Computing Real Integrals with Residues

In this lesson you’ll learn the four standard contour techniques for real integrals: the semicircle for rational functions, the semicircle with Jordan’s lemma for oscillatory integrands, the unit circle for trigonometric integrals over a full period, and the keyhole for branch cuts.

Every method in this lesson follows the same four steps.

1. Choose a contour that includes the real integral you want as one of its pieces.
2. Apply the residue theorem to get the total around the closed contour.
3. Show the unwanted pieces vanish, usually by the ML inequality.
4. Solve for the piece you wanted.

The art is entirely in step 1. Steps 2 to 4 are mechanical.

For $\int_{-\infty}^\infty \frac{p(x)}{q(x)}dx$ with $q$ having no real zeros and $\deg q \ge \deg p + 2$, use the semicircular contour in the upper half plane:

$$\int_{-\infty}^{\infty}\frac{p}{q}\,dx = 2\pi i\sum_{\text{poles in UHP}}\operatorname{Res}\frac{p}{q}$$

The degree condition is what makes the arc vanish. On $|z|=R$ the integrand is $O(R^{-2})$ while the arc length is $\pi R$, so the ML bound is $O(R^{-1})\to0$. With only one degree of decay the bound would be $O(1)$ and the method fails.

The figure carries out the whole argument for $\int\frac{dx}{1+x^4}$, computing the residues, the numerical value of the real integral, and the ML bound on the arc at two radii. All three agree on $\frac{\pi}{\sqrt2}$.

You could close in the lower half plane instead, picking up those residues with a sign flip from the reversed orientation. The answer is the same, which is a good consistency check.

For $\int_{-\infty}^\infty f(x)\cos(ax)\,dx$ or with $\sin$, replace the trigonometric factor by $e^{iaz}$ and take the appropriate part at the end. The tool that makes the arc vanish is stronger than ML alone:

Jordan’s lemma. If $|f(z)|\to0$ uniformly as $|z|\to\infty$ in the upper half plane, and $a>0$, then

$$\int_{\Gamma_R}f(z)e^{iaz}\,dz \to 0$$

on the semicircular arc $\Gamma_R$ in the upper half plane.

The reason is that $|e^{iaz}| = e^{-a\operatorname{Im}z}$ is exponentially small over most of the upper arc. This buys you a full degree: only $f\to0$ is needed, rather than $O(R^{-2})$ decay.

The sign of $a$ decides the half plane. For $a>0$ close upward, for $a<0$ close downward, because that is where $e^{iaz}$ decays.

For $\int_0^{2\pi}R(\cos\theta,\sin\theta)\,d\theta$, substitute $z=e^{i\theta}$ so the interval becomes the unit circle:

$$\cos\theta = \frac{z+z^{-1}}{2}, \qquad \sin\theta = \frac{z-z^{-1}}{2i}, \qquad d\theta = \frac{dz}{iz}$$

The integral becomes a contour integral of a rational function around $|z|=1$, evaluated by residues at whichever poles land inside. A real trigonometric integral becomes rational algebra, which is usually a large simplification.

Integrals like $\int_0^\infty \frac{x^{a-1}}{1+x}dx$ involve a non-integer power, so the integrand has a branch point at 0. The standard contour is the keyhole: out along one side of the positive real axis, around a large circle, back along the other side, and around a small circle at the origin.

Because the two straight pieces sit on opposite sides of the cut, the multi-valued factor differs between them by $e^{2\pi ia}$, and they do not cancel. Instead they combine to give a multiple of the integral you want. That deliberate use of the branch jump is the whole trick.

If the integrand has a pole on the real axis, the integral only exists as a Cauchy principal value, and the contour is indented with a small semicircle around the pole. A small semicircle over a simple pole contributes exactly $\pi i$ times the residue - half of the full $2\pi i$, as the half-turn suggests. This is how $\int_0^\infty\frac{\sin x}{x}dx = \frac{\pi}{2}$ is evaluated, and the $\frac12$ in the answer comes from precisely that half-loop.

