Evaluation of Real Integrals by Contour Integration

Many real integrals that defeat elementary antiderivatives — such as ∫₀^∞ dx/(1 + x⁴) or ∫_{−∞}^{∞} cos(x)/(x² + 1) dx — yield to contour integration: extend the real integral into the complex plane, close the real line into a loop with a large arc, and apply the residue theorem to the closed contour.

The strategy works because the contribution of the closing arc can often be shown to vanish as its radius grows, leaving the desired real integral equal to 2πi times the residues of the integrand's complex extension at the poles caught inside the loop.

Key Concepts

• Closing the contour: integrate f(z) along [−R, R] plus the semicircle of radius R in the upper half-plane, then let R → ∞.
• ML-estimate: the arc integral is bounded by (max |f| on arc) × (arc length πR); if this tends to 0, the arc contributes nothing.
• Jordan's lemma: for a > 0, the integral of e^{iaz}g(z) over the large upper semicircle tends to 0 as R → ∞ when g(z) → 0, justifying the closure for Fourier-type integrals.
• Indented contours: when a pole sits on the real axis, detour around it with a small semicircle; a simple pole on the contour contributes half its residue (iπ instead of 2πi).
• Keyhole contours: used when the integrand has a branch cut, e.g. for ∫₀^∞ x^{a−1}/(1 + x) dx.

Formulas

• Basic recipe: ∫_{−∞}^{∞} f(x) dx = 2πi × (sum of residues of f in the upper half-plane), provided the large-arc integral vanishes.
• Fourier type: ∫_{−∞}^{∞} e^{ikx}g(x) dx = 2πi × (sum of upper half-plane residues of e^{ikz}g(z)) for k > 0, by Jordan's lemma.
• Pole on contour: indent around a simple pole on the real axis; it contributes πi × Res instead of 2πi × Res (principal value integral).
• Even functions: ∫₀^∞ f(x) dx = half of ∫_{−∞}^{∞} f(x) dx when f is even.

Worked Example

Evaluate I = ∫_{−∞}^{∞} dx/(x² + 1).

1. Consider f(z) = 1/(z² + 1) on the closed contour: real segment [−R, R] plus the upper semicircle |z| = R.
2. Poles of f are at z = ±i; only z = i lies in the upper half-plane (inside for R > 1).
3. Residue at z = i: lim (z → i) (z − i)/(z² + 1) = 1/(i + i) = 1/(2i).
4. Arc estimate: on |z| = R, |f| ≤ 1/(R² − 1), arc length πR, so |arc integral| ≤ πR/(R² − 1) → 0 as R → ∞.
5. Hence I = 2πi × 1/(2i) = π. Check: ∫ dx/(x²+1) = arctan x, and arctan(∞) − arctan(−∞) = π/2 − (−π/2) = π. It matches.

Common Misconceptions

• "Always close in the upper half-plane." Only when the exponential decays there: for e^{ikz} with k > 0 close upward; with k < 0 close downward, picking up a minus sign from orientation.
• "The arc integral always vanishes." It must be verified with an estimate. For ∫ dx/x²-type integrands it works, but for e.g. ∫ e^{iz²} dx a semicircle fails and a different contour (a wedge) is needed.
• "Poles on the real axis can be ignored." Never — the contour must be indented around them, and each simple pole on the path contributes exactly half its residue.

Real-World Applications

• Fourier analysis: transform pairs in signal processing and quantum mechanics are evaluated by contour methods.
• Scattering theory: physics integrals with oscillatory kernels are computed via Jordan's lemma closures.
• Probability: characteristic functions of distributions (e.g. the Cauchy distribution) are inverted with residues.

Contour integration evaluates real integrals by promoting them to complex contour integrals: close the contour, verify the arc vanishes, and sum 2πi times the enclosed residues. The worked example shows the full pipeline — and the antiderivative check confirms the answer π.