At z = 0 the function e^{1/z} has:

A pole of infinite order that is still a pole

A Gaussian pulse e^{−ax²} is made narrower by increasing a. Its Fourier transform:

Stays Gaussian and becomes wider in k

Stays Gaussian and becomes narrower in k

A square wave is expanded in a Fourier series. Which statements are correct?

At a jump the full series converges to the mid-point of the jump

The overshoot of a truncated series near a jump does not shrink to zero as more terms are added

Parseval’s theorem gives Σ 1/(2k − 1)² = π²/8

At a jump the full series converges to the mid-point of the jump

The overshoot of a truncated series near a jump does not shrink to zero as more terms are added

Parseval’s theorem gives Σ 1/(2k − 1)² = π²/8

The coefficients fall off as 1/n², because the function is bounded

Mathematical Physics II: Singularities, Residue Integrals, Fourier Analysis and the Delta Function

The second chapter for Section 2 of the GATE Physics paper. The shared Complex Variables chapter teaches the Cauchy–Riemann conditions, Cauchy’s theorem and formula, Laurent series and the residue theorem; the shared Transforms chapter teaches Fourier series by symmetry and the Fourier transform. Section 2 names “singularities, residue theorem and applications” and “Fourier analysis”, and a physics paper asks both more deeply than an engineering one. So this chapter classifies singularities — removable, poles, essential and branch points — and then does what “applications” means: it evaluates real integrals by contour integration, rational, trigonometric and Fourier-type with Jordan’s lemma. It closes with the Fourier analysis physics uses: the complex series and Parseval’s theorem, the transform of a Gaussian, the width–bandwidth product, convolution, and the Dirac delta function, which the quantum chapters meet as a potential.

1. Classifying singularities

A point z₀ where f fails to be analytic is an isolated singularity if f is analytic in a punctured disc around it. The Laurent series about z₀ then classifies it by its negative powers. If there are none, the singularity is removable — sin z/z at z = 0, whose series 1 − z²/6 + … is finite once f(0) is defined as 1. If the negative powers stop at (z − z₀)^−m, it is a pole of order m. If they never stop, it is an essential singularity — e^1/z = Σ z⁻ⁿ/n! at z = 0 — near which f takes almost every complex value.

The four kinds, and how to recognise them
Kind	Example at z = 0	Test
Removable	sin z / z	lim f(z) exists and is finite; residue 0
Pole of order m	1/z³, cos z / z²	(z − z₀)^m f → finite non-zero
Essential	e^1/z, sin(1/z)	infinitely many negative powers
Branch point (not isolated)	√z, ln z	f is multivalued around it; needs a branch cut

The residue at an essential singularity is still the coefficient of z⁻¹, read from the series. For z²e^1/z = z² Σ z⁻ⁿ/n!, the z⁻¹ term comes from n = 3, so the residue is 1/3! = 1/6. A point such as z = 0 for 1/sin(1/z) is not isolated at all: poles at z = 1/(nπ) crowd into it, and no Laurent series about it exists.

2. Real integrals by residues

Rational functions on the whole line. If P/Q has no real poles and deg Q ≥ deg P + 2, close the contour with a large semicircle in the upper half-plane; the arc contributes nothing, and ∫_−∞^∞ P/Q dx = 2πi Σ (residues in the upper half-plane). Thus ∫dx/(1 + x²) = 2πi · 1/(2i) = π, and ∫dx/(1 + x²)² = 2πi · (−i/4) = π/2 (a double pole at z = i).
Trigonometric integrals over a period. Put z = e^iθ: cos θ = (z + 1/z)/2, dθ = dz/(iz), and the integral becomes a contour integral round |z| = 1. The standard result is ∫₀^2π dθ/(a + b cos θ) = 2π/√(a² − b²) for a > |b|, so with a = 5, b = 4 it is 2π/3.
Fourier-type integrals and Jordan’s lemma. For ∫ f(x)e^ikx dx with k > 0 and f → 0 at infinity, the upper arc vanishes even if f falls only as 1/|z| (Jordan’s lemma). So ∫_−∞^∞ cos kx/(x² + a²) dx = Re[2πi · e^−ka/(2ia)] = (π/a)e^−ka, which is π/e ≈ 1.156 for k = a = 1.

⚠️ Close the contour on the side where the exponential decays

For e^ikx with k > 0, e^ikz = e^ikxe^−ky decays only for y > 0, so the contour closes upward and the poles counted are those with Im z > 0. With k < 0 it must close downward, picking up a minus sign from the clockwise sense. Taking cos kx itself onto the arc fails outright, because cos(kz) grows exponentially in both half-planes — write it as the real part of e^ikz first.

3. Fourier series in complex form, and Parseval

On (−π, π) a periodic f is f(x) = Σₙ cₙe^inx, with cₙ = (1/2π)∫ f(x)e^−inx dx; for real f, c₋ₙ = cₙ*, and cₙ = (aₙ − ibₙ)/2 links it to the sine–cosine form. Parseval’s theorem is completeness applied to this basis: (1/2π)∫|f|² dx = Σ|cₙ|² = a₀²/4 + ½Σ(aₙ² + bₙ²). It turns a Fourier series into a sum of a numerical series.

