Residues applied to real integrals
There are real integrals with no elementary antiderivative that come out in two lines if you take them off the real axis and close them in the complex plane. Among other things, that is where the noise bandwidth of a filter comes from.
01A real integral, solved from the outside
Some real integrals have no elementary antiderivative and yet come out in two lines if you take them off the real axis and close them in the complex plane. The maneuver is always the same:
- The real variable is replaced by a complex one and an arc is added to close the path.
- The residue theorem is applied to the closed contour.
- It is shown —with the ML bound— that the arc contributes zero as its radius grows.
- What remains is the integral being sought.
The third step is the one that needs care, and it is the one the lab lets you see: if the integrand does not decay fast enough, the arc does not vanish and the whole method collapses.
02First type: integrals of rational functions
For \( \int_{-\infty}^{\infty} \dfrac{P(x)}{Q(x)}\,dx \) with \( Q \) having no real roots, the condition is that the denominator wins by at least two degrees:
The lead example is the simplest possible one. With \( f(z) = 1/(1+z^2) \), the only pole in the upper half-plane is at \( z = j \), with residue \( 1/(2j) \):
The appeal shows up when the antiderivative does not exist. With \( 1/(1+x^4) \) no arctangent will do, but the poles in the upper half-plane are \( e^{j\pi/4} \) and \( e^{j3\pi/4} \), and the computation is just as short: the result is \( \pi/\sqrt{2} \).
03The arc that has to fade out
On the left, the contour: the segment of the real axis plus the semicircle of radius \( R \). On the right, how each contribution evolves as \( R \) grows. If the arc fades out, the segment converges to the value of the integral.
04Second type: Fourier integrals
These are the ones that appear when transforming. They have the form \( \int R(x)\,e^{j a x}\,dx \) and the exponential factor changes the analysis of the arc, because \( |e^{jaz}| = e^{-a y} \) decays for \( y \gt 0 \) when \( a \gt 0 \). The result that formalizes this is Jordan’s lemma:
With that, the transform of a Lorentzian pulse comes out in one line:
And taking the real part, \( \int \cos(ax)/(1+x^2)\,dx = \pi e^{-a} \), which is the one in the lab. This family is the one that justifies the title of this topic in the curriculum: they are the Fourier integrals, and the residue method is the standard way of solving them.
05Third type: integrals of sines and cosines
For \( \int_0^{2\pi} R(\cos\theta, \operatorname{sin}\theta)\,d\theta \) the trick is different and more elegant: there is no need to invent an arc, because the path is already closed. Setting \( z = e^{j\theta} \), the interval \( [0, 2\pi] \) traces exactly the unit circle:
Applied to a specific case, \( \int_0^{2\pi} \dfrac{d\theta}{2 + \cos\theta} \), it becomes a contour integral with two poles at \( z = -2 \pm \sqrt{3} \), of which only one falls inside the circle, and the result is \( 2\pi/\sqrt{3} \). No trigonometric substitutions or tables.
\( \int_{-\infty}^{\infty} \dfrac{\operatorname{sin} x}{x}\,dx \) has a problem at \( x = 0 \). The way out is to go around it with a tiny semicircle and take the principal value. A simple pole surrounded by half an arc contributes \( j\pi \) times its residue —half of \( 2\pi j \)— and that is where the value \( \pi \) comes from. That is the result behind the overshoot in the step response of an ideal filter: the Gibbs phenomenon.
06Where it is used in electronics
| Problem | Integral | What it gives |
|---|---|---|
| Inverse transform | \( \frac{1}{2\pi j}\int F(s)e^{st}ds \) | The time response, pole by pole |
| Noise bandwidth | \( \int_0^{\infty} |H(j\omega)|^2 d\omega \) | The total noise at the output of a filter |
| Signal energy | \( \int |X(j\omega)|^2 d\omega \) | Parseval, from the frequency side |
| Kramers-Kronig | Principal value over the axis | The real part of a response from the imaginary part |
The second row is the most everyday one. The equivalent noise bandwidth of a first-order low-pass filter comes from integrating \( 1/(1 + (\omega/\omega_c)^2) \) from zero to infinity, which is half of the integral of the first example on this page. The result, \( B_n = \frac{\pi}{2}f_c \), says that a real filter lets through 57 % more noise than an ideal one of the same width, and it is the number used to compute the noise of an amplifier.
07In the lab
Solve \( \int_{-\infty}^{\infty} \dfrac{dx}{1+x^4} \) by locating the four poles, keeping the two in the upper half-plane and adding their residues. Check that it gives \( \pi/\sqrt{2} \approx 2.221 \) and compare it with a numerical integration.
Build an RC low-pass filter with \( f_c = 1 \) kHz, inject white noise and measure the RMS voltage at the output. Compute the equivalent bandwidth from the measurement and compare it with \( \frac{\pi}{2}f_c = 1571 \) Hz.
Solve \( \int_0^{2\pi} \dfrac{d\theta}{a + \cos\theta} \) for \( a \gt 1 \) with the substitution \( z = e^{j\theta} \). Check that the result, \( 2\pi/\sqrt{a^2-1} \), tends to infinity when \( a \to 1 \), and explain what happens to the pole.
08Common mistakes
- Closing the contour without checking the arc. It is the step that makes everything else valid; the lab shows a case where it fails.
- Adding up the residues of all four poles. When you close above, only those in the upper half-plane count.
- Closing above with \( a \lt 0 \). In Fourier integrals the sign of the exponent decides the half-plane.
- Ignoring poles on the real axis. The contour has to be indented and the principal value kept.
- Forgetting the \( 1/(jz) \) when substituting \( d\theta \) in trigonometric integrals.
09Self-assessment
What condition on the degrees does the first type require?
That the denominator exceed the numerator by at least two degrees, so that the arc contributes zero.
What is \( \int_{-\infty}^{\infty} dx/(1+x^2) \) and why?
\( \pi \). The only pole above is \( j \), with residue \( 1/(2j) \), and \( 2\pi j \cdot 1/(2j) = \pi \).
What does Jordan’s lemma add?
It allows the contour to be closed for integrands that decay more slowly, as long as there is a factor \( e^{jaz} \) with the right sign.
How is \( \int_0^{2\pi} \) turned into a contour integral?
With \( z = e^{j\theta} \): the interval traces the unit circle, the sines and cosines are written in terms of \( z \) and \( d\theta = dz/(jz) \).
What does a simple pole surrounded by half an arc contribute?
\( j\pi \) times its residue: half of what a full contour would contribute.
What is the noise bandwidth of a first-order RC?
\( B_n = \frac{\pi}{2}f_c \approx 1.57\,f_c \). It is the integral of \( |H|^2 \), which comes out by residues.
10Further reading
- Ruel V. Churchill and James W. Brown. Complex Variables and Applications. 7th ed. (Spanish edition, “Variable compleja y aplicaciones”), McGraw-Hill, 2004. The chapter on improper integrals, with Jordan’s lemma and the principal value.
- Murray R. Spiegel. Complex Variables. Schaum’s Outline Series (Spanish edition, “Variable compleja”), McGraw-Hill, 2011. The three types with dozens of worked examples, classified by the form of the integrand.
- C. D. Motchenbacher and J. A. Connelly. Low-Noise Electronic System Design. Wiley, 1993. The equivalent noise bandwidth computed for each type of filter, with the integrals worked out.