Leave the real line, collect the poles, return, integrals from easy to advanced lesson

The integrand $1/(1 + x^4)$ is positive and even, and it is at most $1/x^4$ for $x > 0$ , so the integral $\int_0^{\infty} \frac{dx}{1 + x^4}$ converges. This page finds its exact value by leaving the real line. In the complex plane the denominator vanishes at four points, a closed semicircle in the upper half-plane encloses two of them, and the residue theorem gives the integral around that contour as $2\pi i$ times the sum of two residues. An estimate shows that the curved arc contributes nothing as its radius grows, and evenness halves the whole-line value to $\pi/(2\sqrt{2})$ .

Key terms

Analytic function

Definition A complex number $z = x + iy$ , with real $x$ and $y$ and $i^2 = -1$ , is drawn as the point $(x, y)$ of the complex plane, whose horizontal axis is the real line. A function $f$ defined on an open set $U$ of the complex plane is analytic on $U$ if at every point $z_0$ of $U$ the complex derivative

f'(z_0) = \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0}

exists, with $z$ allowed to approach $z_0$ from every direction in the plane. Polynomials are analytic on the whole plane, and sums, products and quotients of analytic functions are analytic wherever no denominator is zero. So a rational function in lowest terms is analytic everywhere except at the zeros of its denominator. A point $p$ is an isolated singularity of $f$ when $f$ is analytic on a punctured disk $0 < \lvert z - p \rvert < r$ but is not analytic, or not defined, at $p$ itself.

In plain words Complex differentiability asks for one limit that is the same whichever way $z$ approaches $z_0$ , a much stronger demand than having partial derivatives in $x$ and $y$ . It has strong consequences. An analytic function has derivatives of every order, and around each point $z_0$ of $U$ it equals its Taylor series on every open disk centered at $z_0$ that lies inside $U$ .

Example The function $f(z) = 1/(1 + z^4)$ is analytic except where $z^4 = -1$ . Every solution has $\lvert z \rvert^4 = 1$ , so $z = e^{i\theta} = \cos\theta + i\sin\theta$ for some angle $\theta$ . The fourth power multiplies the angle by $4$ , and $-1 = e^{i\pi}$ , so $4\theta = \pi + 2\pi k$ and the solutions are

z_k = e^{i(\pi/4 + k\pi/2)}

for $k = 0, 1, 2, 3$ , the points $(\pm 1 \pm i)/\sqrt{2}$ on the unit circle. Two of them, $e^{i\pi/4}$ and $e^{3i\pi/4}$ , lie in the upper half-plane. By contrast, the conjugate $f(z) = \bar{z} = x - iy$ is analytic nowhere, because its difference quotient $\bar{h}/h$ equals $1$ for real $h$ and $-1$ for imaginary $h$ .

Contour integration

Definition A contour $C$ is a piecewise smooth curve $z = \gamma(t)$ for $a \le t \le b$ , traversed in the direction of increasing $t$ . For $f$ continuous on $C$ , the contour integral of $f$ along $C$ is

\begin{aligned} &\int_C f(z)\,dz \\[0.4em] &= \int_a^b f(\gamma(t))\, \gamma'(t)\,dt. \end{aligned}

Its value does not depend on how $C$ is parametrized, as long as the direction is kept, and reversing the direction changes its sign. A closed contour is simple if it does not cross itself, and positively oriented if it runs counterclockwise, keeping its interior on the left. The ML estimate bounds every contour integral. If $\lvert f(z) \rvert \le M$ on $C$ and $C$ has length $L$ , then

\left\lvert \int_C f(z)\,dz \right\rvert \le ML.

As a method, contour integration evaluates a real integral by making the real interval part of a closed contour and computing the integral around that contour in another way.

In plain words A contour integral adds up values of $f$ times small complex steps $dz$ along a directed path, much as a line integral adds up a field along a curve. The ML estimate says that the integral is no larger in size than the largest size of the integrand times the length of the path.

Example Fix a point $p$ and a radius $\rho > 0$ , and go once counterclockwise around the circle $\gamma(\theta) = p + \rho e^{i\theta}$ with $0 \le \theta \le 2\pi$ . Then $\gamma'(\theta) = i\rho e^{i\theta}$ , and

\begin{aligned} \oint_{\gamma} \frac{dz}{z - p} &= \int_0^{2\pi} \frac{i\rho e^{i\theta}}{\rho e^{i\theta}}\,d\theta \\[0.4em] &= 2\pi i \end{aligned}

for every radius. For every integer $n \ne -1$ the same computation gives $\oint_{\gamma} (z - p)^n\,dz = 0$ , because $\int_0^{2\pi} e^{i(n+1)\theta}\,d\theta = 0$ .

