Improper Multiple Integrals and Order of Integration, multivariable and vector calculus lesson

The bell curve $e^{-x^2}$ has no elementary antiderivative, yet its total area is exactly $\sqrt{\pi}$ . The proof squares the integral, turns it into a double integral over the whole plane, and switches to polar coordinates. Each of those steps needs a reason, because the plane is unbounded and ordinary double integrals live on bounded regions. This lesson supplies the reasons, and shows what goes wrong when they are missing.

Key terms

Unbounded regions and singular integrands

Definition The double integral of earlier lessons needs a bounded closed region and a function that is continuous, hence bounded, on it. An improper double integral is one where this fails in one of two ways. Either the region is unbounded, such as the whole plane or a quadrant, or the region is bounded and the integrand is continuous on it but unbounded near certain boundary points or curves that are left out of the region. Those points are singularities of the integrand. In both cases $\iint_D f \, dA$ is not yet defined and must be defined by a limit.

In plain words Either the region runs off to infinity, or the function blows up somewhere. Either way the usual Riemann sums do not apply directly.

Example The function $f(x, y) = \frac{1}{x^2 + y^2}$ on the region outside the unit disk is a problem of the first kind, since the region is unbounded. The function $g(x, y) = \frac{1}{\sqrt{x^2 + y^2}}$ on the unit disk with the origin removed is a problem of the second kind, since $g$ grows without bound near the origin.

Exhaustion by regions

Definition Let $D$ be a set in the plane and let $f$ be continuous on $D$ . A regular exhaustion of $D$ is a sequence of bounded closed regions $K_1 \subseteq K_2 \subseteq K_3 \subseteq \cdots$ , each with a piecewise smooth boundary and each contained in $D$ , such that every bounded closed set contained in $D$ lies inside some $K_n$ . For each $n$ the integral $\iint_{K_n} f \, dA$ is an ordinary double integral. The improper integral is

\iint_D f \, dA = \lim_{n \to \infty} \iint_{K_n} f \, dA,

provided this limit exists and has the same value for every regular exhaustion of $D$ . The same definition works in space with solids in place of regions.

In plain words Integrate over larger and larger nice pieces that eventually reach every part of the region, then let the pieces grow. The answer only counts if it does not depend on how you chose the pieces.

Example The disks $K_n$ of radius $n$ exhaust the plane, and so do the squares $[-n, n] \times [-n, n]$ . To exhaust the punctured unit disk $0 < x^2 + y^2 \le 1$ , use the annuli $\frac{1}{n} \le \sqrt{x^2 + y^2} \le 1$ , which leave a shrinking hole around the singularity at the origin. For $g = \frac{1}{\sqrt{x^2 + y^2}}$ on these annuli, polar coordinates give

\int_0^{2\pi} \int_{1/n}^1 \frac{1}{r} \, r \, dr \, d\theta = 2\pi \left( 1 - \frac{1}{n} \right),

which tends to $2\pi$ .

Continuous nonnegative integrands

Definition Let $f$ be continuous and nonnegative on $D$ , and let $K_n$ be any regular exhaustion of $D$ . The numbers $\iint_{K_n} f \, dA$ increase with $n$ , so they have a limit in $[0, \infty]$ . This limit is the same for every regular exhaustion. We set $\iint_D f \, dA$ equal to it and say the integral converges when it is finite and diverges when it is $\infty$ . A continuous $f$ of any sign is called absolutely integrable on $D$ when $\iint_D \lvert f \rvert \, dA$ is finite. Then the positive and negative parts $\max(f, 0)$ and $\max(-f, 0)$ have finite integrals, every regular exhaustion gives the same limit for $\iint f \, dA$ , and that limit is the difference of the two parts.

In plain words With no negative values there is no cancellation, so the totals can only grow and any reasonable way of growing the region gives the same answer. For a signed function, check the integral of its absolute value first.

