Integration by the residue theorem¶

Method -> "Residue" evaluates the classical families of improper and periodic real integrals that complex analysis dispatches by summing residues. The idea is Cauchy's: close the real integration path into a contour in the complex plane, and the integral equals \(2\pi i\) times the sum of the residues the contour encloses.

In[1]:= Integrate[1/(x^2 + 1), {x, -Infinity, Infinity}, Method -> "Residue"]
Out[1]= Pi

The integrand \(1/(x^2+1)\) has poles at \(x = \pm i\); closing into the upper half-plane encircles the single pole \(x = i\), whose residue is \(1/(2i)\), and \(2\pi i \cdot \tfrac{1}{2i} = \pi\).

This method is the engine behind Mathilda's IntegrateContourResiduebuiltin. It runs in the automatic cascade (just before the Newton–Leibniz antiderivative path); pinningMethod -> "Residue"` forces it and skips the rest.

1. The mathematical idea¶

The residue theorem¶

If \(f\) is meromorphic inside and on a positively-oriented simple closed contour \(C\), with no poles on \(C\), then

\[ \oint_C f(z)\,dz \;=\; 2\pi i \sum_k \operatorname{Res}(f, z_k), \]

the sum running over the poles \(z_k\) enclosed by \(C\). The residue at a pole \(z_0\) is the coefficient of \((z - z_0)^{-1}\) in the Laurent expansion of \(f\) about \(z_0\); Mathilda computes it symbolically with the same primitive that backs the Residue builtin.

To turn a real integral over the line or a period into a contour integral, you add an arc (or another edge) that closes the path, then argue that the added piece contributes nothing — or a known amount — in the limit. Which contour you add is what distinguishes the families below.

Jordan's lemma and closing the contour¶

For a rational integrand that decays like \(1/z^2\) or faster, a large semicircle of radius \(R\) in the upper half-plane contributes \(O(1/R) \to 0\), so the line integral is \(2\pi i\) times the residues in the upper half-plane.

For a Fourier integrand \(R(x)\, e^{i a x}\) with \(a > 0\), Jordan's lemma guarantees the upper semicircle vanishes even when \(R\) decays only like \(1/z\), because \(|e^{i a z}| = e^{-a\,\operatorname{Im} z}\) is exponentially small in the upper half-plane. This is what makes oscillatory integrals like \(\int \cos(x)/(x^2+1)\,dx\) tractable.

Other integrands call for other contours: a wedge of angle \(2\pi/n\) for \(1/(1+x^n)\), a keyhole wrapping the branch cut of \(x^p\) along \([0,\infty)\), or a rectangle of height \(2\pi i\) for integrands periodic in the imaginary direction.

References¶

E. T. Whittaker, G. N. Watson, A Course of Modern Analysis, 4th ed., Cambridge, 1927 — Ch. VI, the calculus of residues.
L. V. Ahlfors, Complex Analysis, 3rd ed., McGraw-Hill, 1979 — Ch. 4–5 (residue theorem, evaluation of definite integrals).
G. F. Carrier, M. Krook, C. E. Pearson, Functions of a Complex Variable: Theory and Technique, McGraw-Hill, 1966 — keyhole and sector contours.
M. J. Ablowitz, A. S. Fokas, Complex Variables: Introduction and Applications, 2nd ed., Cambridge, 2003 — Ch. 4, Jordan's lemma.
NIST Digital Library of Mathematical Functions, §1.10(iv) (the residue theorem), dlmf.nist.gov/1.10.

2. How Mathilda realises it¶

The method is a set of narrow, conjunctive recognizers — one per standard contour. Each recognizer (i) checks the integration bounds, (ii) reduces the integrand to a rational function whose poles it finds with Solve, (iii) classifies each pole against the family's contour (upper half-plane, unit disk, or real axis) by the numeric sign of \(\operatorname{Im}\) or the value of \(|\cdot|\), (iv) sums the enclosed residues, and (v) closes the sum to a scalar.

