Evaluating real integrals with residues

One of the first great successes of complex analysis, and still one of its most used, is the evaluation of definite real integrals. Integrals such as \(\int_{-\infty}^{\infty}\frac{dx}{1 + x^4}\), \(\int_0^{\infty}\frac{\sin x}{x}\,dx\) or \(\int_0^{2\pi}\frac{d\theta}{2 + \cos\theta}\) have no elementary antiderivative or a very awkward one, yet the residue theorem gives their values in a few lines. The method is always the same: choose a complex function and a closed contour so that part of the contour integral is the real integral you want, compute the contour integral by residues, and show that the remaining parts vanish or are related to the original. This lesson works through the four standard types and checks every answer against numerical quadrature.

Type 1: trigonometric integrals over a full period

For an integral \(\int_0^{2\pi}R(\cos\theta, \sin\theta)\,d\theta\) with \(R\) a rational function, put \(z = e^{i\theta}\). As \(\theta\) runs from 0 to \(2\pi\), \(z\) traverses the unit circle once, and

\[\cos\theta = \frac{1}{2}\Big(z + \frac{1}{z}\Big), \qquad \sin\theta = \frac{1}{2i}\Big(z - \frac{1}{z}\Big), \qquad d\theta = \frac{dz}{iz}.\]

The integral becomes the integral of a rational function of \(z\) round the unit circle, which is \(2\pi i\) times the sum of its residues inside.

Type 2: rational functions on the real line

Let \(f = \frac{p}{q}\) be a rational function with \(q\) having no real zeros and \(\deg q \ge \deg p + 2\). Then \(\int_{-\infty}^{\infty}f(x)\,dx\) converges absolutely. Take the contour consisting of the segment \([-R, R]\) and the semicircle \(S_R\): \(z = Re^{it}\), \(0 \le t \le \pi\), in the upper half-plane. For \(R\) large enough to enclose all poles of \(f\) in the upper half-plane,

\[\int_{-R}^{R}f(x)\,dx + \int_{S_R}f(z)\,dz = 2\pi i\sum_{\operatorname{Im} z_k > 0}\operatorname{Res}(f, z_k).\]

The degree condition gives \(|f(z)| \le \frac{K}{R^2}\) on \(S_R\) for large \(R\), so by the ML inequality \(|\int_{S_R}f\,dz| \le \frac{K}{R^2}\pi R \to 0\). Letting \(R \to \infty\),

\[\int_{-\infty}^{\infty}f(x)\,dx = 2\pi i\sum_{\operatorname{Im} z_k > 0}\operatorname{Res}(f, z_k).\]

Type 3: Fourier integrals and Jordan's lemma

Integrals \(\int_{-\infty}^{\infty}f(x)\cos(ax)\,dx\) and \(\int_{-\infty}^{\infty}f(x)\sin(ax)\,dx\), with \(a > 0\), are the real and imaginary parts of \(\int_{-\infty}^{\infty}f(x)e^{iax}\,dx\). We use \(e^{iaz}\), not \(\cos az\), because \(|e^{iaz}| = e^{-ay}\) is at most 1 in the upper half-plane, whereas \(\cos az\) grows exponentially there. When \(\deg q \ge \deg p + 2\), the semicircle argument above works unchanged. When \(\deg q = \deg p + 1\) the ML bound is not enough, and we need a sharper estimate.

Jordan's lemma. If \(|f(z)| \le M_R\) on \(S_R\) and \(M_R \to 0\), then \(\int_{S_R}f(z)e^{iaz}\,dz \to 0\) for \(a > 0\).

