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Fernando Garcia

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Crunching $d$-dimensional integrals down to 1 dimensional integrals

July 19th, 2024.

Introduction

Symmetry appears everything in physics. It is sometimes used to argue the construction of an object (say, constructing a spin 1/2 Lagrangian (density) through Lorentz invariance/symmetry), but other times we use it to break things down such as integrals.

If we have an integral over all $d$-dimensions and the integrand depends only on the radial distance, then all of the angular integration follows almost trivially.

In this post we will find a formula to do this quickly. The first to note is that the angular integration gives none other than the surface area of a $d$-dimensional unit sphere. We will show that this area is equal to

$$\begin{array}{}\text{(i)}& \text{Area}=\frac{2{\pi }^{d/2}}{\mathrm{\Gamma }\left(\frac{1}{2}d\right)}\end{array}$$

Where $d$, again, is the number of dimensions, and $\mathrm{\Gamma }$ is the Euler gamma function:

$$\mathrm{\Gamma }(z)={\int }_0^{\mathrm{\infty }}{t}^{z-1}{e}^{-t}dt$$

This blog post serves a similar purpose the last blog (number 10). The advantage of using Eq. (i) to break down a $d$-dimensional integral is to get a 1-dimensional (radial) integral, which can be computed with less complications.

The surface of the unit $d$-dimensional sphere

We will find the desired expression by performing the integral $\int d^{d}x{e}^{-x^2}$ both in Cartesian and spherical coordinates.

We start by recalling that

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}dx=\sqrt{\pi }$$

So the Cartesian integral is:

$$\begin{array}{rl}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{d}^{d}x{e}^{-x^2}& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{d}^{d}x{e}^{-x_2^2-\cdots -x_d^2}\\ \text{(ii)}& & ={\sqrt{\pi }}^d\end{array}$$

For the spherical integral, we recall that

$$d^{d}x\to d\mathrm{\Omega }dr{r}^{d-1}$$

Where the $dr$ integral goes from $0$ to $\mathrm{\infty }$. Further, $x^2\to r^2$. We see that the integral in spherical coordinates is:

$$\begin{array}{rl}\int d\mathrm{\Omega }dr{r}^{d-1}{e}^{-r^2}& =\int d\mathrm{\Omega }{\int }_0^{\mathrm{\infty }}dr{r}^{d-1}{e}^{-r^2}\\ & =\mathrm{\Omega }{\int }_0^{\mathrm{\infty }}dr{r}^{d-1}{e}^{-r^2}\\ \text{let }{r}^2=u\text{, so }du=2rdr& =\mathrm{\Omega }{\int }_0^{\mathrm{\infty }}du\frac{1}{2}{r}^{d-2}{e}^{-u}\\ & =\frac{\mathrm{\Omega }}{2}{\int }_0^{\mathrm{\infty }}du(r^2{)}^{\frac{d-2}{2}}{e}^{-u}\\ & =\frac{\mathrm{\Omega }}{2}{\int }_0^{\mathrm{\infty }}du{u}^{\frac{d-2}{2}}{e}^{-u}\\ & =\frac{\mathrm{\Omega }}{2}{\int }_0^{\mathrm{\infty }}du{u}^{\frac{d}{2}-1}{e}^{-u}\\ \text{(iii)}& & =\frac{\mathrm{\Omega }}{2}\mathrm{\Gamma }(d/2)\end{array}$$

Where $\mathrm{\Omega }$ denotes the area of the unit $d$-dimensional sphere.

We now equate equations (ii) and (iii) to find:

$$\text{Area}\equiv \mathrm{\Omega }=\frac{2{\pi }^{d/2}}{\mathrm{\Gamma }\left(\frac{1}{2}d\right)}$$

As stated in the introduction.

How to use the formula

So far we only have the area of the unit $d$-dimensional sphere, but what do we do with this? Consider an integral of the form:

$$\int d^{d}xf(x^2)$$

Where $f$ only depends on the radial part. Then we turn the $d$ integrals $\int d{x}_i$ into a single integral over the variable $k^2$:

$$d^{d}x=\frac{{\pi }^{d/2}}{\mathrm{\Gamma }(d/2)}(x^2{)}^{\frac{d}{2}-1}d(x^2)$$

Where the integral over $x^2$ goes from $0$ to $\mathrm{\infty }$. It is often useful to make the trivial substitution $x^2\to v$ or some other variable to avoid confusion with the 2 in the exponent acting just as a label rather than as an actual number.

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