How to express the Laplace transform of $X$ by the Bessel function

Question

Following this question: Bessel function, characteristic function, semicircle distribution.

We know that for $X\sim \mu_{sc}(dx)=\frac{1}{2\pi}\sqrt{2-x^2}dx$, the characteristic function of $X$ is expressed as $$ E[e^{itX}]=\frac{1}{t}J_1(2t) $$ where $ J_n(x)$ is the Bessel function of order $n$, that is $$J_1(z) = \dfrac{1}{iπ} \int_{0}^π e^{iz\cos θ}\cos θ dθ $$

But how to express the Laplace transform of $X$ by the Bessel function, which is $$ E[e^{tX}]=? $$

Answer 1

The probability distribution being compactly supported, its characteristic function is analytic over $\mathbb C$. This well known result is a consequence of complex differentiation under the integral.

Complex differentiation under the integral also shows readily that $J_1$ is analytic over $\mathbb C$ because its integral definition involves a compactly supported continuous function.

Now $t\mapsto \mathbb E[e^{itX}]$ and $t\mapsto \dfrac 1t J_1(t)$ are two analytic functions over $\mathbb C^*$, which you know to be equal on $\mathbb R^*$. By uniqueness of analytic continuation they are equal all over $\mathbb C^*$.

Now just substitute $t$ by $-it$.