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 [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: $$\begin{align} & \int_{-\infty}^{\infty} \frac{dx}{\sqrt{\pi}} \, e^{-iax} \, e^{-x^2/2} \, e^{-(x-b)^2/2} = \int_{-\infty}^{\infty} \frac{dx}{\sqrt{\pi}} \, e^{-x^2+(b-ia)x-b^2/2} \\ & = \int_{-\infty}^{\infty} \frac{dx}{\sqrt{\pi}} \, e^{-(x-(b-ia)/2)^2 -(a^2+b^2+2iab)/4} \\ & = \int_{-\infty}^{\infty} \frac{dy}{\sqrt{\pi}} \, e^{-y^2 -(a^2+b^2+2iab)/4} = \frac{\pi^{-1/2}}{e^{(a^2+b^2)/4}}\ e^{-i a b/2}, \end{align}$$ 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 $b$ can be given any desired phase so you can't say anything about the sign.