Because the residue theorem makes each family's value correct by construction once its structural gates hold, there is no numerical-quadrature crosscheck anywhere in the method. The only post-hoc gate is a self-consistency check:

the closed form must reduce to a scalar (no surviving x or Root), and
for the real-valued families, its imaginary part must vanish — a residual \(\operatorname{Im}\) betrays a mis-classified pole.

Any failure returns unevaluated, and the automatic cascade's Newton–Leibniz path takes over; no existing result is weakened.

Symbolic parameters. When you pass Assumptions -> ..., the recognizers classify parameter-dependent poles and frequencies by reading their sign at a single generic point of the region the assumptions pin, while the residue arithmetic stays symbolic — so \(\int \cos(kx)/(x^2+a^2)\,dx\) comes out as a function of \(a\) and \(k\). A parameter the assumptions leave two-sided unbounded is refused, since its sign is undetermined.

3. Invoking the method¶

Pin the method with Method -> "Residue". The integral must be over the whole line {x, -Infinity, Infinity}, a half-line {x, 0, Infinity}, or a full period such as {x, 0, 2 Pi} — these are the ranges the standard contours close.

A finite, generic range has no closing contour, so the method declines and returns the integral unevaluated (in the automatic cascade, an antiderivative method would handle it instead):

In[1]:= Integrate[1/(x^2 + 1), {x, 0, 5}, Method -> "Residue"]
Out[1]= Integrate[1/(1 + x^2), {x, 0, 5}, Method -> "Residue"]

Supply Assumptions to pin the signs of any free parameters — see §5.

4. Introductory examples¶

4.1 Rational integrands (Family A)¶

A rational function decaying at least as fast as \(1/z^2\) closes into the upper half-plane; the value is \(2\pi i\) times the residues there.

In[1]:= Integrate[1/(x^4 + 1), {x, -Infinity, Infinity}, Method -> "Residue"]
Out[1]= Pi/Sqrt[2]

In[2]:= Integrate[1/(x^6 + 1), {x, -Infinity, Infinity}, Method -> "Residue"]
Out[2]= 2/3 Pi

When the integrand is even, the half-line integral is exactly half the full-line value, and the method uses that shortcut:

In[1]:= Integrate[1/(x^2 + 1), {x, 0, Infinity}, Method -> "Residue"]
Out[1]= 1/2 Pi

In[2]:= Integrate[1/(x^4 + 1), {x, 0, Infinity}, Method -> "Residue"]
Out[2]= (1/2 Pi)/Sqrt[2]

4.2 Fourier / Jordan integrands (Family B)¶

For \(R(x)\) rational times \(\cos\), \(\sin\), or \(e^{i\,\cdot}\) with a positive frequency, Jordan's lemma closes the contour upward. Mathilda evaluates the complex integral \(J = 2\pi i \sum_{\text{UHP}} \operatorname{Res}[R(z)\,e^{i a z}]\) and takes \(\operatorname{Re}[J]\) for a cosine kernel, \(\operatorname{Im}[J]\) for a sine kernel:

In[1]:= Integrate[Cos[x]/(x^2 + 1), {x, -Infinity, Infinity}, Method -> "Residue"]
Out[1]= Pi/E

In[2]:= Integrate[x Sin[x]/(x^2 + 1), {x, -Infinity, Infinity}, Method -> "Residue"]
Out[2]= Pi/E

4.3 Trigonometric integrands over a period (Family C)¶

A function rational in \(\sin x\) and \(\cos x\), integrated over a full period, maps under \(z = e^{i x}\) to a rational contour integral around the unit circle; the value is \(2\pi i\) times the residues inside the disk.

