t)\exp(-\eta |t|-st + 5\eta t)| = |\chi(t)\exp(-\eta |t|-(\Re(s)-5\eta+i\Im(s))t)| = \\ |\chi(t)||\exp(-\eta |t| - (\Re(s)-5\eta)t)| \le \exp(-\eta |t| - (\Re(s)-5\eta)t)$$ holds and the integral of RHS is: $$\int_{-\infty}^{\infty} \exp(-\eta |t| - (\Re(s) -5\eta)t)dt = \int_{-\infty}^{0}\exp(\underbrace{(6\eta - \Re(s))}_{\gt 0 \ \text{when} \ \Re(s) \lt 6\eta}t)dt + \int_{0}^{+\infty}\exp(\underbrace{(4\eta - \Re(s))}_{\lt 0 \ \text{when} \ \Re(s) \gt 4\eta}t)dt$$ This shows that the Laplace transform converges when $\Re(s) \in (4\eta , 6\eta)$. In this case the Laplace transform can be written as: $$\mathcal{L}\{\exp(-\eta|t| + 5\eta t)\chi(t)\}(s) = \sum_{k=1}^{\infty}\int_{(-2)^k-\epsilon}^{(-2)^k+\epsilon} \exp(-(s-5\eta)t-\eta|t|)dt = \sum_{k=1}^{\infty}\left(\frac{\exp((\eta-(s-5\eta))t)}{\eta-(s-5\eta)}\Bigr|_{t = (-2)^{2k-1}-\epsilon}^{t = (-2)^{2k-1}+\epsilon}\right) + \sum_{k=1}^{\infty}\left(\frac{\exp((-\eta-(s-5\eta))t)}{-\eta-(s-5\eta)}\Bigr|_{t = (-2)^{2k}-\epsilon}^{t = (-2)^{2k}+\epsilon}\right)= \left( \frac{\exp((\eta - (s-5\eta))\epsilon) - \exp(-(\eta - (s-5\eta))\epsilon)}{\eta - (s-5\eta)}\right) \times \sum_{k=1}^{\infty}\exp((\eta - (s-5\eta))(-2)^{2k-1}) + \left( \frac{\exp((-\eta - (s-5\eta))\epsilon) - \exp(-(-\eta - (s-5\eta))\epsilon)}{-\eta - (s-5\eta)}\right) \times \sum_{k=1}^{\infty}\exp((-\eta - (s-5\eta))(-2)^{2k})$$ It's clear that the first series diverges when $\Re(s) \ge 6\eta$ and the second series diverges when $\Re(s) \le 4\eta$.
Question: Is this reasoning correct? Also is there any shorter proof?