Square the question, reveal a circle, integrals from easy to advanced lesson

The function $e^{-x^2}$ has no antiderivative built from elementary functions, which is why its integral from $0$ to $1$ is found numerically, as about $0.7468$ . The area under the whole curve, from $-\infty$ to $\infty$ , is a different question, and it has an exact answer. This page multiplies the integral by a copy of itself in a second variable, reads the product as the volume under a round surface over the whole plane and evaluates that volume in polar coordinates. The result is $\sqrt{\pi}$ , and each step comes with the condition that makes it valid.

Key terms

Gaussian integral

Definition The Gaussian integral is the improper integral of $e^{-x^2}$ over the whole real line,

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

Both ends are infinite, so it means the sum of two limits,

\begin{aligned} I &= \lim_{a \to -\infty} \int_a^0 e^{-x^2}\,dx \\[0.4em] &\quad + \lim_{b \to \infty} \int_0^b e^{-x^2}\,dx, \end{aligned}

and it converges when both limits exist as finite numbers. They do, and $I = \sqrt{\pi} \approx 1.7725$ . More generally, for every constant $c > 0$ the substitution $u = \sqrt{c}\,x$ gives

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

In plain words The two tails of the bell curve never reach zero, but they thin out so quickly that the total area is finite. Since $e^{-x^2}$ has no elementary antiderivative, the exact value cannot come from evaluating a formula for an antiderivative at the ends. It comes from a detour through two dimensions, described in the next cards.

Example Finiteness comes first. For $\lvert x \rvert \ge 1$ we have $x^2 \ge \lvert x \rvert$ , so

e^{-x^2} \le e^{-\lvert x \rvert} \qquad (\lvert x \rvert \ge 1).

Each exponential tail has finite area, because $\int_1^b e^{-x}\,dx = e^{-1} - e^{-b}$ tends to $1/e$ as $b \to \infty$ , and on $[-1, 1]$ the integrand is at most $1$ . So $0 < I \le 2 + 2/e \approx 2.74$ , and the value $\sqrt{\pi} \approx 1.7725$ fits. The integrand is even, so the half-line integral is

\int_0^{\infty} e^{-x^2}\,dx = \frac{\sqrt{\pi}}{2},

about $0.8862$ . It exceeds $\int_0^1 e^{-x^2}\,dx \approx 0.7468$ by the area of the tail beyond $1$ , about $0.1394$ .

Squaring the integral

Definition Squaring the integral is a method for a convergent integral $I$ of a nonnegative function. Write $I^2$ as the product of two copies of $I$ with different integration variables, read the product as a double integral, evaluate the double integral in coordinates that fit its integrand, and recover $I$ from $I^2$ . The method treats $I$ as a finite number, so convergence is checked before squaring, and it uses $I \ge 0$ to choose the nonnegative square root.

In plain words A single copy of the Gaussian has no factor that a substitution could use. Two copies in independent variables multiply to $e^{-(x^2+y^2)}$ , a height that depends only on the distance from the origin, and circular symmetry is what polar coordinates handle well. Squaring loses the sign, and the sign of the original integral restores it.

Example For the Gaussian integral, the product of the two copies is a double integral over the plane,

I^2 = \iint_{\mathbb{R}^2} e^{-(x^2+y^2)}\,dA,

and in polar coordinates, with $dA = r\,dr\,d\theta$ , it becomes

\begin{aligned} I^2 &= \int_0^{2\pi} \int_0^{\infty} e^{-r^2}\, r\,dr\,d\theta \\[0.4em] &= 2\pi \int_0^{\infty} r e^{-r^2}\,dr. \end{aligned}

The substitution $u = r^2$ , with $du = 2r\,dr$ , evaluates the radial integral as a limit,

\begin{aligned} \int_0^{b} r e^{-r^2}\,dr &= \frac{1}{2} \int_0^{b^2} e^{-u}\,du \\[0.4em] &= \frac{1 - e^{-b^2}}{2}, \end{aligned}

which tends to $\tfrac12$ as $b \to \infty$ . So $I^2 = 2\pi \cdot \tfrac12 = \pi$ . Both $\sqrt{\pi}$ and $-\sqrt{\pi}$ square to $\pi$ , and the Gaussian integral adds positive heights, so $I > 0$ and

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

Product of independent integrals

Definition Let $f$ be continuous and nonnegative on an interval $J$ , and $g$ continuous and nonnegative on an interval $K$ , where either interval may be unbounded and both integrals converge. Then the product of the two integrals is a double integral over the rectangle $J \times K$ ,

\begin{aligned} &\int_J f(x)\,dx \int_K g(y)\,dy \\[0.4em] &\qquad = \iint_{J \times K} f(x)\, g(y)\,dA. \end{aligned}

The same holds for functions that change sign when $\int_J \lvert f(x) \rvert\,dx$ and $\int_K \lvert g(y) \rvert\,dy$ are finite. The two variables must be different. The product is in general not $\int_J f(x)\, g(x)\,dx$ , which pairs the two factors at the same point.