In[1]:= Integrate[1/(2 + Cos[x]), {x, 0, 2 Pi}, Method -> "Residue"]
Out[1]= (2 Pi)/Sqrt[3]

In[2]:= Integrate[1/(5 - 4 Cos[x]), {x, 0, 2 Pi}, Method -> "Residue"]
Out[2]= 2/3 Pi

5. Worked examples¶

The set below exercises every contour the method knows — the rational and Fourier families, a principal-value indentation, the rectangular strip, the sector wedge (with a symbolic exponent), and the keyhole branch cut. Each transcript is reproduced from the current build.

Rational, upper half-plane. The simplest residue integral: one pole at \(x = i\).

In[1]:= Integrate[1/(x^2 + 1), {x, -Infinity, Infinity}, Method -> "Residue"]
Out[1]= Pi

Rational, four poles. The two upper-half-plane roots of \(x^4 = -1\) contribute.

In[1]:= Integrate[1/(x^4 + 1), {x, -Infinity, Infinity}, Method -> "Residue"]
Out[1]= Pi/Sqrt[2]

Fourier integrand with parameters. With \(a > 0\) and \(k > 0\) the contour closes upward; the single pole at \(x = i a\) gives the exponential decay \(e^{-a k}\) — a function of both parameters, produced by symbolic residue arithmetic:

In[1]:= Integrate[Cos[k*x]/(x^2 + a^2), {x, -Infinity, Infinity},
          Assumptions -> {a > 0, k > 0}, Method -> "Residue"]
Out[1]= (Pi E^(-a k))/a

Principal value — a removable pole on the axis. Sin[x]/x has a removable singularity at \(x = 0\); the indented contour skirts it with a half-circle, so the pole contributes a half residue \(\pi i \operatorname{Res}\). The result is Dirichlet's integral:

In[1]:= Integrate[Sin[x]/x, {x, -Infinity, Infinity}, Method -> "Residue"]
Out[1]= Pi

Rectangular strip — quasi-periodic integrand. \(e^{a x}/(e^x + 1)\) is periodic in the imaginary direction with period \(2\pi i\). A rectangle of that height reduces (via \(w = e^x\)) to a single enclosed pole at \(w = -1\), and the two horizontal edges differ by the constant factor \(1 - e^{2\pi i a}\). With \(0 < a < 1\) the value is the reflection formula:

In[1]:= Integrate[Exp[a*x]/(Exp[x] + 1), {x, -Infinity, Infinity},
          Assumptions -> 0 < a < 1, Method -> "Residue"]
Out[1]= Pi Csc[Pi a]

Sector wedge — a symbolic exponent. \(1/(1+x^n)\) on the half-line closes into a wedge of angle \(2\pi/n\). This is the one family that admits a symbolic exponent \(n\): with the convergence assumption \(n > 1\) the value comes out in closed form in \(n\):

In[1]:= Integrate[1/(1 + x^n), {x, 0, Infinity}, Assumptions -> n > 1, Method -> "Residue"]
Out[1]= (Pi Csc[Pi/n])/n

Keyhole — a branch-power integrand. \(x^{1/3}/(x^2+1)\) carries the branch cut of \(x^{1/3}\) along \([0,\infty)\). A keyhole contour wraps that cut; the jump across it lands the value on a cosecant (here evaluating to a surd):

In[1]:= Integrate[x^(1/3)/(x^2 + 1), {x, 0, Infinity}, Method -> "Residue"]
Out[1]= Pi/Sqrt[3]

The keyhole and sector families share one residue engine, \(\int_0^\infty x^p R(x)\,dx\) reducing in both cases to a Mellin-type sum over the poles of \(R\) weighted by the branch value \(x^{s-1}\); two more half-line branch-power integrals:

In[1]:= Integrate[Sqrt[x]/(x^2 + 1), {x, 0, Infinity}, Method -> "Residue"]
Out[1]= Pi/Sqrt[2]

In[2]:= Integrate[1/(1 + x^3), {x, 0, Infinity}, Method -> "Residue"]
Out[2]= (2/3 Pi)/Sqrt[3]

