Residues and real-integral applications

Unless otherwise stated, circles and simple boundaries are positively oriented.

1. The one coefficient that survives integration

For an isolated singularity $a$, the residue $\operatorname{Res}(f,a)$ is the coefficient $c_{-1}$ in its Laurent expansion on a punctured neighbourhood. Termwise circle integration gives

$$\int_{|z-a|=r}f(z)\,dz=2\pi i\operatorname{Res}(f,a)$$

for sufficiently small positive $r$. Other negative powers integrate to zero. Thus a pole of order two can have zero residue, as $1/(z-a)^2$ does; residue is not pole order.

Residue theorem, region version. Let $\Omega$ be a bounded region with a positively oriented piecewise smooth simple outer boundary, or finitely many such boundary components with holes oriented clockwise. Suppose $f$ is holomorphic on a neighbourhood of $\overline\Omega$ except at finitely many isolated singularities $a_1,\ldots,a_m$ in $\Omega$, with none on its boundary. Then

$$\int_{\partial\Omega}f(z)\,dz=2\pi i\sum_{j=1}^m\operatorname{Res}(f,a_j).$$

For the boundary-cancellation proof, remove small disjoint disks around the singularities. On the remaining compact region $f$ is holomorphic on a neighbourhood. Cut it into simply connected pieces, or use a finite sufficiently fine subdivision with local primitives; internal edges and cut sides cancel. The boundary total is the original boundary minus the positively oriented small circles, and equals zero. Each small-circle integral is the residue contribution. This proves the theorem without assuming poles rather than essential singularities.

A more general closed-curve version uses winding numbers as multiplicities. It requires the appropriate cycle to bound within the holomorphy domain. We do not replace those hypotheses by the vague phrase “sum all poles inside” for arbitrary self-intersecting contours.

2. Efficient residue calculations

For a pole of order at most $m$, let $h(z)=(z-a)^mf(z)$ extend holomorphically near $a$. Its Taylor expansion gives

$$\operatorname{Res}(f,a)=\frac{h^{(m-1)}(a)}{(m-1)!}.$$

The formula remains valid if cancellations make the actual order smaller. For a simple pole, it reduces to $\lim_{z\to a}(z-a)f(z)$.

If $f=g/h$ with $g,h$ holomorphic, $h(a)=0$ and $h'(a)\ne0$, then

$$\operatorname{Res}(g/h,a)=g(a)/h'(a).$$

Factor $h(z)=(z-a)h_1(z)$ with $h_1(a)=h'(a)$ to prove it. If $g(a)=0$ the singularity is removable and the residue is zero; checking cancellation first may be simplest.

Worked examples. At $z=2$, $e^z/(z-2)^3$ has residue $(e^z)''|_{z=2}/2=e^2/2$. At zero, $e^{1/z}$ has residue one, although it has no finite-order pole. For $1/(z^4+1)$, every root $a$ of the denominator is simple and its residue is $1/(4a^3)=-a/4$, because $a^4=-1$.

3. Rational real integrals: locate poles, estimate the arc, then take limits

Consider

$$I=\int_{-\infty}^{\infty}\frac{dx}{x^4+1}.$$

The real integral is absolutely convergent. Close $[-R,R]$ with the upper semicircle, taking $R>2$. The upper poles are $a=e^{i\pi/4}$ and $b=e^{3i\pi/4}$. Their sum is $i\sqrt2$, so the residue sum is $-i\sqrt2/4$. On the semicircle,

$$|z^4+1|\ge R^4-1,\qquad \left|\int_{\text{arc}}\frac{dz}{z^4+1}\right|\le\frac{\pi R}{R^4-1}\to0.$$

Consequently

$$I=2\pi i\left(-\frac{i\sqrt2}{4}\right)=\frac\pi{\sqrt2}.$$

Evenness gives $\int_0^\infty dx/(x^4+1)=\pi/(2\sqrt2)$.

For a rational function without real poles and with denominator degree at least two more than numerator degree, a similar large-semicircle estimate usually vanishes. One must still check actual decay after cancellation, the finite poles enclosed, and absolute convergence on the real axis. A real pole requires indentation or a principal-value statement, not direct use of this recipe.