In plain words In the double integral, $x$ and $y$ vary independently over the whole rectangle. For each fixed $x$ , the inner integral over $y$ is the constant $\int_K g(y)\,dy$ , which comes out as a factor and leaves the product. The result is a volume under a surface over the plane, not an area under the single curve $f(x)\,g(x)$ .

Example The integrals of $x$ over $[0, 1]$ and of $y^2$ over $[0, 2]$ are $\tfrac12$ and $\tfrac83$ , and their product is

\int_0^1 \int_0^2 x y^2\,dy\,dx = \frac{4}{3},

while the integral of the pointwise product $x \cdot x^2$ over $[0, 1]$ is only $\tfrac14$ . For the Gaussian the same mistake gives

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

about $1.2533$ , which is not $I^2 = \pi \approx 3.1416$ .

Improper double integral

Definition Let $f$ be continuous and nonnegative on the plane. Take bounded regions $D_1 \subseteq D_2 \subseteq D_3 \subseteq \cdots$ , such as disks or squares centered at the origin, so that every bounded set lies inside some $D_n$ . The improper double integral of $f$ over the plane is

\iint_{\mathbb{R}^2} f\,dA = \lim_{n \to \infty} \iint_{D_n} f\,dA.

Because $f \ge 0$ , the integrals over $D_n$ increase with $n$ , so the limit exists as a number or as $\infty$ , and it is the same for every such sequence of regions. By Tonelli’s theorem, the iterated integrals in either order give this same value. When $f$ changes sign, the integral over the plane, and Fubini’s theorem for swapping the order, need absolute integrability, $\iint_{\mathbb{R}^2} \lvert f \rvert\,dA < \infty$ .

In plain words Integrate over larger and larger bounded pieces of the plane and take the limit. With a nonnegative integrand nothing can cancel, so neither the shape of the pieces nor the order of integration can change the answer. With a sign-changing integrand, positive and negative parts can balance differently on disks and on squares, and the answer may then depend on the shape unless the absolute value has a finite integral.

Example Take $f(x, y) = e^{-(x^2+y^2)}$ , the disk $D_\rho$ of radius $\rho$ and the square $S_n = [-n, n] \times [-n, n]$ . In polar coordinates over a disk,

\iint_{D_\rho} f\,dA = \pi\left(1 - e^{-\rho^2}\right),

and over the square the double integral splits as $I_n^2$ , where $I_n = \int_{-n}^{n} e^{-x^2}\,dx$ . The disk $D_n$ fits inside $S_n$ , which fits inside $D_{n\sqrt{2}}$ , and $f \ge 0$ , so

\begin{aligned} &\pi\left(1 - e^{-n^2}\right) \le I_n^2 \\[0.4em] &\quad \le \pi\left(1 - e^{-2n^2}\right). \end{aligned}

Both bounds tend to $\pi$ , so the squares and the disks give the same limit $I^2 = \pi$ . For $n = 1$ the three numbers are about $1.986$ , $2.231$ and $2.716$ .

Polar coordinates on the whole plane

Definition Polar coordinates write $x = r\cos\theta$ and $y = r\sin\theta$ . The whole plane is covered by $0 \le r < \infty$ and a full turn $0 \le \theta \le 2\pi$ , and the map is one-to-one from $0 < r < \infty$ , $0 \le \theta < 2\pi$ onto the plane without the origin. The leftover sets, the origin and the repeated ray $\theta = 2\pi$ , have zero area. The Jacobian determinant of $(r, \theta) \mapsto (x, y)$ is $r$ , and its absolute value gives $dA = r\,dr\,d\theta$ . For $f$ continuous on the plane and either nonnegative or absolutely integrable, and with $g(r, \theta) = f(r\cos\theta, r\sin\theta)$ ,

\begin{aligned} &\iint_{\mathbb{R}^2} f(x, y)\,dA \\[0.4em] &= \int_0^{2\pi} \int_0^{\infty} g(r, \theta)\, r\,dr\,d\theta. \end{aligned}

When $f(x, y) = h(x^2 + y^2)$ depends only on the distance from the origin, the angular integral is the factor $2\pi$ , and

\begin{aligned} &\iint_{\mathbb{R}^2} f\,dA \\[0.4em] &= 2\pi \int_0^{\infty} h(r^2)\, r\,dr. \end{aligned}

In plain words Polar coordinates cut the plane into thin rings. The ring between radii $r$ and $r + dr$ has circumference $2\pi r$ and width $dr$ , so its area is about $2\pi r\,dr$ . That factor $r$ is the one the one-variable Gaussian lacked, since the substitution $u = x^2$ needs $du = 2x\,dx$ and the integrand $e^{-x^2}$ has no factor $x$ . Covering the whole plane means every radius from $0$ to $\infty$ and a full turn of angles, not a quadrant or a disk.

