Gaussian Integrals in Higher Dimensions

Two Routes to a Higher-Dimensional Gaussian

For $a>0$, the function $x\mapsto e^{-a|x|^2}$ on $\mathbb{R}^n$ is positive and depends only on the distance from the origin. Its integral can therefore be approached in two ways. In Cartesian coordinates it is a product of one-dimensional Gaussians; in radial coordinates it is a one-variable integral weighted by the size of each spherical shell. Comparing the answers determines the geometric constant in the radial formula.

Throughout, $n$ is a positive integer, $a>0$, and $m$ is Lebesgue measure on $\mathbb{R}^n$ or on the indicated Euclidean space. The measure space $(\mathbb{R}^n,\mathcal{L},m)$ is sigma-finite, and the functions integrated below are nonnegative and measurable. All displayed integral equalities are equalities of Lebesgue integrals; we will show these particular integrals are finite.

One-dimensional input: The one-dimensional Gaussian integral, with Lebesgue measure on $\mathbb{R}$, is $\int_{\mathbb{R}}e^{-a t^2}\,dt=\sqrt{\pi/a}$ for $a>0$. We use this earlier result, including its finiteness, rather than re-deriving it here.

Product Integration

Write $x=(x_1,\ldots,x_n)$, so that $|x|^2=x_1^2+\cdots+x_n^2$. The integrand factors into one-variable functions:

e^{-a|x|^2}=\prod_{j=1}^n e^{-a x_j^2}.

Tonelli’s theorem applies because the product Lebesgue measure is sigma-finite and the function is nonnegative and measurable. It permits iterated integration, even before finiteness has been established. Each one-dimensional integral is $\sqrt{\pi/a}$, so successive integrations give

\int_{\mathbb{R}^n}e^{-a|x|^2}\,dm(x) =\prod_{j=1}^n\int_{\mathbb{R}}e^{-a t^2}\,dt =\left(\sqrt{\frac{\pi}{a}}\right)^n =\frac{\pi^{n/2}}{a^{n/2}}.

The resulting value is finite, so the nonnegative function is integrable. This is a use of Tonelli, not an unqualified interchange of integrals: its nonnegativity and the sigma-finiteness of Lebesgue measure supply the required hypotheses.

Worked Example: The Gaussian Integral in Two Dimensions

Evaluate $\int_{\mathbb{R}^2}e^{-3(x_1^2+x_2^2)}\,dm(x)$ by product integration. The integrand is nonnegative and measurable, and Lebesgue measure on $\mathbb{R}^2$ is sigma-finite, so Tonelli’s theorem applies. For each coordinate, the one-dimensional value at $a=3$ is $\sqrt{\pi/3}$. Therefore

\int_{\mathbb{R}^2}e^{-3(x_1^2+x_2^2)}\,dm(x) =\left(\sqrt{\frac{\pi}{3}}\right)\left(\sqrt{\frac{\pi}{3}}\right) =\frac{\pi}{3}.

Check: multiplying the two factors gives $(\pi/3)^{1/2}(\pi/3)^{1/2}=\pi/3$, as required. In the radial calculation below, the same integral will be $2\pi\cdot(1/6)=\pi/3$.

Radial Integration and the Gamma Integral

Let $B(0,r)=\{x\in\mathbb{R}^n:|x|\le r\}$, and write $V_n=m(B(0,1))$ for the volume of the unit ball. Scaling Lebesgue measure gives $m(B(0,r))=V_n r^n$ for $r\ge0$. The radial integration formula follows by recording how much volume lies at each radius.

Radial integration formula: If $g:[0,\infty)\to[0,\infty]$ is Borel measurable, then, with $m$ Lebesgue measure on $\mathbb{R}^n$, $$ \int_{\mathbb{R}^n}g(|x|)\,dm(x) =nV_n\int_0^\infty g(r)r^{n-1}\,dr. $$ Both sides are allowed to be infinite.