Worked: f(x) = x on (−π, π) is odd, so aₙ = 0 and bₙ = 2(−1)ⁿ⁺¹/n. The left side is (1/2π)∫x² dx = π²/3; the right side is ½Σ4/n² = 2Σ1/n². So Σ 1/n² = π²/6 ≈ 1.645. The square wave that is ±1 gives Σ 1/(2k − 1)² = π²/8 in the same way. Near a jump a truncated series overshoots by about 9% of the jump however many terms are kept — the Gibbs phenomenon — while the full series converges to the mid-point of the jump.

4. The Fourier transform in physics, and the Dirac delta

Physics usually writes the symmetric pair F(k) = (1/√2π)∫ f(x)e^−ikx dx and f(x) = (1/√2π)∫ F(k)e^ikx dk; engineering conventions move the 2π or flip the sign, and a question states its own. Four properties do most of the work: differentiation becomes multiplication, f′ ↔ ikF; a shift becomes a phase, f(x − a) ↔ e^−ikaF; convolution becomes a product, (f∗g) ↔ √2π FG; and Plancherel ∫|f|² dx = ∫|F|² dk keeps the norm.

The Gaussian is its own transform in shape: e^−ax² ↦ (1/√(2a)) e^−k²/4a. A narrow Gaussian (large a) has a wide transform, and the product of the widths is fixed — the mathematical root of Δx Δk ≥ ½ and so of the uncertainty principle. The Dirac delta δ(x) is the limit of ever narrower unit-area Gaussians; its transform is the constant 1/√2π, which gives δ(x) = (1/2π)∫e^ikx dk.

Delta-function rules
Rule	Worked
∫ f(x) δ(x − a) dx = f(a)	∫ (x³ + 1) δ(x − 2) dx = 9
δ(ax) = δ(x)/\|a\|	∫ x² δ(2x − 4) dx = ∫ x² δ(x − 2)/2 dx = 2
δ(g(x)) = Σ δ(x − xᵢ)/\|g′(xᵢ)\|	δ(x² − 1) = [δ(x − 1) + δ(x + 1)]/2
∫ f(x) δ′(x) dx = −f′(0)	by parts; the boundary term vanishes

⚠️ The scaling rule is the one examiners set

δ(2x − 4) is not δ(x − 2). Substituting u = 2x shows that it carries half the weight: ∫ f(x) δ(2x − 4) dx = f(2)/2. An answer of f(2) is the option placed there for anyone who drops the 1/|a|.

Key takeaways

Removable, pole or essential is read from the negative powers of the Laurent series; a branch point is not an isolated singularity, and nor is a point where poles accumulate.
∫_−∞^∞ P/Q dx = 2πi × (upper-half-plane residues) when deg Q ≥ deg P + 2; ∫₀^2π dθ/(a + b cos θ) = 2π/√(a² − b²).
For e^ikx with k > 0 close upward (Jordan’s lemma): ∫ cos kx/(x² + a²) dx = (π/a)e^−ka. Never put cos kz on the arc.
Parseval turns a Fourier series into a numerical sum: f = x gives Σ1/n² = π²/6.
A Gaussian transforms to a Gaussian of inverse width; δ(ax) = δ(x)/|a|, and ∫ f δ(x − a) dx = f(a).

