Functional determinants I wish to know what is the result of this Gaussian Functional Integral
$$Z[\chi] = \int[\mathcal{D}\phi]~e^{-i\int d^dx ~\phi^2\chi}$$ where $\phi, \chi$ are position dependent fields. Now, my question is whether
$$Z[\chi] = (\det\chi)^{-1/2} ?$$
But, since, $\chi$ is not an operator just a scalar field is $Z[\chi] = \chi^{-1/2}$ then? What is the correct answer?
 A: $\det \chi$ is a slight abuse of notation. You are actually computing the determinant of the multiplication operator $m_\chi:\phi \mapsto\chi\phi$. As @CosmasZachos pointed out in the comment, this is the product integral :
$$\det m_\chi = \prod_{x\in\mathbb R^d} \chi(x)^{\text d^dx} = \exp\left(\int_{x\in\mathbb R^d}\ln(\chi(x))\text d^dx\right) \tag 1$$
Edit
The multiplication operator is diagonal, as it can be written as an integral kernel proportional to $\delta(x-y)$ :
$$m_\chi(x)\phi(x) = \int_{\mathbb R^d} \text d^dy \ m_\chi(x)\delta^{(d)}(x-y)\phi(y)$$
Then, the determinant is just the product of the diagonal entries, which gives equation $(1)$ above.
Another way to see this, more rigorously, is to discretize space on a finite lattice $\Lambda$ of size $L^d$ and lattice step $a \to 0$ and take the limit $a\to 0, L\to \infty$.
Then, a scalar field is just an element of $\mathbb R^{L^d}$ and a multiplication operator is a diagonal $L^d \times L^d$ matrix. Therefore :
$$\det m_\chi |_{\Lambda,a} = \prod_{x\in \Lambda}\chi(x)$$
To get a regular limit, we take it to the power $a$, to get :
\begin{align}
\det m_\chi &= \lim_{a\to 0,\Lambda \to \infty}\det m_\chi|_{\Lambda,a } \\
&= \lim \exp\left(a\sum_{x\in\Lambda}\ln \chi(x) \right) \\
&= \exp\left(\int_{x\in\mathbb R^d}\ln\chi(x)\text d^dx\right)
\end{align}
