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Contour integrals and the residue theorem

Cauchy's theorem, the integral formula, and evaluating real integrals with residues.

The integral of a holomorphic function around a closed loop is zero; around a singularity it is 2πi times the residue. Real integrals like ∫dx/(1 + x²) close in the upper half-plane and pick up the residue at i. Picture it: the contour along the real axis closed by a great semicircle enclosing the pole at i. Think it: Cauchy's theorem is Green's theorem plus the Cauchy-Riemann equations.

Exemplo resolvido · integrate 1/(1+x^2) dx from -oo to oo

Calculus · Definite-Integrals

Integrate 1/(x^2 + 1) from -oo to oo

\int_{-\infty}^{\infty} \frac{1}{x^{2} + 1}\, dx

Passo a passo

  1. \int_{-\infty}^{\infty} \frac{1}{x^{2} + 1}\, dx

    First find an antiderivative F, then evaluate F(b) − F(a).

  2. \int \frac{1}{x^{2} + 1}\, dx = \operatorname{atan}{\left(x \right)}

    Recognise an inverse-trig / inverse-hyperbolic form.

  3. F(\infty) - F(-\infty) = \left(\frac{\pi}{2}\right) - \left(- \frac{\pi}{2}\right)

    Fundamental theorem of calculus: plug in the limits.

  4. = \pi \approx 3.1416

    Simplify.

Revelar a resposta
\pi

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Como: Contour integrals and the residue theorem

  1. First find an antiderivative F, then evaluate F(b) − F(a).
  2. Recognise an inverse-trig / inverse-hyperbolic form.
  3. Fundamental theorem of calculus: plug in the limits.
  4. Simplify.

Perguntas que as pessoas fazem

Why is complex differentiability so much stronger than real?

The limit must be the same from every direction in the plane, not just two. That forces the Cauchy-Riemann equations, which in turn force infinitely many derivatives and a convergent Taylor series.

What is a residue?

The coefficient of 1/(z − a) in the Laurent series at a singularity a. The residue theorem says a contour integral equals 2πi times the sum of the residues inside, which evaluates many real integrals in one line.

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