Example For $p > 0$ and the region $D$ outside the unit disk, integrating $\frac{1}{(x^2 + y^2)^p}$ over the annuli $1 \le r \le n$ gives

\begin{aligned} &\int_0^{2\pi} \int_1^n r^{-2p} \, r \, dr \, d\theta \\[0.4em] &= 2\pi \int_1^n r^{1 - 2p} \, dr. \end{aligned}

As $n \to \infty$ this is finite exactly when $p > 1$ , and then it equals $\frac{\pi}{p - 1}$ . For $p = 2$ the total is $\pi$ . For $p = 1$ it is $2\pi \ln n$ , which grows without bound, so the integral diverges. In the plane the cutoff for convergence at infinity is the power $r^{-2}$ , not $r^{-1}$ as on a line.

Elementary improper integral form of Tonelli

Definition Let $D$ be a region that can be described in both ways

\begin{gathered} a < x < b, \quad g_1(x) < y < g_2(x), \\[0.4em] c < y < d, \quad h_1(y) < x < h_2(y), \end{gathered}

where $a$ and $c$ may be $-\infty$ , $b$ and $d$ may be $\infty$ , and the boundary functions are continuous, with $g_1$ or $h_1$ allowed to be the constant $-\infty$ and $g_2$ or $h_2$ the constant $\infty$ . Any of the strict inequalities may be replaced by $\le$ where the boundary point is finite and belongs to $D$ . Let $f$ be continuous and nonnegative on $D$ . Read every one-variable integral as an improper integral wherever an endpoint is infinite or $f$ is unbounded near it, and assume that in each order the inner integral is finite and is a continuous function of the outer variable. Then Tonelli’s theorem says

\begin{aligned} \iint_D f \, dA &= \int_a^b \int_{g_1(x)}^{g_2(x)} f \, dy \, dx \\[0.4em] &= \int_c^d \int_{h_1(y)}^{h_2(y)} f \, dx \, dy, \end{aligned}

as values in $[0, \infty]$ . Either all three are the same finite number or all three are infinite.

In plain words For a nonnegative continuous integrand you may compute in either order and the answer is the true total, even when it is infinite. Once the inner integrals behave, no check for cancellation is needed.

Example Consider

\int_0^\infty \int_x^\infty e^{-y^2} \, dy \, dx.

The inner integral has no elementary formula, but it is finite and continuous in $x$ . The integrand is continuous and positive on the region $0 \le x \le y$ , and in the other order, with $y$ from $0$ to $\infty$ and $x$ from $0$ to $y$ , the inner integral is $y e^{-y^2}$ , also continuous. So Tonelli allows the swap,

\begin{aligned} \int_0^\infty \int_0^y e^{-y^2} \, dx \, dy &= \int_0^\infty y e^{-y^2} \, dy \\[0.4em] &= \frac{1}{2}. \end{aligned}

Fubini under absolute integrability

Definition Let $D$ be described in both ways as above and let $f$ be continuous on $D$ , of any sign. Assume that in each order the inner integrals of $\lvert f \rvert$ are finite, so the inner integrals of $f$ converge absolutely, and that the inner integrals of $f$ and of $\lvert f \rvert$ are continuous functions of the outer variable. If $f$ is absolutely integrable, that is

\iint_D \lvert f \rvert \, dA < \infty,

then the improper integral $\iint_D f \, dA$ exists and both iterated integrals of $f$ converge and equal it. In practice you check absolute integrability by applying Tonelli to $\lvert f \rvert$ , computing or bounding an iterated integral of $\lvert f \rvert$ in whichever order is easier.

In plain words Before swapping the order for a function that changes sign, show that the total of its absolute value is finite. Then cancellation cannot play tricks and both orders agree.

Example On the first quadrant let $f(x, y) = e^{-x - y} \sin x$ . Since $\lvert f \rvert \le e^{-x} e^{-y}$ and

\int_0^\infty \int_0^\infty e^{-x} e^{-y} \, dy \, dx = 1,

the function is absolutely integrable. The inner integrals $e^{-x} \sin x$ and $\frac{1}{2} e^{-y}$ , and those of $\lvert f \rvert$ , are continuous. So either order is valid, and

\begin{aligned} &\int_0^\infty e^{-y} \, dy \cdot \int_0^\infty e^{-x} \sin x \, dx \\[0.4em] &= 1 \cdot \frac{1}{2} = \frac{1}{2}. \end{aligned}

Failures without hypotheses

Definition Without absolute integrability, the two iterated integrals of a continuous function can both exist and be different, and the limit over a regular exhaustion can depend on which exhaustion is used, or fail to exist. In such cases the improper double integral is not defined, and no single number deserves the name.