Practice questions (13)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

At z = 0 the function sin z / z has:
1. A removable singularity
2. A simple pole
3. An essential singularity
4. A branch point
Show answer
Answer: A — A removable singularity
sin z/z = 1 − z²/6 + z⁴/120 − …, with no negative powers, and the limit at 0 is 1. Defining f(0) = 1 removes the singularity. The 1/z in the formula tempts "simple pole", but the zero of sin z cancels it.
At z = 0 the function e^1/z has:
1. An essential singularity
2. A pole of infinite order that is still a pole
3. A removable singularity
4. A branch point
Show answer
Answer: A — An essential singularity
e^1/z = Σ z⁻ⁿ/n! has infinitely many negative powers, which defines an essential singularity; near it the function takes almost every complex value. A pole needs the negative powers to stop at a finite order, and the function is single-valued, so it is not a branch point.
The residue of z² e^1/z at z = 0, to three decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.167
z² e^1/z = z²(1 + 1/z + 1/(2z²) + 1/(6z³) + …); the z⁻¹ term is z²/(6z³), so the residue is 1/6 = 0.167. The residue formula for poles cannot be used at an essential singularity, and taking the 1/(2z²) term instead gives 0.5.
The value of ∫_−∞^∞ dx/(x² + 1)², to three decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 1.571
Close upward; z = i is a double pole. Res = d/dz[1/(z + i)²] at z = i = −2/(2i)³ = −2/(−8i) = −i/4. The integral is 2πi(−i/4) = π/2 = 1.571. Treating the pole as simple, lim (z − i)f, is undefined here; using the residue of 1/(1 + z²) gives π instead.
The value of ∫₀^2π dθ/(5 + 4 cos θ), to three decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 2.094
With z = e^iθ the integrand becomes dz/[i(2z² + 5z + 2)] = dz/[2i(z + ½)(z + 2)]; only z = −½ lies inside |z| = 1, with residue 1/[2i · (3/2)] = 1/(3i). So the integral is 2πi/(3i) = 2π/3 = 2.094, which matches 2π/√(25 − 16). Including the pole at −2 gives 0.
The value of ∫_−∞^∞ cos x/(x² + 1) dx, to three decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 1.156
Write cos x = Re e^ix and close upward (Jordan’s lemma). Residue of e^iz/(z² + 1) at z = i is e⁻¹/(2i), so the integral is Re[2πi · e⁻¹/(2i)] = π/e = 1.156. Dropping the exponential gives π = 3.142, the answer for 1/(x² + 1) alone.
At z = 0, which of the following functions have an ISOLATED singularity?
1. 1/z²
2. √z
3. e^1/z
4. 1/sin(1/z)
Show answer
Answer: A — 1/z²; C — e^{1/z}
1/z² has a double pole and e^1/z an essential singularity, both isolated. √z is multivalued round 0 — a branch point, not an isolated singularity. 1/sin(1/z) has poles at z = 1/(nπ) that accumulate at 0, so no punctured disc around 0 is free of singularities.
The Fourier series of f(x) = x on (−π, π) has bₙ = 2(−1)ⁿ⁺¹/n. Using Parseval’s theorem, the value of Σ_n=1^∞ 1/n², to three decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 1.645
(1/π)∫_−π^π x² dx = 2π²/3 must equal Σbₙ² = 4Σ1/n², so Σ1/n² = π²/6 = 1.645. Using (1/2π) on the left with Σbₙ² on the right (mixing the two normalisations) gives π²/12 = 0.822.
A Gaussian pulse e^−ax² is made narrower by increasing a. Its Fourier transform:
1. Stays Gaussian and becomes wider in k
2. Stays Gaussian and becomes narrower in k
3. Becomes a sinc function
4. Becomes a Lorentzian
Show answer
Answer: A — Stays Gaussian and becomes wider in k
e^−ax² ↦ (1/√(2a))e^−k²/4a: the transform is Gaussian with width ∝ √a, so a narrower pulse has a wider spectrum and the width product is fixed. A sinc belongs to a rectangular pulse and a Lorentzian to e^−a|x|.
The value of ∫_−∞^∞ (x³ + 1) δ(x − 2) dx is ____.
Numerical answer — type the value.
Show answer
Answer: 9
The delta function picks out the value of the rest of the integrand at x = 2: 2³ + 1 = 9. Evaluating at x = 0 (reading δ(x − 2) as δ(x)) gives 1, and evaluating at x = −2 gives −7.
The value of ∫_−∞^∞ x² δ(2x − 4) dx is ____.
Numerical answer — type the value.
Show answer
Answer: 2
δ(2x − 4) = δ(2(x − 2)) = δ(x − 2)/2, so the integral is 2²/2 = 2. Ignoring the scaling rule δ(ax) = δ(x)/|a| gives 4, and it is the most common wrong answer.
With F(k) the Fourier transform of f(x) in the convention F(k) = (1/√2π)∫f e^−ikx dx, the transform of df/dx is:
1. ik F(k)
2. −ik F(k)
3. dF/dk
4. F(k)/(ik)
Show answer
Answer: A — ik F(k)
Integrate by parts: (1/√2π)∫f′e^−ikx dx = [boundary term, zero] + ik(1/√2π)∫f e^−ikx dx = ikF. The sign −ik belongs to the opposite convention e^+ikx; dF/dk is the transform of −ixf, and F/(ik) is integration.
A square wave is expanded in a Fourier series. Which statements are correct?
1. At a jump the full series converges to the mid-point of the jump
2. The overshoot of a truncated series near a jump does not shrink to zero as more terms are added
3. Parseval’s theorem gives Σ 1/(2k − 1)² = π²/8
4. The coefficients fall off as 1/n², because the function is bounded
Show answer
Answer: A — At a jump the full series converges to the mid-point of the jump; B — The overshoot of a truncated series near a jump does not shrink to zero as more terms are added; C — Parseval’s theorem gives Σ 1/(2k − 1)² = π²/8
The series converges to the average of the one-sided limits at a jump, and the Gibbs overshoot stays near 9% of the jump while moving closer to it. For ±1, bₙ = 4/(nπ) for odd n, and Parseval gives (16/π²)·½Σ1/(2k − 1)² = 1, i.e. π²/8. Those coefficients fall as 1/n: a discontinuity forces 1/n, and 1/n² needs a continuous function.