Residue theorem
Objective. Evaluate contour and real integrals.
- 1. Learn
- 2. Worked example
- 3. Practice
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Learn: the key idea
The residue theorem states that for f holomorphic inside and on a simple closed contour γ except for finitely many isolated singularities z₁,...,zₙ inside γ, ∮_γ f(z)dz = 2πi Σ Res(f,zₖ). This converts contour integration into algebra: locate poles inside the contour, sum their residues, multiply by 2πi. The theorem also evaluates real integrals, such as ∫_{-∞}^{∞}dx/(1+x²), by closing the contour with a large semicircle in the upper half-plane and showing the arc contribution vanishes, leaving only the residue at z=i, giving π. This technique turns otherwise hard real integrals into residue calculations.
Worked example
Evaluate the contour integral ∮_{|z|=2} 1/(z²+1) dz using the residue theorem.
- 1Poles of 1/(z²+1) are at z=i and z=−i, both inside |z|=2
- 2Res at z=i is 1/(2i); Res at z=−i is 1/(−2i)
- 3Sum of residues = 1/(2i) + 1/(−2i) = 0
- 4Integral = 2πi × 0 = 0
Answer: 0
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Practice
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Start practisingCommon mistake. Students include poles outside the contour or forget to check the semicircular arc vanishes (Jordan's lemma) before closing contours; omitting this check can wrongly include contributions that do not actually vanish for integrands that don't decay fast enough.
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