difference of characteristic function for measure and random variable

Question

Suppose random variable $X$ follow a certain (known) distribution. And I denote the probability measure $\mu$ as the distribution (pushforward measure) of $X$. Is there any difference between $\hat{\mu}(t)=\int_{\mathbb{R}}{e^{itx}\mu(dx)}$ (Fourier transformation of $\mu$) and $\phi_{X}(t)$ (the characteristic function of $X$)

Answer 1

The characteristic function of a real random variable $X$ is the characteristic function of its probability distribution, and it is defined as $$ t\mapsto \operatorname{E}(e^{itX}) $$ where $i=\sqrt{-1}$ is the imaginary unit. The characteristic function is equal to $$ t\mapsto \int_{\mathbb R} e^{itx}\,d\mu(x) $$ or it may be denoted by $$ t\mapsto \int_{\mathbb R} e^{itx}\,\mu(dx). $$ If there's a difference it's because someone is following a convention according to which the Fourier transform of $\mu$ is $$ t\mapsto \int_{\mathbb R} e^{-itx}\,\mu(dx) $$ where there is a minus sign in the exponent. Or perhaps they're using that times some constant or the integral without the minus sign times some constant.