The “integration by induction” (where we integrate over each internal position in turn, seeing that we get a result that can be summarised in a general formula) as done in Sec. 5.3↑ above is only one of the many ways to calculate the general form of a Gaussian integral. We now want to investigate a more robust, but slightly more complicated alternative. The basic relation we want to use is (let us start assuming that $\alpha$ and $N$ are real; we shall generalise to complex $\alpha$ later) $$\int\prod_{i=1}^{n}dq_{i}\exp\left(-\frac{1}{2}\alpha\sum_{ij}q_{i}N_{ij}q_{j}\right)=\left(\frac{2\pi}{\alpha}\right)^{n/2}\left(\det N\right)^{-1/2}.$$ This is straightforward to prove. Note that the matrix $N$ can be assumed to be symmetric, and thus we can diagonalise this matrix $N$ , i.e., transform it to a diagonal form, $$O^{T}NO=\diag(\lambda_{1},\ldots,\lambda_{N})$$ where $O$ is the orthogonal matrix of eigenvectors, which we assume to be normalised. Thus it satisfies the “ortho-normalisation” [W]  [W] A combination of orthogonality and normalisation. condition $$\begin{equation} O^{T}O=I.\label{eq:PI:OTO}\end{equation}$$ We now transform $$\vec{q}=O\vec{x},$$ and use the fact that we can make a change of integration variables without any additional factor, $$\prod_{i=1}^{n}dq_{i}=\prod_{i=1}^{n}dx_{i},$$ since the change in volume element, the Jacobian of this transformation, is just the absolute value of the determinant of $O$ , which is known to be $1$  [X]  [X] Trivially, from (↓) we find that $\det(O^{T}O)=\left(\det O\right)^{2}=1$ , and thus $|\det O|=1$ .. We now write $$\begin{align*} \int\prod_{i=1}^{n}dq_{i}\exp\left(-\frac{1}{2}\alpha\sum_{ij}q_{i}N_{ij}q_{j}\right) & =\int\prod_{i=1}^{n}dx_{i}\exp\left(-\frac{1}{2}\alpha\sum_{ij}\sum_{k}O_{ik}x_{k}N_{ij}\sum_{l}O_{jl}x_{l}\right)\\ & =\int\prod_{i=1}^{n}dx_{i}\exp\left(-\frac{1}{2}\alpha\sum_{kl}x_{k}x_{l}\sum_{ij}O_{ik}N_{ij}O_{jl}\right)\\ & =\int\prod_{i=1}^{n}dx_{i}\exp\left(-\frac{1}{2}\alpha\sum_{k}x_{k}x_{k}\lambda_{k}\right)\\ & =\prod_{i=1}^{n}\left(\frac{2\pi}{\alpha\lambda_{i}}\right)^{1/2}\\ & =\left(\frac{2\pi}{\alpha}\right)^{n/2}\prod_{i=1}^{n}\lambda_{i}^{-1/2}=\left(\frac{2\pi}{\alpha}\right)^{n/2}\left(\det N\right)^{-1/2},\end{align*}$$ using $\det N=\prod_{i}\lambda_{i}$ .