The density of the distribution whose Laplace transform is the following

Question

Is anybody aware of the density of the distribution whose Laplace transform is the following.

\begin{equation} \mathbb{E}[e^{tX}] = \frac{e^{t/2}-1}{t/2} \end{equation}

Note: $X$ is a continuous random variable, most probably bounded between $0$ and $1$.

Answer 1

If $X \sim U([0,\frac 12])$, that is $X$ is uniformly distributed on $[0,\frac 12]$, we have \begin{align*} \def\E{\mathbf E} \E[e^{tX}] &= 2\int_0^{\frac 12} \exp(tx)\, dx\\ &= \frac 2t \exp(tx)\bigl|_0^{\frac 12}\\ &= \frac{\exp(t/2) - 1}{t/2} \end{align*} So the density with respect to Lebesgue measure on $\mathbf R$ is $2\chi_{[0,\frac 12]}$.