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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, the expression is in general not positive or negative.

Still, if this phase $e^{-i a b/2}$ would be present for any wave function $\Psi$, we could still say that the "complex sign" of this expression is known in advance, independent of $\Psi$. This is, however, surely not the case, since we can always define another wave function $\chi(x) = \Psi(x-c)$ and then the result for $\chi$ would be: $$\begin{align} & \int_{-\infty}^{\infty} dx\ e^{-iax}\ \chi(x)\ \chi^*(x-b) = \int_{-\infty}^{\infty} dx\ e^{-iax}\ \Psi(x-c)\ \Psi^*(x-c-b) \\ & = \int_{-\infty}^{\infty} dy\ e^{-ia(y+c)}\ \Psi(y)\ \Psi^*(y-b) \\ & = e^{-iac}\ \int_{-\infty}^{\infty} dy\ e^{-iay}\ \Psi(y)\ \Psi^*(y-b). \end{align}$$ So by choosing $c$ we can make the results for $\chi$ differ by an arbitrary phase $e^{-iac}$ from the result for $\Psi$, without having to change $a$ or $b$. So the result cannot be predicted based on knowledge of $a$ and $b$.

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, the expression is in general not positive or negative.

Still, if this phase $e^{-i a b/2}$ would be present for any wave function $\Psi$, we could still say that the "complex sign" of this expression is known in advance, independent of $\Psi$. This is, however, surely not the case, since we can always define another wave function $\chi(x) = \Psi(x-c)$ and then the result for $\chi$ would be: $$\begin{align} & \int_{-\infty}^{\infty} dx\ e^{-iax}\ \chi(x)\ \chi^*(x-b) = \int_{-\infty}^{\infty} dx\ e^{-iax}\ \Psi(x-c)\ \Psi^*(x-c-b) \\ & = \int_{-\infty}^{\infty} dy\ e^{-ia(y+c)}\ \Psi(y)\ \Psi^*(y-b) \\ & = e^{-iac}\ \int_{-\infty}^{\infty} dy\ e^{-iay}\ \Psi(y)\ \Psi^*(y-b). \end{align}$$ So by choosing $c$ we can make the result for $\chi$ differ by an arbitrary phase $e^{-iac}$ from the result for $\Psi$, without having to change $a$ or $b$. So the result cannot be predicted based on knowledge of $a$ and $b$.

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.

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, the expression is in general not positive or negative.

Still, if this phase $e^{-i a b/2}$ would be present for any wave function $\Psi$, we could still say that the "complex sign" of this expression is known in advance, independent of $\Psi$. This is, however, surely not the case, since we can always define another wave function $\chi(x) = \Psi(x-c)$ and then the result for $\chi$ would be: $$\begin{align} & \int_{-\infty}^{\infty} dx\ e^{-iax}\ \chi(x)\ \chi^*(x-b) = \int_{-\infty}^{\infty} dx\ e^{-iax}\ \Psi(x-c)\ \Psi^*(x-c-b) \\ & = \int_{-\infty}^{\infty} dy\ e^{-ia(y+c)}\ \Psi(y)\ \Psi^*(y-b) \\ & = e^{-iac}\ \int_{-\infty}^{\infty} dy\ e^{-iay}\ \Psi(y)\ \Psi^*(y-b). \end{align}$$ So by choosing $c$ we can make the results for $\chi$ differ by an arbitrary phase $e^{-iac}$ from the result for $\Psi$, without having to change $a$ or $b$. So the result cannot be predicted based on knowledge of $a$ and $b$.

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

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.