Special integrals: M-dimensional Gaussian

Regular readers of this blog will know that I love my integrals. For those that share the same enjoyment, I recommend this channel that often covers integrals from the MIT integration bee. For example, see this case of writing an integral in terms of itself. Very cool.

In this post, I want to think about the m-dimensional Gaussian. It is a common integral in QFT. Here’s how we compute it when of the form,

    \[ z = \int_{\mathbb{R}^n} d^n xe^{-\frac{1}{2} x^{T}Ax} \]

Two comments: A is a real symmetric matrix, A = A^T \in \mathbb{R}^{n \times n} \Leftrightarrow A_{ij} = A_{ji}. X is a column vector,

    \[ x = \begin{pmatrix} x_1 \\ x_2 \\ . \\ . \\ . \\ x_n \\ \end{pmatrix} \]

Since A is real and symmetric, we make use of a result from spectral theorem. In particular, consider the following spectral theorem proposition:

    \[ A = A^T \in \mathbb{R}^{n \times n} \]

This implies A has eigenvalues \lambda_i \in  \mathbb{R}. It can also be diagonalised into a matrix D = diag(\lambda_1, \lambda_2, ... , \lambda_n) by an orthogonal matrix O, such that

    \[ OAO^T = D = \begin{pmatrix} \lambda & 0 & ... & 0 \\ 0 & \lambda_2 & ... & 0 \\ . & . & ... & . \\ . & . & ... & . \\ 0 & 0 & ... & \lambda_n \\ \end{pmatrix} \Leftrightarrow A = O^T DO \]

Here, it should be highlighted O is an n \times m matrix. From these considerations,

    \[ \implies z = \int_{\mathbb{R}^n} d^n xe^{-\frac{1}{2} x^{T} Ax} \]

    \[ = \int_{\mathbb{R}^n} d^n xe^{-\frac{1}{2} x^{T} O^T DO_x} \]

    \[ = \int_{\mathbb{R}^n} d^n xe^{-\frac{1}{2} (O_x)^{T} D(O_x)} \]

At this point, we perform variable substitution. Notice,

    \[ y \ : = O_{x} \]

    \[ d^{n} y = det (\frac{dy}{dx}) d^{n} x \]

    \[ \implies d^{n} x = \frac{1}{det (O)} d^{n} y \]

    \[ \therefore z = \int_{\mathbb{R}^n} d^{n} y \frac{1}{det (O)} e^{-\frac{1}{2} y^{T} Dy} \]

    \[ =   \int_{\mathbb{R}^n} d^{n} y \frac{1}{det (O)} e^{-\frac{1}{2} \sum_{i=1}^{n} \lambda_i y_{i}^2} \]

    \[ \frac{1}{det (O)} \prod_{i=1}^{n} \int_{- \infty}^{\infty} dy_{i} e^{-\frac{1}{2} \lambda_i y_{i}^2 } \]

From the 1-dimensional case, we know: e^{-\frac{1}{2} \lambda_i y_{i}^2} = \sqrt{\frac{2 \pi}{ \lambda_i}}. So,

    \[ z = \frac{1}{det(O)}  \prod_{i=1}^{n} \sqrt{\frac{2 \pi}{ \lambda_i}} \]

Now, recall that:

    \[ D = \begin{pmatrix} \lambda & 0 & ... & 0 \\ 0 & \lambda_2 & ... & 0 \\ . & . & ... & . \\ . & . & ... & . \\ 0 & 0 & ... & \lambda_n \\ \end{pmatrix} \]

From this, we can simplify our last result

    \[ z = \frac{1}{\det (O)} \frac{(2 \pi)^{\frac{n}{2}}}{\sqrt{\det (D)}} \]

    \[ = \frac{(2 \pi)^{\frac{n}{2}}}{\sqrt{\det (O^T DO)}} \]

    \[ \therefore z = \frac{(2 \pi)^{\frac{n}{2}}}{\sqrt{\det (A)}} \]