Proof. On \(S_R\), \(|e^{iaz}| = e^{-aR\sin t}\), so the integral is at most \(M_RR\int_0^\pi e^{-aR\sin t}\,dt = 2M_RR\int_0^{\pi/2}e^{-aR\sin t}\,dt\). On \([0, \frac{\pi}{2}]\), concavity gives \(\sin t \ge \frac{2t}{\pi}\), so the last integral is at most \(\int_0^{\infty}e^{-2aRt/\pi}\,dt = \frac{\pi}{2aR}\). The bound is \(\frac{\pi M_R}{a} \to 0\). \(\square\)

Indented contours: the Dirichlet integral

Sometimes the integrand has a pole on the real axis. The classical example is \(\int_0^{\infty}\frac{\sin x}{x}\,dx\), which Euler had evaluated by other means and which now carries Dirichlet's name. Consider \(\frac{e^{iz}}{z}\), which has a simple pole at 0. Use the contour: \([-R, -\varepsilon]\), the small semicircle \(C_\varepsilon\) of radius \(\varepsilon\) above 0 traversed clockwise, \([\varepsilon, R]\), and the large semicircle \(S_R\). No poles are inside, so the total is 0.

By Jordan's lemma the integral over \(S_R\) tends to 0. On the small semicircle, \(\frac{e^{iz}}{z} = \frac{1}{z} + g(z)\) with \(g\) analytic, so the integral of \(g\) tends to 0 (ML), and \(\int\frac{dz}{z}\) over a clockwise half-circle is \(-i\pi\). Hence

\[\lim\Big(\int_{-R}^{-\varepsilon} + \int_{\varepsilon}^{R}\Big)\frac{e^{ix}}{x}\,dx = i\pi.\]

The real part of the integrand, \(\frac{\cos x}{x}\), is odd and cancels; the imaginary part gives \(2\int_0^{\infty}\frac{\sin x}{x}\,dx = \pi\). So

\[\int_0^{\infty}\frac{\sin x}{x}\,dx = \frac{\pi}{2}.\]

In general, a small semicircle round a simple pole contributes \(-i\pi\) times the residue when traversed clockwise, half of what a full circle would give.

The same idea gives Cauchy principal values of integrals that do not converge in the ordinary sense. For \(f\) rational with simple poles \(x_j\) on the real axis, poles \(z_k\) in the upper half-plane, and \(\deg q \ge \deg p + 2\), indenting round each real pole gives

\[\operatorname{PV}\int_{-\infty}^{\infty}f(x)\,dx = 2\pi i\sum_k\operatorname{Res}(f, z_k) + \pi i\sum_j\operatorname{Res}(f, x_j).\]

For example, \(f(z) = \frac{1}{z(z^2 + 1)}\) has a simple pole at 0 with residue 1 and a pole at \(i\) with residue \(\frac{1}{i \cdot 2i} = -\frac{1}{2}\). The formula gives \(2\pi i(-\frac{1}{2}) + \pi i(1) = 0\), which is correct, since the integrand is odd.

Type 4: branch cuts and the keyhole contour

For integrals of the form \(\int_0^{\infty}x^{a-1}f(x)\,dx\) with non-integer \(a\), or with a logarithm, the integrand has a branch point at 0. Cut the plane along the positive real axis, using the branch of \(z^{a-1} = e^{(a-1)\log z}\) with \(0 < \arg z < 2\pi\). The keyhole contour runs along the top of the cut from \(\varepsilon\) to \(R\), round the large circle counterclockwise, back along the bottom of the cut from \(R\) to \(\varepsilon\), and round a small circle clockwise.

Rectangular contours

Integrands involving exponentials are often periodic in the imaginary direction, and then a rectangle replaces the semicircle. Consider \(I = \int_{-\infty}^{\infty}\frac{e^{ax}}{1 + e^x}\,dx\) for \(0 < a < 1\), which converges because the integrand decays like \(e^{ax}\) as \(x \to -\infty\) and like \(e^{(a-1)x}\) as \(x \to \infty\). Take \(f(z) = \frac{e^{az}}{1 + e^z}\) and the rectangle with vertices \(\pm R\) and \(\pm R + 2\pi i\).

On the top edge, \(z = x + 2\pi i\), the denominator is unchanged because \(e^{2\pi i} = 1\), while the numerator gains a factor \(e^{2\pi ia}\). Traversed from right to left, the top edge contributes \(-e^{2\pi ia}\int_{-R}^{R}\frac{e^{ax}}{1 + e^x}\,dx\). On the vertical sides the integrand is bounded by \(\frac{e^{aR}}{e^R - 1}\) or by \(\frac{e^{-aR}}{1 - e^{-R}}\), and the sides have length \(2\pi\), so they vanish as \(R \to \infty\).