Cauchy’s theorem

Definition Let $C$ be a simple closed, piecewise smooth contour, and let $f$ be analytic on an open set that contains $C$ and its interior. Then

\oint_C f(z)\,dz = 0.

Analyticity is needed at every point inside $C$ , not only on $C$ itself.

In plain words An analytic function has no circulation around a loop that encloses no singular points. When $f'$ is continuous, Green’s theorem shows why. Writing $f = u + iv$ , the real and imaginary parts of $\oint_C f(z)\,dz$ are circulation integrals whose scalar curls, $-v_x - u_y$ and $u_x - v_y$ , vanish by the Cauchy-Riemann equations $u_x = v_y$ and $u_y = -v_x$ .

Example All four singular points of $1/(1 + z^4)$ lie on the unit circle, so they are at distance at least $2$ from the point $3$ . The function is therefore analytic on an open set containing the closed disk $\lvert z - 3 \rvert \le 1$ , and

\oint_{\lvert z - 3 \rvert = 1} \frac{dz}{1 + z^4} = 0.

The condition matters. The function $1/z$ is analytic everywhere except at $0$ , yet its integral around the unit circle is $2\pi i$ , not $0$ , because the singular point $0$ lies inside.

Simple pole and its residue

Definition Let $f$ be analytic on a punctured disk $0 < \lvert z - p \rvert < r$ . The point $p$ is a simple pole of $f$ if the limit

A = \lim_{z \to p} (z - p) f(z)

exists and is not $0$ . Then $f(z) = A/(z - p) + g(z)$ , where $g$ is analytic on the whole disk $\lvert z - p \rvert < r$ once $g(p)$ is defined by its limit, and the coefficient $A$ is the residue, written $\operatorname{Res}(f, p)$ . At any isolated singularity, the residue is the coefficient of $1/(z - p)$ in the Laurent series of $f$ around $p$ , and it equals $\frac{1}{2\pi i} \oint f(z)\,dz$ around every counterclockwise circle $\lvert z - p \rvert = \rho$ with $0 < \rho < r$ . If $f = P/Q$ with $P$ and $Q$ analytic at $p$ , $P(p) \ne 0$ , $Q(p) = 0$ and $Q'(p) \ne 0$ , then $p$ is a simple pole and

\operatorname{Res}(f, p) = \frac{P(p)}{Q'(p)}.

In plain words Near a simple pole, $f$ looks like $A/(z - p)$ plus a well-behaved part. A small counterclockwise loop around $p$ gets nothing from the well-behaved part, by Cauchy’s theorem, and $2\pi i A$ from the singular part, by the circle computation. The residue is the one number that a small loop around the pole can detect.

Example For $f(z) = 1/(1 + z^4)$ take $P = 1$ and $Q = 1 + z^4$ . At each root $p$ of $Q$ the derivative $Q'(p) = 4p^3$ is not $0$ , so every pole is simple. Since $p^4 = -1$ , it follows that $1/p^3 = p/p^4 = -p$ , and

\operatorname{Res}(f, p) = \frac{1}{4p^3} = -\frac{p}{4}.

At $p_1 = e^{i\pi/4} = (1 + i)/\sqrt{2}$ the residue is $-(1 + i)/(4\sqrt{2})$ .

Residue theorem

Definition Let $C$ be a positively oriented, simple closed, piecewise smooth contour. Let $f$ be analytic on an open set containing $C$ and its interior, except at finitely many points $p_1, \dots, p_m$ inside $C$ , none of them on $C$ . Then

\begin{aligned} &\oint_C f(z)\,dz \\[0.4em] &= 2\pi i \sum_{k=1}^{m} \operatorname{Res}(f, p_k). \end{aligned}

Singular points outside $C$ contribute nothing.

In plain words Cut a small disk around each singular point out of the inside of $C$ , and join each small circle to $C$ by a pair of cuts. Each cut is traversed once in each direction, so the two passes cancel, as shared edges cancel in Green’s theorem, and Cauchy’s theorem applies to the region that is left. So the integral around $C$ equals the sum of the counterclockwise integrals around the small circles, and each of those is $2\pi i$ times a residue.

