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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)}} ]

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