Example 1: Evaluate $\displaystyle\int_{-\infty}^{\infty}\frac{dx}{1+x^2}$.

Solution. Poles at $\pm i$; only $i$ is in the upper half plane. With $q = 1+z^2$ and $q'=2z$,

$$\operatorname{Res}_{z=i} = \frac{1}{2i}$$ $$\int_{-\infty}^\infty\frac{dx}{1+x^2} = 2\pi i\cdot\frac{1}{2i} = \pi$$

✓ Consistent with $\arctan x$ evaluated from $-\infty$ to $\infty$.

The degree condition holds since $\deg q = 2 = \deg p + 2$.

Example 2: Evaluate $\displaystyle\int_{-\infty}^{\infty}\frac{dx}{1+x^4}$.

Solution. The poles are the fourth roots of $-1$, namely $e^{i\pi/4}$, $e^{3i\pi/4}$, $e^{5i\pi/4}$, $e^{7i\pi/4}$. The first two are in the upper half plane.

Since $q = 1+z^4$ has $q'=4z^3$, and $z^4=-1$ at each pole,

$$\operatorname{Res} = \frac{1}{4z^3} = \frac{z}{4z^4} = \frac{z}{-4} = -\frac{z}{4}$$

Summing over the two upper poles:

$$-\frac14\left(e^{i\pi/4}+e^{3i\pi/4}\right) = -\frac14\left(\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i-\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\right) = -\frac{i\sqrt2}{4}$$ $$\int_{-\infty}^{\infty}\frac{dx}{1+x^4} = 2\pi i\left(-\frac{i\sqrt2}{4}\right) = \frac{\pi\sqrt2}{2} = \frac{\pi}{\sqrt2}\approx2.2214$$

∎

Try this with real methods. Partial fractions over $\mathbb{R}$ requires factoring $1+x^4$ into two irreducible quadratics with irrational coefficients, and the resulting integrals are unpleasant. The contour method does not care about the degree.

Example 3: Evaluate $\displaystyle\int_{-\infty}^{\infty}\frac{\cos x}{1+x^2}dx$.

Solution. Consider $\frac{e^{iz}}{1+z^2}$ on the upper semicircle. Since $a = 1 > 0$, Jordan’s lemma applies and the arc vanishes.

The only pole in the upper half plane is $z=i$, with

$$\operatorname{Res}_{z=i}\frac{e^{iz}}{1+z^2} = \frac{e^{i\cdot i}}{2i} = \frac{e^{-1}}{2i}$$

So

$$\int_{-\infty}^\infty\frac{e^{ix}}{1+x^2}dx = 2\pi i\cdot\frac{e^{-1}}{2i} = \frac{\pi}{e}$$

The result is real, so taking real parts,

$$\int_{-\infty}^{\infty}\frac{\cos x}{1+x^2}dx = \frac{\pi}{e}\approx1.1557$$

∎

And as a bonus, the imaginary part gives $\int\frac{\sin x}{1+x^2}dx = 0$, which is also obvious by oddness. Two integrals for the price of one is typical of this method.

Example 4: Evaluate $\displaystyle\int_0^{2\pi}\frac{d\theta}{2+\cos\theta}$.

Solution. Substitute $z=e^{i\theta}$, so $\cos\theta = \frac{z+z^{-1}}{2}$ and $d\theta = \frac{dz}{iz}$:

$$\oint_{|z|=1}\frac{1}{2+\frac{z+z^{-1}}{2}}\cdot\frac{dz}{iz} = \oint\frac{2}{4z+z^2+1}\cdot\frac{dz}{i}$$

The denominator $z^2+4z+1$ has roots $z=-2\pm\sqrt3$. Only $-2+\sqrt3\approx-0.268$ is inside the unit circle.

$$\operatorname{Res} = \frac{2}{i}\cdot\frac{1}{2z+4}\bigg|_{z=-2+\sqrt3} = \frac{2}{i}\cdot\frac{1}{2\sqrt3} = \frac{1}{i\sqrt3}$$ $$\int_0^{2\pi}\frac{d\theta}{2+\cos\theta} = 2\pi i\cdot\frac{1}{i\sqrt3} = \frac{2\pi}{\sqrt3}\approx3.6276$$

∎

Checking which root is inside is the crucial step, and it is where most errors in this type occur.

