At the end of the eighteenth century, Euler and Laplace computed hard real integrals with a daring trick: they replaced the real variable with a complex one, did the calculations and returned to the real. Laplace called this “the passage from the real to the imaginary”. The results checked out, but nobody could say when the trick was legitimate. It was to answer this question that Cauchy, at 25, submitted a memoir on definite integrals to the Academy in 1814. It was only published in 1827, and it was only the beginning.

Functions differentiable in the plane

A complex function $f(z)$, with $z = x + iy$, can be written as $f = u + iv$, where $u$ and $v$ are real functions of $x$ and $y$. For $f$ to have a derivative in the complex sense, $u$ and $v$ must satisfy the Cauchy–Riemann equations:

$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \qquad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}.$$

They had already appeared in d’Alembert and Euler; Cauchy put them at the centre of the theory, and Riemann would use them thirty years later as his starting point. For $f(z) = z^2 = (x^2 - y^2) + i\,(2xy)$: $u_x = 2x = v_y$ and $u_y = -2y = -v_x$.

The integral theorem

In 1825 Cauchy published the Mémoire sur les intégrales définies prises entre des limites imaginaires. Integrating from a point $z_0$ to a point $z_1$ in the plane requires choosing a path between them. The central theorem says that the path does not matter, as long as the function stays finite and continuous in the region swept out when one path is deformed into the other. Put another way: the integral around any closed curve that does not go around a troublesome point is zero,

$$\oint f(z)\,dz = 0.$$

The most important example is the one where the theorem does not apply. Integrate $\frac{1}{z}$ around the circle of radius 1, traversed counterclockwise. With $z = e^{it}$, $dz = i e^{it}\,dt$, and

$$\oint \frac{dz}{z} = \int_0^{2\pi} \frac{i e^{it}}{e^{it}}\,dt = \int_0^{2\pi} i\,dt = 2\pi i.$$

The result is not zero because the circle goes around $z = 0$, where $\frac{1}{z}$ blows up.

Residues

In 1826, in his Exercices de mathématiques, Cauchy turned this calculation into a method, the calculus of residues. If near a point $a$ the function behaves like $\frac{R}{z - a}$ plus a well-behaved part, the number $R$ is the residue of $f$ at $a$. A closed curve that goes once around $a$ gives

$$\oint f(z)\,dz = 2\pi i \cdot R,$$

and if it goes around several such points, the residues are added. In the figure, the circle can be moved and enlarged. The function $\frac{1}{1 + z^2}$ blows up at $z = i$ and $z = -i$, and the integral depends only on which of these points lie inside:

A real integral

Here is the method at work on a real integral:

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

Since $1 + z^2 = (z - i)(z + i)$, near $z = i$ the function is approximately $\frac{1}{2i} \cdot \frac{1}{z - i}$: the residue at $i$ is $\frac{1}{2i}$. Take the closed path formed by the segment from $-R$ to $R$ on the real axis and the semicircle of radius $R$ in the upper half-plane. For $R > 1$ it goes around the point $i$ only, and

$$\int_{-R}^{R} \frac{dx}{1 + x^2} + \int_{\text{arc}} \frac{dz}{1 + z^2} = 2\pi i \cdot \frac{1}{2i} = \pi.$$

On the arc, $|1 + z^2| \ge R^2 - 1$, and its length is $\pi R$, so the integral over the arc is at most $\frac{\pi R}{R^2 - 1}$, which goes to zero. What remains is

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

It is the same result one gets with $\arctan x$, but the method of residues handles in the same way integrals for which no antiderivative is known.

The integral formula

In 1831, already in exile in Turin, Cauchy took the next step, the integral formula: if $f$ is well behaved inside a circle and $a$ is an interior point,

$$f(a) = \frac{1}{2\pi i} \oint \frac{f(z)}{z - a}\,dz.$$

The values of the function on the boundary determine all its values inside. From the formula Cauchy deduced that every function differentiable in the complex sense has a power series, and that this series converges in the largest circle containing no troublesome point. This explains an old mystery: the series $\frac{1}{1 + x^2} = 1 - x^2 + x^4 - \cdots$ converges only for $|x| < 1$, although the function is perfectly smooth along the whole real line. The reason lies off the line: the points $\pm i$, at distance 1 from the origin.

Exercise: the complex exponential

Check the Cauchy–Riemann equations for $e^z = e^x \cos y + i\,e^x \sin y$.

$u = e^x \cos y$ and $v = e^x \sin y$. Then $u_x = e^x \cos y = v_y$ and $u_y = -e^x \sin y = -v_x$. The equations hold over the whole plane: the exponential is differentiable everywhere, and its series $1 + z + \frac{z^2}{2} + \cdots$ converges for every $z$, because there is no troublesome point.

Exercise: the radius of convergence

Add the first terms of $1 - x^2 + x^4 - x^6 + \cdots$ for $x = \frac{1}{2}$ and for $x = 2$, and compare with $\frac{1}{1 + x^2}$.

For $x = \frac{1}{2}$: $1 - 0.25 + 0.0625 - 0.015625 = 0.796875$, and $\frac{1}{1.25} = 0.8$: the series converges. For $x = 2$: $1 - 4 + 16 - 64 = -51$, and the terms only grow, while $\frac{1}{5} = 0.2$. By Cauchy’s theorem, the radius of convergence is the distance from the origin to $\pm i$, which is 1, and $\frac{1}{2} < 1 < 2$.

Sources: Augustin-Louis Cauchy, “Mémoire sur les intégrales définies” (1814, published 1827) · Augustin-Louis Cauchy, Mémoire sur les intégrales définies prises entre des limites imaginaires (1825) · Augustin-Louis Cauchy, “Sur le calcul des résidus”, Exercices de mathématiques 1 (1826) · Frank Smithies, Cauchy and the Creation of Complex Function Theory (1997) · Gauss’s letter to Bessel, 18 December 1811, in Briefwechsel zwischen Gauss und Bessel (1880).

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