There is a mistake in your probability result (which should be clear by the fact that it is unbounded). Using the interval $\text{CI}(X) = [X-b, X+c]$ you should have the coverage probability:

$$\begin{align} \mathbb{P}(\theta \in \text{CI}(X)) &= \mathbb{P}(X-b \leqslant \theta \leqslant X+c) \\[6pt] &= \mathbb{P}(\theta-c \leqslant X \leqslant \theta+b) \\[6pt] &= \int \limits_{\theta-c}^{\theta+b} \text{Laplace}(x|\theta,1) \ dx \\[6pt] &= \frac{1}{2} \int \limits_{\theta-c}^{\theta+b} e^{-|x-\theta|} \ dx \\[6pt] &= \frac{1}{2} \Bigg[ \ \int \limits_{\theta}^{\theta+b} e^{-x+\theta} \ dx - \int \limits_{\theta}^{\theta+c} e^{-x+\theta} \ dx \Bigg] \\[6pt] &= \frac{1}{2} \Bigg[ (1-e^{-b}) - (1-e^{-c}) \Bigg] \\[6pt] &= \frac{e^{-c} - e^{-b}}{2}. \\[6pt] \end{align}$$

(Observe that, unlike your result, this approaches one when $b \rightarrow \infty$ or $c \rightarrow \infty$.) Thus, finding the optimal confidence interval of this form requires you to solve the following optimisation problem:

$$\text{Minimise } b+c \quad \text{ subject to } \quad e^{-c} - e^{-b} = 2(1-\alpha).$$

With a bit of work, it should be possible for you to show that the optima occurs when $b=c$, so that the optimal confidence interval is one with midpoint at $x$. This is unsurprising, given that the Laplace distribution is symmetric around the mean parameter $\theta$.

There is a mistake in your probability result (which should be clear by the fact that it is unbounded). Using the interval $\text{CI}(X) = [X-b, X+c]$ you should have the coverage probability:

$$\begin{align} \mathbb{P}(\theta \in \text{CI}(X)) &= \mathbb{P}(X-b \leqslant \theta \leqslant X+c) \\[6pt] &= \mathbb{P}(\theta-c \leqslant X \leqslant \theta+b) \\[6pt] &= \int \limits_{\theta-c}^{\theta+b} \text{Laplace}(x|\theta,1) \ dx \\[6pt] &= \frac{1}{2} \int \limits_{\theta-c}^{\theta+b} e^{-|x-\theta|} \ dx \\[6pt] &= \frac{1}{2} \Bigg[ \ \int \limits_{\theta}^{\theta+b} e^{-x+\theta} \ dx + \int \limits_{\theta-c}^{\theta} e^{x-\theta} \ dx \Bigg] \\[6pt] &= \frac{1}{2} \Bigg[ \ \int \limits_{0}^{b} e^{-r} \ dr + \int \limits_{-c}^{0} e^{r} \ dr \Bigg] \\[6pt] &= \frac{1}{2} \Bigg[ (1-e^{-b}) + (1-e^{-c}) \Bigg] \\[6pt] &= 1 - \frac{e^{-c} + e^{-b}}{2}. \\[6pt] \end{align}$$

(Observe that, unlike your result, this approaches one when $b \rightarrow \infty$ and $c \rightarrow \infty$.) Thus, finding the optimal confidence interval of this form requires you to solve the following optimisation problem:

$$\text{Minimise } b+c \quad \text{ subject to } \quad e^{-c} + e^{-b} \leqslant 2 \alpha.$$

With a bit of work, it should be possible for you to show that the optima occurs when $b=c$, so that the optimal confidence interval is one with midpoint at $x$. This is unsurprising, given that the Laplace distribution is symmetric around the mean parameter $\theta$.

There is a mistake in your probability result (which should be clear by the fact that it is unbounded). Using the interval $\text{CI}(X) = [X-b, X+c]$ you should have the coverage probability:

$$\begin{align} \mathbb{P}(\theta \in \text{CI}(X)) &= \mathbb{P}(X-b \leqslant \theta \leqslant X+c) \\[6pt] &= \mathbb{P}(\theta-c \leqslant X \leqslant \theta+b) \\[6pt] &= \int \limits_{\theta-c}^{\theta+b} \text{Laplace}(x|\theta,1) \ dx \\[6pt] &= \frac{1}{2} \int \limits_{\theta-c}^{\theta+b} e^{-|x-\theta|} \ dx \\[6pt] &= \frac{1}{2} \Bigg[ \ \int \limits_{\theta}^{\theta+b} e^{-x+\theta} \ dx - \int \limits_{\theta}^{\theta+c} e^{-x+\theta} \ dx \Bigg] \\[6pt] &= \frac{1}{2} \Bigg[ (1-e^{-b}) - (1-e^{-c}) \Bigg] \\[6pt] &= \frac{e^{-c} - e^{-b}}{2}. \\[6pt] \end{align}$$

(Observe that, unlike your result, this approaches one when $b \rightarrow \infty$ or $c \rightarrow \infty$.) Thus, finding the optimal confidence interval of this form requires you to solve the following optimisation problem:

$$\text{Minimise } b+c \quad \text{ subject to } \quad e^{-c} - e^{-b} = 2(1-\alpha).$$

With a bit of work, it should be possible for you to show that the optima occurs when $b=c$, so that the optmial confidence interval is one with midpoint at $x$. This is unsurprising, given that the Laplace distribution is symmetric around the mean parameter $\theta$.

There is a mistake in your probability result (which should be clear by the fact that it is unbounded). Using the interval $\text{CI}(X) = [X-b, X+c]$ you should have the coverage probability:

$$\begin{align} \mathbb{P}(\theta \in \text{CI}(X)) &= \mathbb{P}(X-b \leqslant \theta \leqslant X+c) \\[6pt] &= \mathbb{P}(\theta-c \leqslant X \leqslant \theta+b) \\[6pt] &= \int \limits_{\theta-c}^{\theta+b} \text{Laplace}(x|\theta,1) \ dx \\[6pt] &= \frac{1}{2} \int \limits_{\theta-c}^{\theta+b} e^{-|x-\theta|} \ dx \\[6pt] &= \frac{1}{2} \Bigg[ \ \int \limits_{\theta}^{\theta+b} e^{-x+\theta} \ dx - \int \limits_{\theta}^{\theta+c} e^{-x+\theta} \ dx \Bigg] \\[6pt] &= \frac{1}{2} \Bigg[ (1-e^{-b}) - (1-e^{-c}) \Bigg] \\[6pt] &= \frac{e^{-c} - e^{-b}}{2}. \\[6pt] \end{align}$$

(Observe that, unlike your result, this approaches one when $b \rightarrow \infty$ or $c \rightarrow \infty$.) Thus, finding the optimal confidence interval of this form requires you to solve the following optimisation problem:

$$\text{Minimise } b+c \quad \text{ subject to } \quad e^{-c} - e^{-b} = 2(1-\alpha).$$

With a bit of work, it should be possible for you to show that the optima occurs when $b=c$, so that the optimal confidence interval is one with midpoint at $x$. This is unsurprising, given that the Laplace distribution is symmetric around the mean parameter $\theta$.