4. Exponential factors and a Fourier-type integral

Let $a>0$ and $b\ge0$. On the upper half-plane, $|e^{ibz}|=e^{-b\operatorname{Im}z}\le1$. For $R>2a$,

$$\left|\int_{\text{upper arc}}\frac{e^{ibz}}{z^2+a^2}\,dz\right| \le\frac{\pi R}{R^2-a^2}\to0.$$

The only enclosed pole is $ia$, with residue $e^{-ab}/(2ia)$. Therefore

$$\int_{-\infty}^{\infty}\frac{e^{ibx}}{x^2+a^2}\,dx=\frac\pi a e^{-ab}.$$

The imaginary part vanishes by oddness; the real part gives the cosine integral. For $b<0$, close in the lower half-plane, whose contour is clockwise, or use the real integral's symmetry in $b$. The general answer is $\pi e^{-a|b|}/a$. Choosing the wrong half-plane would make the exponential grow rather than decay.

This example needs no separate Jordan lemma because the rational factor already decays as $R^{-2}$. It is important not to cite a stronger asymptotic tool when the elementary bound suffices.

5. Trigonometric integrals become rational contour integrals

Set $z=e^{it}$, so $dt=dz/(iz)$ and $\cos t=(z+z^{-1})/2$. For real $a>|b|$ and $b\ne0$,

$$\int_0^{2\pi}\frac{dt}{a+b\cos t} =\int_{|z|=1}\frac{2\,dz}{i(bz^2+2az+b)}.$$

Write $d=\sqrt{a^2-b^2}>0$. The roots are $r_\pm=(-a\pm d)/b$. Their product is one, and

$$|r_+|=\frac{|b|}{a+d}<1,\qquad |r_-|>1.$$

The residue at $r_+$ is $2/[i(2br_++2a)]=1/(id)$. The result is

$$\int_0^{2\pi}\frac{dt}{a+b\cos t}=\frac{2\pi}{\sqrt{a^2-b^2}}.$$

For $b=0$, the integral is $2\pi/a$, agreeing with the formula. The condition $a>|b|$ keeps the real denominator strictly positive and the contour poles off the unit circle. Other parameter regimes require a different analysis.

6. Principal value is not the same as an ordinary improper integral

For a real singularity at zero, the symmetric Cauchy principal value is

$$\operatorname{PV}\int_{-\infty}^{\infty}f(x)dx =\lim_{R\to\infty}\lim_{\varepsilon\downarrow0} \left(\int_{-R}^{-\varepsilon}f(x)dx+\int_\varepsilon^Rf(x)dx\right),$$

when these indicated limits exist. Other simultaneous limit conventions must be stated explicitly. The ordinary improper integral requires separate one-sided convergence at every singular endpoint and at each infinity. Symmetric cancellation can exist when those separate limits do not.

Worked indentation: $e^{iz}/z$. Use the upper semicircle, with a small upper semicircle around zero traversed clockwise from $-\varepsilon$ to $\varepsilon$. The enclosed region contains no pole. The small arc tends to $-i\pi$, since $e^{iz}=1+O(\varepsilon)$ there and $\int dz/z=-i\pi$.

The large arc does vanish, but the crude estimate $|e^{iz}|\le1$ gives only $\pi$, not a vanishing bound. Parameterization gives instead

$$\left|\int_{\text{large arc}}\frac{e^{iz}}z\,dz\right| \le\int_0^\pi e^{-R\sin t}dt \le2\int_0^{\pi/2}e^{-2Rt/\pi}dt\le\pi/R.$$

Here $\sin t\ge2t/\pi$ on $[0,\pi/2]$, and symmetry handles the other half. The contour equation yields

$$\operatorname{PV}\int_{-\infty}^{\infty}\frac{e^{ix}}x\,dx=i\pi.$$

Its real part, $\cos x/x$, has no ordinary integral across zero; its symmetric principal value is zero. The imaginary part $\sin x/x$ extends continuously at zero and has convergent tails: integration by parts bounds $|\int_A^B\sin x/x\,dx|$ by $3/A$ for $B>A>0$. Hence its ordinary improper integral exists and

$$\int_0^\infty\frac{\sin x}{x}\,dx=\pi/2.$$

The continuous extension at zero and tail convergence, not the principal-value notation by itself, justify this final ordinary integral.