6. Convergence, principal values, and divergence¶

Genuine axis poles diverge. A pure rational integrand with a non-removable pole on the real axis has no convergent improper integral (only a principal value would exist, which plain Integrate does not compute). The method detects this and reports divergence rather than returning a wrong finite number.
Removable axis poles give a principal value. A simple axis pole whose kernel supplies a matching zero — like Sin[x]/x at \(0\) — is admitted as an indented contour and contributes a half residue, as in §5.
The value must close to a real scalar. For the real-valued families the method verifies that the summed residues produce a scalar with vanishing imaginary part. If a pole was mis-classified the check fails and the integral is left for another method — the method never returns an unverified value.

7. The contour table¶

Family	Surface form	Contour	Value
A — rational	\(R(x)\) on \((-\infty, \infty)\)	upper semicircle	\(2\pi i \sum_{\text{UHP}} \operatorname{Res}\)
B — Fourier / Jordan	\(R(x)\,\{\cos,\sin,e^{i\cdot}\}(a x)\) on \((-\infty, \infty)\)	upper semicircle (Jordan)	\(\operatorname{Re}/\operatorname{Im}\) of \(2\pi i \sum_{\text{UHP}} \operatorname{Res}[R\,e^{i a x}]\)
C — trigonometric	rational in \(\sin x, \cos x\) over a period	unit circle, \(z = e^{i x}\)	\(2\pi i \sum \operatorname{Res}\) inside the disk
Principal value	simple removable pole on the axis (A/B)	indented axis	\(+\, \pi i \operatorname{Res}\) at the axis pole
Half-line (even)	even \(R(x)\) on \([0, \infty)\)	—	\(\tfrac12 \times\) the full-line value
Sector	\(x^m / (c + x^n)\) on \([0, \infty)\), \(n\) possibly symbolic	wedge of angle \(2\pi/n\)	\(\tfrac{\pi}{n} c^{s/n-1} \csc(\pi s/n)\), \(s = m{+}1\)
Keyhole / Mellin	\(x^p R(x)\) on \([0, \infty)\), \(p\) non-integer	keyhole around \([0,\infty)\)	Mellin sum over poles of \(R\), \(\times\, \tfrac{1}{1 - e^{2\pi i s}}\)
Rectangular strip	\(e^{c x} R(e^x)\) on \((-\infty, \infty)\)	rectangle, height \(2\pi i\)	reduced to the keyhole core by \(w = e^x\)

All the algebraic families close their residue sums with RootReduce; the Fourier family closes via the \(\operatorname{Re}\)/\(\operatorname{Im}\) of the Jordan integral.

8. Limitations¶

Improper or periodic ranges only. The method needs a closing contour, so a generic finite range is declined (see §3).
Logarithmic integrands are largely out of scope. The keyhole engine carries a single branch power \(x^p\), not the \(\log\)-weighted integrands whose keyhole picks up a \((\log)^2\) or a squared pole. Both of these return unevaluated:

In[1]:= Integrate[Log[x]/(x^2 + 1)^2, {x, 0, Infinity}, Method -> "Residue"]
Out[1]= Integrate[Log[x]/(1 + x^2)^2, {x, 0, Infinity}, Method -> "Residue"]

In[2]:= Integrate[Log[1 + x^2]/(1 + x^2), {x, 0, Infinity}, Method -> "Residue"]
Out[2]= Integrate[Log[1 + x^2]/(1 + x^2), {x, 0, Infinity}, Method -> "Residue"]

The first pairs a logarithm with a double pole; the second has branch points at \(x = \pm i\) on the closing contour. Both are classical residue exercises, but outside the families this method recognises — a natural direction for a future extension.

Undetermined parameter signs are refused. A symbolic parameter that Assumptions leaves two-sided unbounded cannot be classified against the contour, so the family declines rather than guess a sign.