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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.

Àwọn ààtò àwọn ìṣàmúlò-ètò · integrate 1/(1+x^2) dx from -oo to oo

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

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

Àwọn ìṣàmúlò-ètò

  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.

Fi àwọn àgbèwọlé hàn
\pi

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Bii o ṣe le: 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.

Àwọn Àtòjọ-ẹ̀yàn

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.

Diẹ̀ nínú Complex Analysis