The third book of the Instituzioni deals with integral calculus: areas under curves, volumes, lengths. The versiera gives a surprising example. The curve never touches the $x$-axis; it stretches out without end on both sides. Even so, the region between the curve and the axis has a finite area.

A well-known antiderivative

With $a = 1$, the curve is $y = \frac{1}{1 + x^2}$. We need a function whose derivative is $\frac{1}{1 + x^2}$. It exists and is an old acquaintance: the arctangent. If $u = \arctan x$, then $x = \tan u$ and

$$\begin{aligned} \frac{dx}{du} &= \frac{1}{\cos^2 u} = 1 + \tan^2 u = 1 + x^2 \\ &\Longrightarrow\quad \frac{du}{dx} = \frac{1}{1 + x^2}. \end{aligned}$$

So the area between $-X$ and $X$ is

$$\int_{-X}^{X} \frac{dx}{1 + x^2} = \arctan X - \arctan(-X) = 2\arctan X.$$

As $X$ grows without limit, $\arctan X$ approaches $\frac{\pi}{2}$, and the total area is

$$\int_{-\infty}^{\infty} \frac{dx}{1 + x^2} = \pi.$$

Four circles

With any diameter $a$, the change of variable $x = a u$ gives

$$\int_{-\infty}^{\infty} \frac{a^3}{x^2 + a^2}\,dx = a^2 \int_{-\infty}^{\infty} \frac{du}{1 + u^2} = \pi a^2.$$

The circle that generates the curve has diameter $a$, that is, radius $\frac{a}{2}$ and area $\pi\frac{a^2}{4}$. The area under the versiera is exactly four times the area of the circle. It is the kind of clean result that eighteenth-century mathematicians loved, and that Agnesi presents as an example of the power of the new calculus.

Leibniz’s series

The same integral explains one of the most beautiful formulas in mathematics. For $|x| < 1$, the sum of the geometric progression gives

$$\frac{1}{1 + x^2} = 1 - x^2 + x^4 - x^6 + \cdots$$

Integrating term by term from 0 to 1, and remembering that $\arctan 1 = \frac{\pi}{4}$:

$$\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots$$

This is Leibniz’s series, published in 1682 and known earlier to the Indian mathematicians of the Kerala school. In terms of the versiera: the area under the curve between 0 and 1 is exactly $\frac{\pi}{4}$, the same as the area of the circle of diameter 1 that generates it.

Exercise: the area between −1 and 1

Compute the area under $y = \frac{1}{1 + x^2}$ between $x = -1$ and $x = 1$. What fraction of the total area is it?

$2\arctan 1 = 2 \cdot \frac{\pi}{4} = \frac{\pi}{2}$. The total area is $\pi$, so the stretch between $-1$ and $1$, which corresponds to $-a$ and $a$, contains half the area. The other half is in the two tails, which stretch out to infinity.

Exercise: Leibniz's series converges slowly

Add the first four terms of the series $1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7}$ and compare with $\frac{\pi}{4} \approx 0.7854$.

$1 - 0.3333 + 0.2 - 0.1429 = 0.7238$. The error is about 0.06, still large. Since the terms decrease only like $\frac{1}{2n+1}$, getting five decimal places right takes about a hundred thousand terms. The series is beautiful, but terrible for computing $\pi$.

Sources: Maria Gaetana Agnesi, Instituzioni analitiche ad uso della gioventù italiana, vol. II (1749), book III · Gottfried Wilhelm Leibniz, “De vera proportione circuli ad quadratum circumscriptum in numeris rationalibus”, Acta Eruditorum (1682) · Ranjan Roy, “The Discovery of the Series Formula for π by Leibniz, Gregory and Nilakantha”, Mathematics Magazine 63 (1990).

All topics on Maria Gaetana Agnesi · ← Maxima, minima and points of inflection · The inverse method of tangents →

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