Contour Integration of Infinite Improper Rational Integrals across (-∞, ∞)
Idea in plain words
Evaluation of Infinite Real Integrals: I = ∫_{-∞}^∞ [P(x) / Q(x)] dx where Q(x) has no real zeros and deg(Q) ≥ deg(P) + 2.
Exam tip
High-yield exam topic (Very High).
Explanation
Improper real integrals extending across the entire real axis (-∞, ∞) cannot be evaluated by ordinary antiderivatives when polynomials lack elementary factorizations. By closing the real axis with a semi-infinite semicircular arc in the upper half complex plane, Jordan's lemma guarantees that the arc integral vanishes to zero as radius R → ∞, converting the infinite integral into the sum of residues at upper-half-plane poles.
Key points
- Contour Construction: Closed contour Γ consisting of the real line segment [-R, R] and the upper semicircle C_R: z = R e^{iθ} (0 ≤ θ ≤ π).
- Jordan's Lemma / Semicircle Vanishing: As R → ∞, ∫_{C_R} f(z) dz → 0 because |f(z)| ≤ M / R².
- Formula: ∫_{-∞}^∞ f(x) dx = 2πi ∑ Res[f(z)] at all poles lying strictly in the UPPER HALF-PLANE (Im(z) > 0).
- Poles with Trigonometric Factors: For ∫_{-∞}^∞ [P(x)/Q(x)] cos(mx) dx, integrate f(z) = [P(z)/Q(z)] e^{imz} and take the Real part of 2πi ∑ Res.
Example
Example
Evaluate ∫{-∞}^∞ dx / (x² + 1)²: Poles at z = ±i (order 2). Upper pole is z = i. Res(i) = d/dz [(z - i)² / (z² + 1)²]|{z=i} = d/dz [1/(z + i)²]|_{z=i} = -2/(2i)³ = -2/(-8i) = 1/(4i). I = 2πi(1/4i) = π/2.
Code examples
// Complete Analytical Contour Solution: Infinite Real Integral
// Evaluate: I = ∫_{-∞}^∞ dx / (x^4 + 1)
// Step 1: Consider complex function f(z) = 1 / (z^4 + 1) over contour Γ = [-R, R] ∪ C_R:
// ∮_Γ f(z) dz = ∫_{-R}^R [dx / (x^4 + 1)] + ∫_{C_R} f(z) dz
// As R -> ∞, ∫_{C_R} f(z) dz -> 0 (since degree denominator - degree numerator = 4 ≥ 2).
// Therefore: ∫_{-∞}^∞ dx / (x^4 + 1) = 2πi ∑ Res[f(z), upper-half poles]
//
// Step 2: Find poles (roots of z^4 + 1 = 0 => z^4 = -1 = e^{i(π + 2kπ)}):
// z_k = e^{i (2k + 1)π / 4} for k = 0, 1, 2, 3
// z0 = e^{i π/4} = cos(π/4) + i sin(π/4) = (1 + i) / √2 [Im > 0: UPPER]
// z1 = e^{i 3π/4} = cos(3π/4) + i sin(3π/4) = (-1 + i) / √2 [Im > 0: UPPER]
// z2 = e^{i 5π/4} [Im < 0: LOWER - Ignore]
// z3 = e^{i 7π/4} [Im < 0: LOWER - Ignore]
//
// Step 3: Compute residues using P(z)/Q'(z) formula:
// Q'(z) = 4 z^3. Since z^4 = -1 => z^3 = -1/z.
// Res(z_k) = 1 / (4 z_k^3) = z_k / (4 z_k^4) = -z_k / 4
// Sum of residues at upper-half poles:
// ∑ Res = Res(z0) + Res(z1) = -(1/4) [ z0 + z1 ]
// = -(1/4) [ (1 + i)/√2 + (-1 + i)/√2 ]
// = -(1/4) [ (2i) / √2 ] = -i / (2 √2)
//
// Step 4: Apply Cauchy's Residue Theorem:
// I = 2πi * [ -i / (2 √2) ]
// = 2π (-i^2) / (2 √2) = π / √2 (Exact University Analytical Solution!)
Exam questions
- [8 Marks] Using contour integration, evaluate: ∫_0^∞ [dx / (x² + a²)] and hence evaluate ∫_0^∞ [dx / (x² + 1)²].
- [8 Marks] Evaluate ∫_{-∞}^∞ [cos(mx) / (x² + a²)] dx (a > 0, m > 0) using contour integration.
- [7 Marks] Show that ∫_0^∞ [dx / (x⁶ + 1)] = π / 3 using Cauchy's residue theorem.