In plain words When positive and negative parts are both infinite, the answer depends on the order in which you add them up, just as for a conditionally convergent series.

Example On $0 < x \le 1$ , $0 < y \le 1$ take

f(x, y) = \frac{x^2 - y^2}{(x^2 + y^2)^2}.

Since the derivative of $\frac{y}{x^2 + y^2}$ with respect to $y$ is $f$ , and the derivative of $\frac{-x}{x^2 + y^2}$ with respect to $x$ is also $f$ ,

\begin{aligned} \int_0^1 \int_0^1 f \, dy \, dx &= \int_0^1 \frac{dx}{1 + x^2} \\[0.4em] &= \frac{\pi}{4}, \\[0.4em] \int_0^1 \int_0^1 f \, dx \, dy &= -\int_0^1 \frac{dy}{1 + y^2} \\[0.4em] &= -\frac{\pi}{4}. \end{aligned}

Fubini does not apply because $f$ is not absolutely integrable. In polar coordinates $\lvert f \rvert = \frac{\lvert \cos 2\theta \rvert}{r^2}$ , and the quarter annulus $\varepsilon \le r \le 1$ , $0 < \theta < \frac{\pi}{2}$ lies in the region, so

\begin{aligned} \iint \lvert f \rvert \, dA &\ge \int_0^{\pi/2} \lvert \cos 2\theta \rvert \, d\theta \int_\varepsilon^1 \frac{dr}{r} \\[0.4em] &= 1 \cdot \ln \frac{1}{\varepsilon}, \end{aligned}

which grows without bound as $\varepsilon \to 0^+$ . Exhaustions disagree too. Over the squares $[\frac{1}{n}, 1] \times [\frac{1}{n}, 1]$ the integral is $0$ for every $n$ , because $f(y, x) = -f(x, y)$ . Over the rectangles $[\frac{1}{n^2}, 1] \times [\frac{1}{n}, 1]$ the integrals tend to $-\frac{\pi}{4}$ .

For an exhaustion example on an unbounded region, take $\sin(x^2 + y^2)$ on the whole plane. Over disks of radius $R$ polar coordinates give

\begin{aligned} &\int_0^{2\pi} \int_0^R \sin(r^2) \, r \, dr \, d\theta \\[0.4em] &= \pi \big( 1 - \cos R^2 \big), \end{aligned}

which keeps oscillating between $0$ and $2\pi$ and has no limit. Over the squares $[-a, a] \times [-a, a]$ the identity

\begin{aligned} &\sin(x^2 + y^2) \\[0.4em] &= \sin(x^2) \cos(y^2) + \cos(x^2) \sin(y^2) \end{aligned}

turns the integral into $2 S(a) C(a)$ , where $S(a) = \int_{-a}^{a} \sin(x^2) \, dx$ and $C(a) = \int_{-a}^{a} \cos(x^2) \, dx$ . The Fresnel integrals converge, though not absolutely, and

\begin{aligned} \int_{-\infty}^{\infty} \sin(x^2) \, dx &= \int_{-\infty}^{\infty} \cos(x^2) \, dx \\[0.4em] &= \sqrt{\frac{\pi}{2}}. \end{aligned}

So the square totals tend to $2 \cdot \frac{\pi}{2} = \pi$ . Two exhaustions, two different behaviors.

Gaussian integral

Definition The Gaussian integral is

\int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi}.

The proof uses only the tools above. The function $e^{-(x^2 + y^2)}$ is continuous and positive on the plane, so its integral over the plane has one value in $[0, \infty]$ , the same for every regular exhaustion.

In plain words Compute the same total twice, once with squares, where it splits into the square of the one-variable integral, and once with disks, where polar coordinates finish the job.