Example 5: Evaluate $\displaystyle\int_0^\infty\frac{\sin x}{x}dx$.

Solution. Consider $\frac{e^{iz}}{z}$ on a contour indented by a small semicircle above the pole at 0. There are no poles enclosed, so the total is 0.

The large arc vanishes by Jordan’s lemma. The small semicircle above a simple pole, traversed clockwise, contributes $-\pi i\operatorname{Res} = -\pi i$.

So the two straight pieces sum to $\pi i$, giving

$$\text{P.V.}\int_{-\infty}^{\infty}\frac{e^{ix}}{x}dx = \pi i$$

Taking imaginary parts, $\int_{-\infty}^\infty\frac{\sin x}{x}dx = \pi$, and by evenness of $\frac{\sin x}{x}$,

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

∎

The $\frac12$ comes from the half-loop around the pole. This integral, the Dirichlet integral, has no elementary antiderivative at all.

Example 6: Evaluate $\displaystyle\int_0^\infty\frac{dx}{1+x^3}$.

Solution. The integrand is not even, so a full-line contour will not work directly. Instead use a sector contour of angle $\frac{2\pi}{3}$, since $e^{2\pi i/3}$ maps the positive real axis to the other ray and multiplies the integrand in a controlled way.

On the returning ray, substituting $z = e^{2\pi i/3}t$ gives $z^3=t^3$, so the integrand is unchanged while $dz$ picks up a factor $e^{2\pi i/3}$. The sector encloses the single pole $z=e^{i\pi/3}$, with residue $\frac{1}{3z^2}$ evaluated there.

Carrying this out gives

$$\int_0^\infty\frac{dx}{1+x^3} = \frac{2\pi}{3\sqrt3}\approx1.2092$$

∎

Match the contour to the symmetry of the integrand. A cube in the denominator suggests a $120°$ sector, just as a fourth power suggests $90°$.

Example 7: A keyhole integral.

Set up $\displaystyle\int_0^\infty\frac{x^{-1/2}}{1+x}dx$.

Solution. The integrand has a branch point at 0 and a pole at $-1$. Use a keyhole around the positive real axis with the branch cut there.

On the lower edge of the cut, $z^{-1/2}$ has picked up a factor $e^{-i\pi}= -1$ relative to the upper edge, so the two straight contributions add rather than cancel, giving $2I$ where $I$ is the wanted integral. The residue at $z=-1$, with the branch chosen so $\arg z\in(0,2\pi)$, is $e^{-i\pi/2} = -i$.

$$2I = 2\pi i(-i) = 2\pi \quad\Longrightarrow\quad I = \pi$$

Verification by real methods: substituting $x=u^2$ turns it into $2\int_0^\infty\frac{du}{1+u^4}$… which is Example 2’s cousin, and indeed gives $\pi$. ✓

Fourier transforms. Most transforms in a table were computed by contour integration, and the transform of $\frac{1}{1+x^2}$ being $\pi e^{-|k|}$ is Example 3 with a parameter. Every filter design that uses a closed-form transform relies on this.

Inverse Laplace transforms. The Bromwich contour integral is evaluated by residues, with each pole contributing an exponential mode. This is the mathematics behind step and impulse responses in control theory.

Diffraction and scattering. The Sommerfeld theory of diffraction, and optical propagation integrals, are evaluated by steepest-descent and residue methods along contours in the complex plane.

Dispersion relations. The Kramers–Kronig relations connect the real and imaginary parts of a material’s response, and they are derived by contour integration with a principal value exactly like Example 5. They are used to check the consistency of measured optical data.

Statistical mechanics. Partition functions and their asymptotics are extracted by contour integration around singularities, which is how phase transitions show up as singularity locations.

Signal processing. The $\operatorname{sinc}$ function’s integral, Example 5, is the reason ideal low-pass filters have the impulse response they do, and its slow decay is why real filters must be windowed.