Unit 7 of 7

Advanced transformations

  • 3 chapters
  • Chapters 28 to 30

Each chapter evaluates an integral in a larger setting, a second dimension for the Gaussian integral, a family of integrals for parameter differentiation and the complex plane for contour integration.

Squaring the Gaussian integral gives an integral over the whole plane, and polar coordinates evaluate it as pi, so the Gaussian integral itself is the square root of pi. The integral of (x minus 1) over ln x from 0 to 1 joins a family whose parameter derivative is elementary, and an integrable bound justifies differentiating under the integral sign. The integral of 1 over (1 plus x to the fourth) from 0 to infinity is half an integral over the whole line, which closes into a semicircle in the upper half-plane, where the residue theorem adds two residues and the arc vanishes in the limit.

Every chapter will be part of one long video. The video is coming soon, and each chapter page already has the definitions and an example for every key term.

The Gaussian integral, parameter differentiation and contour integration

  1. Square the question, reveal a circle

    Why the Gaussian integral of e to the minus x squared over the whole real line equals the square root of pi, found by squaring the integral, reading it as an improper double integral and using polar coordinates.

    Chapter 28

  2. Make a family, then differentiate the family

    How differentiation under the integral sign evaluates the integral of (x minus 1) over ln x from 0 to 1 as ln 2, using a parameter family, the Leibniz integral rule, an integrable bound and a known value for the constant.

    Chapter 29

  3. Leave the real line, collect the poles, return

    How contour integration evaluates the integral of 1 over (1 plus x to the fourth) from 0 to infinity, with analytic functions, simple poles and residues, Cauchy's theorem, the residue theorem and an upper half-plane semicircle.

    Chapter 30