7. Exercises

P01 - Foundation. Find residues of $z/(z^2+1)$ at $i$ and $-i$.

P02 - Core. Find the residue at zero of $\sin z/z^4$.

P03 - Core. Evaluate $\int_{|z|=2}e^z/[z(z-1)]\,dz$ counterclockwise.

P04 - Core. Use a semicircle to evaluate $\int_{-\infty}^{\infty}dx/(x^2+4)$, including its arc estimate.

P05 - Core. Evaluate $\int_0^{2\pi}dt/(3+2\cos t)$ and verify that its denominator never vanishes.

P06 - Diagnosis. Why does a double pole not automatically contribute a nonzero contour integral? Give an example.

P07 - Proof. If $f$ is holomorphic near $a$, prove that the residue of its derivative at $a$ is zero. Extend the assertion to a derivative of a function holomorphic on a punctured disk.

P08 - Extension. For $b>0$, evaluate $\int_{-\infty}^{\infty}\cos(bx)/(x^2+9)\,dx$ and explain the permitted half-plane.

P09 - Diagnosis. Explain why $\operatorname{PV}\int_{-1}^1dx/x=0$ does not prove that the ordinary improper integral exists.

8. Full solutions

P01. At either simple zero $a$ of $z^2+1$, the ratio formula gives $a/(2a)=1/2$.

P02. $\sin z/z^4=z^{-3}-\tfrac16z^{-1}+\cdots$, so the residue is $-1/6$. Its pole order is three.

P03. The enclosed simple poles are zero and one. Their residues are $-1$ and $e$. The integral is $2\pi i(e-1)$.

P04. The upper pole is $2i$ with residue $1/(4i)$. On the upper arc $|z^2+4|\ge R^2-4$, giving an arc bound $\pi R/(R^2-4)\to0$. Thus the integral is $2\pi i/(4i)=\pi/2$. The real integrand is continuous and behaves as $x^{-2}$, proving ordinary absolute convergence.

P05. The denominator is at least one, since $-1\le\cos t\le1$. The formula gives $2\pi/\sqrt{9-4}=2\pi/\sqrt5$.

P06. Integration detects the $z^{-1}$ coefficient, not every negative coefficient. The function $1/(z-a)^2$ has a double pole and residue zero; its integral on a small positively oriented circle is zero. It even has primitive $-1/(z-a)$ on the punctured plane.

P07. If $f=\sum c_n(z-a)^n$ is a Laurent series, its derivative is $\sum n c_n(z-a)^{n-1}$ on compact subannuli. A $(z-a)^{-1}$ term would require $n=0$, but its coefficient is $0c_0=0$. This proves the punctured-disk assertion and includes the holomorphic case.

P08. Use $e^{ibz}$ in the upper half-plane, where its modulus is at most one. The upper pole is $3i$, residue $e^{-3b}/(6i)$, and the arc estimate is $\pi R/(R^2-9)\to0$. Taking real parts yields $\pi e^{-3b}/3$. Closing in the lower half-plane without changing the exponential would introduce growth and invalidate this estimate.

P09. The symmetric truncated sum cancels exactly. But $\int_\varepsilon^1dx/x=-\log\varepsilon\to+\infty$ and $\int_{-1}^{-\varepsilon}dx/x=\log\varepsilon\to-\infty$. Separate finite one-sided limits fail; an undefined sum of infinities is not an ordinary integral.

9. Summary

Residues compress a contour integral into local Laurent data. A valid real-integral calculation must also establish which poles are enclosed, their orientation, vanishing auxiliary arcs and the correct real convergence notion. The next chapter applies residues of logarithmic derivatives to count zeros rather than integrate explicitly.