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This integral appears in Ashok p. 82-83.

We have the integral

$$ I = \int \prod_{i,j}d\theta^*_id\theta_j e^{-(\theta_i^*M_{ij}\theta_j + c_i^*\theta_i + \theta_i^*c_i)}, $$

and if the inverse of $M$ exists and if we define the change of variables

\begin{equation} \begin{split} \theta'_i &= M_{ij}\theta_j + c_i,\\ \theta_i^{*'} &= \theta_i^* + c_j^* M_{ij}^{-1} \end{split} \end{equation}

then we can compute that the integral gives

\begin{equation} \begin{split} I &= \operatorname{det}M_{ij}\int \prod_{ij}d\theta^*_id\theta_je^{-((\theta_i^* + c_j^*M_{ij}^{-1}) \theta_i' - c_i^*M_{ij}^{-1}c_j)}\\ &= \operatorname{det}M_{ij}\int \prod_{ij}d\theta^*_id\theta_je^{-\theta_i^{*'}\theta_i' + c_i^*M_{ij}^{-1}c_j}. \end{split} \end{equation}

I'm certainly missing something, but why the change of variables $\theta_j^*\rightarrow\theta_j^{*'}$ does not carry and extra $\operatorname{det}M_{ij}$?

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    $\begingroup$ Isn't $c_j$ a "constant"? Think about the $1\mathrm{D}$ changes of variables $x' = \alpha x + \beta$ and $y' = y + \beta/ \alpha$. The former would give you a factor $1/\alpha$, but the latter would not add any factor to the integral. Also I think that the answer should give you a $1/\mathrm{det}(M_{ij})$ in front instead of the $\mathrm{det}(M_{ij})$ $\endgroup$ – QuantumApple Feb 26 at 0:44
  • $\begingroup$ Yes is a constant. Damn you are completely right, the transformation matrix for the second case is just the identity matrix. $\endgroup$ – user2820579 Feb 26 at 0:47

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