If we assume that $\Psi(x)$ is just a scalar function and the integral $\int dx$ is over the real axis, then the result can be any complex number. And if we consider more complicated cases, as in the description of Bochner's theorem that you mention (see Wikipedia,) then this will in general not be different.
For the simplest case, take a very simple (normalized) wave function: $\Psi(x) = \pi^{-1/4}\,e^{-x^2/2}$. The integral in question then gives us: $$\begin{align} & \int_{-\infty}^{\infty} \frac{dx}{\sqrt{\pi}} \, e^{-iax} \, e^{-x^2/2} \, e^{-(x-b)^2/2} = \int_{-\infty}^{\infty} \frac{dx}{\sqrt{\pi}} \, e^{-x^2+(b-ia)x-b^2/2} \\ & = \int_{-\infty}^{\infty} \frac{dx}{\sqrt{\pi}} \, e^{-(x-(b-ia)/2)^2 -(a^2+b^2+2iab)/4} \\ & \quad = \int_{-\infty}^{\infty} \frac{dy}{\sqrt{\pi}} \, e^{-y^2 -(a^2+b^2+2iab)/4} = \frac{\pi^{-1/2}}{e^{(a^2+b^2)/4}}\ e^{-i a b/2}, \end{align}$$ where in the second line we just complete the square for $x$ and subsequently use the fact that a change in integration variable $x\rightarrow x-(b-ia)/2$ does not change the result for the integral from $-\infty$ to $\infty$. So this gives us a positive real number times a phase factor $e^{-i a b/2}$, which by choice of $a$ and/or $b$ can be given any desired phase so you can't say anything about the sign.