Example Write $I = \int_{-\infty}^{\infty} e^{-x^2} \, dx$ , which converges because $e^{-x^2} \le e^{-\lvert x \rvert}$ for $\lvert x \rvert \ge 1$ . Over the square $[-a, a] \times [-a, a]$ the integrand factors, so

\begin{aligned} &\iint_{[-a, a]^2} e^{-(x^2 + y^2)} \, dA \\[0.4em] &= \left( \int_{-a}^{a} e^{-x^2} \, dx \right)^2, \end{aligned}

which tends to $I^2$ . Over the disk of radius $R$ , a bounded region where the polar change of variables is valid,

\begin{aligned} &\int_0^{2\pi} \int_0^R e^{-r^2} \, r \, dr \, d\theta \\[0.4em] &= \pi \big( 1 - e^{-R^2} \big), \end{aligned}

which tends to $\pi$ . Both are regular exhaustions of the plane and the integrand is positive, so the limits agree. Hence $I^2 = \pi$ , and since $I > 0$ , $I = \sqrt{\pi}$ .

Distinction from measure-theoretic generalizations

Definition Everything in this lesson concerns continuous integrands, regions with piecewise smooth boundaries, and limits over regular exhaustions, all built on Riemann integrals. Advanced courses replace this with the Lebesgue integral from measure theory, which integrates far more general functions over far more general sets. There, a function is integrable exactly when the integral of its absolute value is finite, Tonelli’s theorem holds for every measurable nonnegative function, and Fubini’s theorem holds for every integrable function, with the inner integrals required to exist only outside a set of measure zero. Those versions drop our extra assumption that the inner integrals are continuous, but they are different theorems and are not used in this course.

In plain words The rules here are the everyday versions for continuous functions. The general theory removes the side conditions but needs a whole new definition of the integral.

Example The function $\sin(x^2 + y^2)$ from above has no improper double integral over the plane in our sense, and it is not Lebesgue integrable either, because the integral of $\lvert \sin(x^2 + y^2) \rvert$ over the disk of radius $R$ grows without bound. On the other hand, $e^{-(x^2 + y^2)}$ is integrable in both senses, with the same value $\pi$ . For continuous, absolutely integrable functions the two theories always give the same number.

Check yourself

Compute the integral of $(x^2 + y^2)^{-3/2}$ over the region $x^2 + y^2 \ge 1$ , or show that it diverges.

Show the answer

The integrand is continuous and positive there, so any regular exhaustion works. With annuli $1 \le r \le n$ ,

\begin{aligned} &\int_0^{2\pi} \int_1^n r^{-3} \, r \, dr \, d\theta \\[0.4em] &= 2\pi \left( 1 - \frac{1}{n} \right) \to 2\pi. \end{aligned}

The integral converges to $2\pi$ .

Does the integral of $\frac{1}{x^2 + y^2}$ over the punctured unit disk $0 < x^2 + y^2 \le 1$ converge?

Show the answer

No. Over the annulus $\varepsilon \le r \le 1$ the integral is

\int_0^{2\pi} \int_\varepsilon^1 \frac{1}{r^2} \, r \, dr \, d\theta = 2\pi \ln \frac{1}{\varepsilon},

which grows without bound as $\varepsilon \to 0^+$ . Since the integrand is positive, the integral diverges.

Why is it wrong to conclude from the two iterated integrals of $\frac{x^2 - y^2}{(x^2 + y^2)^2}$ over the unit square that the double integral equals $\frac{\pi}{4}$ ?

Show the answer

The integrand changes sign and is not absolutely integrable near the origin, so Fubini’s theorem does not apply. The other order gives $-\frac{\pi}{4}$ , and the exhaustion by squares $[\frac{1}{n}, 1] \times [\frac{1}{n}, 1]$ gives $0$ . Different exhaustions give different limits, so the improper double integral is not defined.

Improper Multiple Integrals and Order of Integration

Key terms

Unbounded regions and singular integrands

Exhaustion by regions

Continuous nonnegative integrands

Elementary improper integral form of Tonelli

Fubini under absolute integrability

Failures without hypotheses

Gaussian integral

Distinction from measure-theoretic generalizations

Check yourself

How this lesson connects