From a One-Dimensional Integral to a Plane Integral
The integral of \(e^{-x^2}\) over \(\mathbb{R}\) does not have an elementary antiderivative. Its value can nevertheless be found exactly by relating it to a two-dimensional integral: squaring the one-dimensional integral produces a nonnegative function on the plane that depends only on distance from the origin. Polar coordinates then reduce that plane integral to a single radial integral.
All integrals in this tutorial are Lebesgue integrals. The measure space \((\mathbb{R},\mathcal{L},m)\), where \(\mathcal{L}\) is the Lebesgue sigma-algebra and \(m\) is Lebesgue measure, is complete and sigma-finite. The product measure \(m\times m\) on \(\mathcal{L}\otimes\mathcal{L}\) is sigma-finite. The functions used below are real-valued, nonnegative, and measurable; any stated integral is allowed to have value \(+\infty\) until finiteness has been established. We use Tonelli's theorem for nonnegative measurable functions on this product space, and the polar-coordinate formula for nonnegative radial functions established in the earlier material on polar integrals.
Set \(g(x)=e^{-x^2}\) and \(I=\int_{\mathbb{R}}g\,dm\), initially as an extended nonnegative number. Since \(g\) is continuous and strictly positive, \(I>0\): on \([-1,1]\), \(g(x)\ge e^{-1}\), so \(I\ge 2e^{-1}>0\). It remains to compute \(I\), including proving that it is finite.
Proof. Define \(F(x,y)=e^{-(x^2+y^2)}\) on \(\mathbb{R}^2\). This function is continuous, hence measurable for the product sigma-algebra, and \(F(x,y)=g(x)g(y)\ge0\). Tonelli's theorem applies because the product measure space is sigma-finite and \(F\) is nonnegative and measurable. Thus $$ \int_{\mathbb{R}^2}F\,d(m\times m) =\int_{\mathbb{R}}\left(\int_{\mathbb{R}}g(x)g(y)\,dm(y)\right)dm(x). $$ Since \(g(x)>0\) for every \(x\), the inner integral is \(g(x)I\), with the extended value \(+\infty\) allowed if \(I=+\infty\). Consequently, the iterated integral is \(I^2\), interpreted as \(+\infty\) if \(I=+\infty\).
We now calculate the plane integral independently. For \(R>0\), let \(B_R=\{(x,y)\in\mathbb{R}^2:x^2+y^2<R^2\}\). The polar-coordinate formula for a nonnegative radial function gives $$ \int_{B_R}F\,d(m\times m) =\int_0^{2\pi}\int_0^R e^{-r^2}r\,dr\,d\theta =2\pi\int_0^R e^{-r^2}r\,dr. $$ The factor \(r\) is the Jacobian in polar coordinates. Substituting \(u=r^2\), so \(du=2r\,dr\), yields $$ 2\pi\int_0^R e^{-r^2}r\,dr =\pi\int_0^{R^2}e^{-u}\,du =\pi(1-e^{-R^2}). $$ For positive integers \(n\), the disks \(B_n\) increase to \(\mathbb{R}^2\). The Monotone Convergence Theorem applies to \(F\chi_{B_n}\): these functions are nonnegative and measurable, increase pointwise to \(F\), and \(F\) is measurable. Therefore $$ \int_{\mathbb{R}^2}F\,d(m\times m) =\lim_{n\to\infty}\pi(1-e^{-n^2})=\pi. $$ In particular, the plane integral is finite. The Tonelli identity then implies \(I^2=\pi\), so \(I\) is finite. Since \(I>0\), its value is the positive square root, \(I=\sqrt{\pi}\). \(\square\)
Checking the Radial Calculation on Disks
The proof uses an integral over the entire plane. The finite-radius calculation also records exactly how much of that integral lies inside a disk. It is useful to keep the disk formula separate from the whole-plane limit: on a bounded disk the radial integral has a finite upper endpoint, while the full plane is obtained by increasing the radius and applying Monotone Convergence.
Worked Example: The Gaussian Integral Inside a Disk
Compute \(\int_{B_{\sqrt{\ln 2}}} e^{-(x^2+y^2)}\,d(m\times m)\), where \(B_R\) is the open disk of radius \(R\) centred at the origin.
The integrand is continuous and nonnegative, so the polar-coordinate formula applies. The radius is \(R=\sqrt{\ln 2}\), hence \(R^2=\ln 2\). Applying the formula and evaluating the radial integral gives $$ \int_{B_{\sqrt{\ln 2}}}e^{-(x^2+y^2)}\,d(m\times m) =2\pi\int_0^{\sqrt{\ln 2}}e^{-r^2}r\,dr =\pi\left(1-e^{-(\sqrt{\ln 2})^2}\right) =\pi(1-e^{-\ln 2}) =\frac{\pi}{2}. $$ Check: the full-plane integral is \(\pi\), and the integral outside this disk, calculated by the same radial formula, is \(\pi e^{-\ln 2}=\pi/2\). The disk and its exterior therefore contribute \(\pi/2+\pi/2=\pi\), as required.
Worked Example: The Gaussian Integral Outside a Disk
For \(R>0\), compute the integral of \(e^{-(x^2+y^2)}\) over the region \(\{(x,y):x^2+y^2\ge R^2\}\). The boundary circle has plane measure zero, so using a strict or non-strict radial inequality gives the same integral.