Proof. Map $x\in\mathbb{R}^n$ to its radius $|x|$, and let $\nu$ be the resulting measure on $[0,\infty)$: for a Borel set $A$, define $\nu(A)=m(\{x:|x|\in A\})$. This measure is locally finite, since $\nu([0,R])=V_nR^n<\infty$. For $0\le s<t$, scaling of balls gives $\nu((s,t])=V_n(t^n-s^n)$; the sphere at radius $s$ has measure zero, as follows by bounding its measure by the volume of arbitrarily thin surrounding annuli. Thus $\nu((s,t])=\int_s^t nV_n r^{n-1}\,dr$. The two locally finite Borel measures agree on intervals, so uniqueness of measures on the generating intervals gives $d\nu(r)=nV_n r^{n-1}\,dr$. By the definition of the image measure, integration of a nonnegative Borel function $g$ against $\nu$ equals $\int_{\mathbb{R}^n}g(|x|)\,dm(x)$. Integrating against the density just obtained proves the formula. $\square$

For $s>0$, define the gamma integral by $\Gamma(s)=\int_0^\infty u^{s-1}e^{-u}\,du$. It is finite: near zero the power $u^{s-1}$ is integrable for $s>0$, and at infinity the exponential decay dominates the power. Substituting $u=ar^2$ into the radial integral gives

\int_0^\infty e^{-ar^2}r^{n-1}\,dr =\frac12a^{-n/2}\Gamma\!\left(\frac n2\right).

Indeed, $r=(u/a)^{1/2}$ and $dr=\frac12a^{-1/2}u^{-1/2}\,du$, so $r^{n-1}dr=\frac12a^{-n/2}u^{n/2-1}\,du$. We also use $\Gamma(s+1)=s\Gamma(s)$ for $s>0$. Integration by parts on finite intervals and passage to the endpoints gives this identity: the boundary term $u^s e^{-u}$ tends to zero both as $u\downarrow0$ and as $u\to\infty$.

Comparing the Evaluations

Theorem: For every positive integer $n$ and every $a>0$, $$ \int_{\mathbb{R}^n}e^{-a|x|^2}\,dm(x) =\frac{\pi^{n/2}}{a^{n/2}} =nV_n\int_0^\infty e^{-ar^2}r^{n-1}\,dr =\frac{nV_n}{2}\Gamma\!\left(\frac n2\right)a^{-n/2}. $$ Consequently $V_n=\pi^{n/2}/\Gamma(n/2+1)$, and the radial coefficient $nV_n$ is $2\pi^{n/2}/\Gamma(n/2)$.

Proof. The product evaluation above gives $\pi^{n/2}a^{-n/2}$. The radial integration formula applies to $g(r)=e^{-ar^2}$, which is nonnegative and Borel measurable. The substitution already computed its radial integral, so the same integral equals $\frac{nV_n}{2}\Gamma(n/2)a^{-n/2}$. Equating these two evaluations and dividing by the positive number $a^{-n/2}$ gives $\pi^{n/2}=\frac{nV_n}{2}\Gamma(n/2)$. Since $\Gamma(n/2+1)=(n/2)\Gamma(n/2)$, this is $V_n=\pi^{n/2}/\Gamma(n/2+1)$. Substituting this value into $nV_n$ gives $2\pi^{n/2}/\Gamma(n/2)$. All quantities are finite: the product evaluation is finite, and the radial expression equals it. $\square$

The comparison explains precisely how the methods agree. The product method supplies the Gaussian integral directly; the radial method supplies the gamma integral multiplied by the measure of spherical shells. Their equality fixes that geometric multiplier. In particular, the radial computation is not obtained by treating the two-dimensional factor $2\pi r$ as dimension-independent.

Worked Example: The Gaussian Integral in Three Dimensions

Evaluate $\int_{\mathbb{R}^3}e^{-2|x|^2}\,dm(x)$ by both methods. Product integration is valid by Tonelli: the function is nonnegative and measurable, and Lebesgue measure is sigma-finite. Each coordinate contributes $\sqrt{\pi/2}$, so

\int_{\mathbb{R}^3}e^{-2|x|^2}\,dm(x) =\left(\sqrt{\frac{\pi}{2}}\right)^3 =\frac{\pi^{3/2}}{2\sqrt{2}}.