Example The circle $\lvert z - i \rvert = 1$ , run counterclockwise, encloses the pole $i$ of $1/(z^2 + 1)$ but not the pole $-i$ , which is at distance $2$ from $i$ . With $P = 1$ and $Q = z^2 + 1$ , the residue at $i$ is $1/(2i)$ , so

\begin{aligned} \oint_{\lvert z - i \rvert = 1} \frac{dz}{z^2 + 1} &= 2\pi i \cdot \frac{1}{2i} \\[0.4em] &= \pi. \end{aligned}

Upper half-plane semicircle

Definition For $R > 0$ , the upper half-plane semicircle $C_R$ runs along the real axis from $-R$ to $R$ and then back along the arc $\Gamma_R$ , given by $z = Re^{i\theta}$ with $0 \le \theta \le \pi$ . It is a positively oriented simple closed contour whose inside is the half-disk of points with $\lvert z \rvert < R$ and $\operatorname{Im} z > 0$ . Let $P$ and $Q$ be polynomials with $P/Q$ in lowest terms, $\deg Q \ge \deg P + 2$ and no real zeros of $Q$ . Then the real integral of $P/Q$ converges absolutely, and

\begin{aligned} &\int_{-\infty}^{\infty} \frac{P(x)}{Q(x)}\,dx \\[0.4em] &= 2\pi i \sum_{\operatorname{Im} p > 0} \operatorname{Res}\!\left(\frac{P}{Q}, p\right), \end{aligned}

where the sum runs over the zeros $p$ of $Q$ in the upper half-plane. The proof takes $R$ larger than every $\lvert p \rvert$ and uses a vanishing arc bound. For large $R$ there is a constant $K$ with $\lvert P(z)/Q(z) \rvert \le K/R^2$ on $\Gamma_R$ , so by the ML estimate the arc integral is at most $\pi R \cdot K/R^2 = \pi K/R$ in absolute value, which tends to $0$ as $R \to \infty$ .

In plain words Closing the real segment with a half circle gives a closed contour, so the residue theorem applies. The price is an extra integral along the arc. When the integrand shrinks faster than $1/R$ on the arc, the growing length $\pi R$ cannot keep up, the arc’s contribution disappears as $R \to \infty$ , and only the real line is left.

Example For $f(z) = 1/(1 + z^4)$ and $R > 1$ , the contour $C_R$ encloses exactly the two upper poles $p_1 = (1 + i)/\sqrt{2}$ and $p_2 = (-1 + i)/\sqrt{2}$ , and no pole lies on it. Their sum is $i\sqrt{2}$ , so the residues $-p/4$ add up to $-i\sqrt{2}/4 = -i/(2\sqrt{2})$ , and

\begin{aligned} &\oint_{C_R} f(z)\,dz \\[0.4em] &= 2\pi i \cdot \left(-\frac{i}{2\sqrt{2}}\right) \\[0.4em] &= \frac{\pi}{\sqrt{2}}. \end{aligned}

On the arc $\lvert z^4 \rvert = R^4$ , so the reverse triangle inequality gives $\lvert 1 + z^4 \rvert \ge R^4 - 1$ , and the ML estimate with length $\pi R$ gives

\left\lvert \int_{\Gamma_R} f(z)\,dz \right\rvert \le \frac{\pi R}{R^4 - 1},

which tends to $0$ as $R \to \infty$ . The real integral converges absolutely, because $0 < f(x) \le 1$ and $f(x) \le 1/x^4$ for $x \ne 0$ . For every $R > 1$ the integral over $[-R, R]$ plus the arc integral equals $\pi/\sqrt{2}$ . As $R \to \infty$ the first tends to the whole-line integral and the second tends to $0$ , so

\int_{-\infty}^{\infty} \frac{dx}{1 + x^4} = \frac{\pi}{\sqrt{2}}.

The integrand is even, so the original half-line integral is half of this,

\int_0^{\infty} \frac{dx}{1 + x^4} = \frac{\pi}{2\sqrt{2}},

about $1.1107$ . As a check, the integral from $0$ to $1$ lies between $\tfrac12$ and $1$ , and the tail beyond $1$ is less than $\int_1^{\infty} x^{-4}\,dx = \tfrac13$ , so the answer has to lie between $\tfrac12$ and $\tfrac43$ , and it does.

Check yourself

Use an upper half-plane semicircle to evaluate $\int_{-\infty}^{\infty} \frac{dx}{x^2 + 1}$ , and check the answer with an antiderivative.