Inside the rectangle the only pole is where \(e^z = -1\), at \(z = i\pi\). It is simple, with residue \(\frac{e^{ai\pi}}{e^{i\pi}} = -e^{i\pi a}\). The residue theorem gives \((1 - e^{2\pi ia})I = -2\pi ie^{i\pi a}\), and so \(I = \frac{\pi}{\sin\pi a}\), the same value as the keyhole integral. That is no coincidence: the substitution \(x = \ln t\) turns one integral into the other.

Choosing a method

In practice the shape of the integrand picks the contour. A full period of a trigonometric expression suggests the unit circle. A rational function on the whole line, possibly multiplied by \(e^{iax}\), suggests a large semicircle, in the upper half-plane for \(a > 0\) and the lower for \(a < 0\), where the orientation reverses the sign. An integral over \([0, \infty)\) of an even function can be doubled into one over the line. A fractional power or a logarithm calls for a keyhole round the branch cut. And an exponential that is periodic in \(y\) calls for a rectangle. In every case, the only real work beyond computing residues is to show that the unwanted pieces vanish, and the ML inequality and Jordan's lemma handle almost every situation.

import cmath, math

def simpson(g, a, b, n=20000):
    h = (b - a) / n
    s = g(a) + g(b)
    for k in range(1, n):
        s += (4 if k % 2 else 2) * g(a + k*h)
    return s * h / 3

# Type 1: map [0, 2*pi]; the integrand is smooth and periodic.
num = simpson(lambda t: 1/(2 + math.cos(t)), 0, 2*math.pi)
print(f"int 1/(2+cos t):      quadrature {num:.10f}   residues {2*math.pi/math.sqrt(3):.10f}")

# Types 2 and 3: map the real line to (-pi/2, pi/2) by x = tan u, dx = sec^2 u du.
def real_line(f, n=20000):
    e = 1e-9
    return simpson(lambda u: f(math.tan(u)) / math.cos(u)**2, -math.pi/2 + e, math.pi/2 - e, n)
print(f"int 1/(1+x^4):        quadrature {real_line(lambda x: 1/(1 + x**4)):.10f}   residues {math.pi/math.sqrt(2):.10f}")
print(f"int cos x/(x^2+1):    quadrature {simpson(lambda x: math.cos(x)/(x*x + 1), -200*math.pi, 200*math.pi, 400000):.8f}     residues {math.pi/math.e:.8f}")

# Dirichlet integral: integrate sin(x)/x over [0, N*pi] and average two partial results.
si = lambda x: math.sin(x)/x if x else 1.0
a = simpson(si, 0, 2000*math.pi, 400000); b = simpson(si, 0, 2001*math.pi, 400000)
print(f"int_0^inf sin x / x:  quadrature {(a + b)/2:.6f}       residues {math.pi/2:.6f}")

# Keyhole example with a = 1/2: substitute x = t^2 to remove the singularity at 0.
val = real_line(lambda t: 2/(1 + t*t)) / 2          # integral over t > 0 of 2/(1+t^2)
print(f"int x^(-1/2)/(1+x):   quadrature {val:.8f}     residues {math.pi/math.sin(math.pi/2):.8f}")

# The residue side computed with cmath, for 1/(1+x^4).
poles = [cmath.exp(1j*math.pi*(2*k + 1)/4) for k in range(4)]
upper = [p for p in poles if p.imag > 0]
print("2*pi*i * sum of residues:", 2j*math.pi*sum(1/(4*p**3) for p in upper))

int 1/(2+cos t):      quadrature 3.6275987285   residues 3.6275987285
int 1/(1+x^4):        quadrature 2.2214414691   residues 2.2214414691
int cos x/(x^2+1):    quadrature 1.15572733     residues 1.15572735
int_0^inf sin x / x:  quadrature 1.570796       residues 1.570796
int x^(-1/2)/(1+x):   quadrature 3.14159265     residues 3.14159265
2*pi*i * sum of residues: (2.221441469079183+5.231802747012948e-16j)

Common mistakes