For the radial evaluation, $\Gamma(1/2)=\sqrt{\pi}$ follows from the one-dimensional Gaussian integral by substituting $u=t^2$ in the gamma integral. The recurrence gives $\Gamma(3/2)=\frac12\sqrt{\pi}$, hence $V_3=\pi^{3/2}/\Gamma(5/2)=4\pi/3$ and $3V_3=4\pi$. Also, substitution in the radial integral gives $\int_0^\infty r^2e^{-2r^2}\,dr=\frac12\,2^{-3/2}\Gamma(3/2)=\sqrt{\pi}/(8\sqrt{2})$. Therefore

\int_{\mathbb{R}^3}e^{-2|x|^2}\,dm(x) =4\pi\frac{\sqrt{\pi}}{8\sqrt{2}} =\frac{\pi^{3/2}}{2\sqrt{2}}.

Check: the product value is $\pi^{3/2}/(2\sqrt{2})$, and the radial value has the same numerator and denominator after $4/8=1/2$. The sphere factor is $4\pi$, not the two-dimensional factor $2\pi$.

Worked Example: A Four-Dimensional Gaussian

Evaluate $\int_{\mathbb{R}^4}e^{-|x|^2/2}\,dm(x)$. The integrand is nonnegative and measurable, so Tonelli applies on the sigma-finite Lebesgue space. Each coordinate contributes $\sqrt{2\pi}$; thus the product calculation is

\int_{\mathbb{R}^4}e^{-|x|^2/2}\,dm(x) =(\sqrt{2\pi})^4 =(2\pi)^2 =4\pi^2.

For a radial check, $\Gamma(2)=1$, since $\Gamma(1)=\int_0^\infty e^{-u}\,du=1$ and the recurrence gives $\Gamma(2)=1\cdot\Gamma(1)$. Also $\Gamma(3)=2$, so $V_4=\pi^2/\Gamma(3)=\pi^2/2$, and $4V_4=2\pi^2$. The radial integral is, using $u=r^2/2$ and $r^3\,dr=2u\,du$,

\int_0^\infty r^3e^{-r^2/2}\,dr =2\int_0^\infty ue^{-u}\,du =2\Gamma(2) =2.

Multiplication by the radial coefficient gives $2\pi^2\cdot2=4\pi^2$. Check: the radial result equals the product result, and the substitution reduces the one-variable integral to $2\Gamma(2)=2$.

Common Pitfall and Scope of the Formula

A common error is to use $2\pi r$ as the shell factor in every dimension. In three dimensions, this would incorrectly give $2\pi\int_0^\infty r^2e^{-2r^2}\,dr$, which equals $2\pi\cdot\sqrt{\pi}/(8\sqrt{2})=\pi^{3/2}/(4\sqrt{2})$, half the correct answer. The correction is to use the three-dimensional coefficient $3V_3=4\pi$, giving $4\pi\cdot\sqrt{\pi}/(8\sqrt{2})=\pi^{3/2}/(2\sqrt{2})$. The shell factor depends on the dimension.

The assumption $a>0$ is essential for a finite Gaussian integral. If $a=0$, the integrand is identically $1$ and its integral over $\mathbb{R}^n$ is infinite. If $a<0$, then $e^{-a|x|^2}\ge1$ everywhere, so its integral is again infinite. Thus the displayed finite-value formula is for $a>0$; it is not a finite-integral formula for nonpositive $a$.

Check Your Understanding

In each question, keep track of the dimension, the parameter $a$, and the radial coefficient.

Use product integration to evaluate $\int_{\mathbb{R}^2}e^{-|x|^2}\,dm(x)$, and then verify it using $V_2=\pi$.
Starting from $V_n=\pi^{n/2}/\Gamma(n/2+1)$, find the radial coefficient $nV_n$ in dimension five.
Use the substitution $u=ar^2$ to express $\int_0^\infty r^{n-1}e^{-ar^2}\,dr$ in terms of $\Gamma$.
Explain why Tonelli’s theorem applies to the Cartesian product evaluation even before the value of the integral is known to be finite.
For $a=0$, identify exactly why the finite Gaussian formula does not apply.