Show the answer

The function $1/(z^2 + 1)$ has simple poles at $i$ and $-i$ . For $R > 1$ the contour $C_R$ encloses only $i$ , where the residue is $1/(2i)$ , so the integral around $C_R$ is $2\pi i \cdot \frac{1}{2i} = \pi$ . On the arc $\lvert z^2 + 1 \rvert \ge R^2 - 1$ , so the arc integral is at most $\pi R/(R^2 - 1)$ in absolute value, which tends to $0$ . The real integral converges, so

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

The antiderivative $\arctan x$ agrees, since $\arctan b \to \pi/2$ as $b \to \infty$ and $\arctan a \to -\pi/2$ as $a \to -\infty$ , a difference of $\pi$ .

To evaluate $\int_{-\infty}^{\infty} \frac{dx}{(x^2 + 1)(x^2 + 4)}$ , a classmate adds the residues at all four poles $\pm i$ and $\pm 2i$ and multiplies the sum by $2\pi i$ . What do they get, what went wrong, and what is the correct value?

Show the answer

Let $f = 1/Q$ with $Q(z) = (z^2 + 1)(z^2 + 4)$ . The derivative is $Q'(z) = 2z(z^2 + 4) + 2z(z^2 + 1)$ , and the residues $1/Q'(p)$ are

\begin{aligned} \operatorname{Res}(f, i) &= -\tfrac{i}{6}, \\[0.4em] \operatorname{Res}(f, 2i) &= \tfrac{i}{12}, \\[0.4em] \operatorname{Res}(f, -i) &= \tfrac{i}{6}, \\[0.4em] \operatorname{Res}(f, -2i) &= -\tfrac{i}{12}. \end{aligned}

All four add up to $0$ , so the classmate gets $0$ , which is impossible for the integral of a positive function. The upper semicircle $C_R$ with $R > 2$ encloses only $i$ and $2i$ , and the lower poles lie outside it and contribute nothing. The enclosed residues add up to $-\tfrac{i}{6} + \tfrac{i}{12} = -\tfrac{i}{12}$ . On the arc the denominator has absolute value at least $(R^2 - 1)(R^2 - 4)$ , so the arc integral is at most $\pi R/\left((R^2 - 1)(R^2 - 4)\right)$ in absolute value and tends to $0$ . Hence

\begin{aligned} &\int_{-\infty}^{\infty} \frac{dx}{(x^2 + 1)(x^2 + 4)} \\[0.4em] &= 2\pi i \cdot \left(-\frac{i}{12}\right) = \frac{\pi}{6}. \end{aligned}

Partial fractions confirm it, since the integrand equals $\tfrac13\left(\frac{1}{x^2 + 1} - \frac{1}{x^2 + 4}\right)$ , whose integral is $\tfrac13\left(\pi - \tfrac{\pi}{2}\right) = \tfrac{\pi}{6}$ .

Evaluate $\int_0^{\infty} \frac{dx}{1 + x^6}$ by contour integration.

Show the answer

The poles of $1/(1 + z^6)$ are the sixth roots of $-1$ , the points $e^{i(\pi/6 + k\pi/3)}$ for $k = 0, \dots, 5$ , all on the unit circle. Each is simple, since the derivative $6p^5$ of the denominator is not $0$ there. For $R > 1$ the semicircle $C_R$ encloses the three upper poles

\frac{\sqrt{3} + i}{2}, \quad i, \quad \frac{-\sqrt{3} + i}{2}.

Because $p^6 = -1$ , each residue is $1/(6p^5) = -p/6$ . The three poles add up to $2i$ , so the residues add up to $-i/3$ , and the integral around $C_R$ is $2\pi i \cdot (-i/3) = 2\pi/3$ . On the arc $\lvert 1 + z^6 \rvert \ge R^6 - 1$ , so the arc integral is at most $\pi R/(R^6 - 1)$ in absolute value and tends to $0$ . The real integral converges, so the whole-line integral is $2\pi/3$ , and by evenness

\int_0^{\infty} \frac{dx}{1 + x^6} = \frac{\pi}{3},

about $1.0472$ .

Leave the real line, collect the poles, return

Key terms

Analytic function

Contour integration

Cauchy’s theorem

Simple pole and its residue

Residue theorem

Upper half-plane semicircle

Check yourself

How this lesson connects

Builds on