For a finite outer radius \(S>R\), the annular region \(R\le\sqrt{x^2+y^2}<S\) has integral $$ 2\pi\int_R^S e^{-r^2}r\,dr =\pi(e^{-R^2}-e^{-S^2}). $$ As \(S\) increases through the positive integers above \(R\), these annuli increase to the region outside the disk, except possibly its boundary. Monotone Convergence applies because the restricted integrands are nonnegative and increase to the restricted Gaussian. Taking \(S\to\infty\) gives the exterior integral \(\pi e^{-R^2}\).
For \(R=\sqrt{\ln 4}\), substitution gives \(e^{-R^2}=e^{-\ln4}=1/4\), so the exterior integral is \(\pi/4\). Check: the integral inside this disk is \(\pi(1-e^{-\ln4})=3\pi/4\). Adding the interior and exterior values gives \(3\pi/4+\pi/4=\pi\), the whole-plane integral.
Scaling the One-Dimensional Integral
Once the unscaled integral is known, a change of variable gives a useful family of exact values. The parameter condition matters: the formula below is for \(a>0\), ensuring that the integrand decays at infinity. The case \(a=0\) has an infinite integral, and for \(a<0\) the integrand grows rather than decays.
Proof. The one-dimensional change-of-variables formula for Lebesgue integrals applies to the linear substitution \(u=\sqrt{a}\,x\), which is a bijection of \(\mathbb{R}\) with positive derivative \(\sqrt a\). Thus \(x=u/\sqrt a\) and \(dx=du/\sqrt a\), and $$ \int_{\mathbb{R}}e^{-ax^2}\,dm(x) =\frac1{\sqrt a}\int_{\mathbb{R}}e^{-u^2}\,dm(u) =\frac1{\sqrt a}\sqrt{\pi} =\sqrt{\frac{\pi}{a}}. $$ The integral is finite because it is a positive constant times the already established finite Gaussian integral. \(\square\)
Worked Example: A Gaussian with a Quadratic Coefficient
Compute \(J=\int_{\mathbb{R}}e^{-4x^2}\,dm(x)\), and verify the result by squaring \(J\) and using polar coordinates.
The scaling result with \(a=4\) gives \(J=\sqrt{\pi/4}=\sqrt{\pi}/2\). For an independent check, \(J\) is positive and Tonelli's theorem applies to the nonnegative continuous function \(e^{-4(x^2+y^2)}\) on the sigma-finite product space. Hence \(J^2\) equals its plane integral. Polar coordinates, followed by \(u=4r^2\), give $$ J^2 =2\pi\int_0^\infty e^{-4r^2}r\,dr =2\pi\cdot\frac18\int_0^\infty e^{-u}\,du =\frac{\pi}{4}. $$ Here \(\int_0^\infty e^{-u}\,du=1\), since \(\int_0^T e^{-u}\,du=1-e^{-T}\) and \(e^{-T}\to0\). Taking the positive square root gives \(J=\sqrt{\pi}/2\), agreeing with the change of variables. Check: squaring the stated value gives \((\sqrt{\pi}/2)^2=\pi/4\), exactly the plane integral.
A Common Polar-Coordinate Pitfall
In polar coordinates, the area element is \(r\,dr\,d\theta\), not \(dr\,d\theta\). Omitting \(r\) changes the integral and destroys the calculation. For example, on a disk of radius \(R\), the correct integral is \(2\pi\int_0^R e^{-r^2}r\,dr=\pi(1-e^{-R^2})\). Omitting the Jacobian would instead produce \(2\pi\int_0^R e^{-r^2}\,dr\), which is not the disk integral.
The discrepancy is already visible for small radii without evaluating the incorrect integral. The correct disk integral divided by \(R^2\) tends to \(\pi\), since \((1-e^{-R^2})/R^2\to1\). The expression with the missing factor, divided by \(R^2\), is at least \(2\pi e^{-R^2}/R\), which tends to \(+\infty\). The two expressions therefore cannot agree for all small positive \(R\). The Jacobian is essential, not a cosmetic part of the notation.
A second point to preserve is the order of the argument. The one-dimensional integral is initially allowed to be infinite. Tonelli's theorem is applicable because the plane integrand is nonnegative; it does not require the desired finiteness in advance. The polar calculation proves that the plane integral is finite, and only then does the Tonelli identity establish that the one-dimensional integral is finite and has the stated value. This avoids assuming the conclusion while squaring the integral.
Check Your Understanding
In each question, keep track of nonnegativity, the measure space, and the polar Jacobian.
- Use the disk formula to compute the integral of \(e^{-(x^2+y^2)}\) over \(B_{\sqrt{\ln 3}}\).
- For \(a>0\), explain why the product function \(e^{-a(x^2+y^2)}\) meets the nonnegativity and measurability hypotheses of Tonelli's theorem on \((\mathbb{R}^2,\mathcal{L}\otimes\mathcal{L},m\times m)\).
- Use the scaling result to evaluate \(\int_{\mathbb{R}}e^{-9x^2}\,dm(x)\), then check by squaring the proposed value.
- Write the polar-coordinate integral over an annulus with inner radius \(R\) and outer radius \(S\), where \(0<R<S<\infty\), and evaluate it.
- Identify the missing factor in the incorrect polar expression \(2\pi\int_0^R e^{-r^2}\,dr\) for the integral over a disk, and state the correct expression.