Example Over the whole plane,

\begin{aligned} &\iint_{\mathbb{R}^2} \frac{dA}{(1 + x^2 + y^2)^2} \\[0.4em] &= 2\pi \int_0^{\infty} \frac{r\,dr}{(1 + r^2)^2} = \pi, \end{aligned}

because $\int_0^b r\,(1 + r^2)^{-2}\,dr = \tfrac12 - \tfrac{1}{2(1 + b^2)}$ tends to $\tfrac12$ as $b \to \infty$ . Leaving out the factor $r$ would give $2\pi \int_0^{\infty} (1 + r^2)^{-2}\,dr = 2\pi \cdot \tfrac{\pi}{4} = \tfrac{\pi^2}{2} \approx 4.93$ , a different and wrong number.

Check yourself

Use squaring the integral to evaluate $\int_{-\infty}^{\infty} e^{-x^2/2}\,dx$ .

Show the answer

Call the integral $J$ . It converges by the same comparison as the Gaussian, since $x^2/2 \ge \lvert x \rvert$ for $\lvert x \rvert \ge 2$ , and it is positive. Two copies in $x$ and $y$ give

\begin{aligned} J^2 &= \iint_{\mathbb{R}^2} e^{-(x^2+y^2)/2}\,dA \\[0.4em] &= 2\pi \int_0^{\infty} r e^{-r^2/2}\,dr. \end{aligned}

With $u = r^2/2$ and $du = r\,dr$ , the radial integral up to $b$ is $1 - e^{-b^2/2}$ , which tends to $1$ as $b \to \infty$ . So $J^2 = 2\pi$ , and since $J > 0$ ,

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

about $2.5066$ . This is the constant that makes the standard normal density $e^{-x^2/2}/\sqrt{2\pi}$ have total area $1$ .

Find $\int_0^{\infty} e^{-x^2}\,dx$ by squaring it directly, without using the value of the full Gaussian integral. Which angles does the polar integral use?

Show the answer

Call the integral $K$ , which is positive and finite. Its square is the integral of $e^{-(x^2+y^2)}$ over the first quadrant $x \ge 0$ , $y \ge 0$ , not over the whole plane. In polar coordinates that quadrant is $0 \le r < \infty$ with $0 \le \theta \le \pi/2$ , so

K^2 = \frac{\pi}{2} \int_0^{\infty} r e^{-r^2}\,dr = \frac{\pi}{4}.

Since $K > 0$ , it follows that $K = \sqrt{\pi}/2 \approx 0.8862$ , half of the Gaussian integral, as symmetry predicts. Using a full turn $0 \le \theta \le 2\pi$ here would give $K^2 = \pi$ , four times too large.

Let $f(x, y) = \sin(x^2 + y^2)$ . Compute the integral of $f$ over the disk $D_\rho$ of radius $\rho$ centered at the origin. Does $f$ have an integral over the whole plane in the sense of this page?

Show the answer

In polar coordinates, with $u = r^2$ ,

\begin{aligned} &\iint_{D_\rho} f\,dA \\[0.4em] &= 2\pi \int_0^{\rho} r \sin(r^2)\,dr \\[0.4em] &= \pi\left(1 - \cos \rho^2\right). \end{aligned}

As $\rho \to \infty$ this keeps oscillating between $0$ and $2\pi$ , so the disks give no limit. The squares $[-n, n] \times [-n, n]$ do give a limit, $\pi$ , which follows from the Fresnel integrals, so the answer depends on the shape of the regions. The method of this page does not apply, because $f$ changes sign and is not absolutely integrable. The integral of $\lvert f \rvert$ over $D_\rho$ is $\pi \int_0^{\rho^2} \lvert \sin u \rvert\,du$ , which equals $2\pi k$ when $\rho^2 = k\pi$ and so grows without bound. The function has no integral over the whole plane.

Square the question, reveal a circle

Key terms

Gaussian integral

Squaring the integral

Product of independent integrals

Improper double integral

Polar coordinates on the whole plane

Check yourself

How this lesson connects

Builds on