SYSTEMATIC MATHEMATICS

Integral Transforms & Special Functions

Why can a complicated wave be described by frequencies, a differential equation become algebra, and circular or spherical boundary problems produce named functions? This course builds those answers from the ground up instead of presenting a table of formulas. Across forty-two visible chapters in eight content-sized units of lengths 5, 6, 6, 5, 5, 6, 6, and 3, it treats transforms as operators with declared variables, kernels, units, domains, convergence regions, inverse hypotheses, and verification routes. Fourier series and transforms, Laplace methods, Sturm–Liouville expansions, Gamma, Beta, Bessel, Legendre, Hermite, Laguerre, Hankel, distribution, Hilbert, Mellin, and Z-transform ideas are derived through exact examples and failure boundaries. The final dossier combines analytic, spectral, and numerical evidence without claiming that a finite plot proves convergence or that a named special function validates a physical model.

Before this course: Completed Single-Variable Calculus I & II, Linear Algebra, Differential Equations & Dynamical Systems, Real Analysis, and the opening complex-exponential and contour chapters of Complex Analysis. The course assumes improper integrals, infinite series, complex exponentials, integration by parts, linear ODEs, eigenvectors, inner products, uniform and L² convergence language, and basic proof. It does not assume Partial Differential Equations, distribution theory, signal-processing software, or a table of transforms.

COURSE FACTSLevel, chapters, units, prerequisite, and outcome
Chapter 1

A transform is an operator, not a magic change of letters

Objective: Why is s not replaced by t after the integral is evaluated?

An integral transform compares a function with a family of kernel shapes and records the resulting coefficients. The transformed variable labels which kernel member was used; it is not the original input renamed. Before using a transform table, name the original variable, transformed variable, kernel, integration or summation region, normalization, units, and convergence assumptions. The exact claim for this chapter is: If the integral exists for f and g, then T(af+bg)=aTf+bTg for scalars a,b.

Here is the object in precise language: For a kernel K(s,t), an integral transform is (Tf)(s)=∫_D K(s,t)f(t)dt whenever the integral exists; t is integrated out, while s indexes the output. A transform is a new representation of the same declared object, not permission to ignore domain, endpoint, regularity, or inverse-existence conditions.

Use the concrete problem “Let K(s,t)=e^{-st} on t≥0. Compute T(2e^{-t}+3e^{-2t})(s) for real s>−1.” to make every symbol observable. Rebuild the three steps, check the result in the original representation, and then test this boundary: Linearity does not guarantee existence: two individually divergent integrals cannot be separated and manipulated as though both were finite.

If the integral exists for f and g, then T(af+bg)=aTf+bTg for scalars a,b.

Insert af(t)+bg(t) into the defining integral.

Use linearity of integration to separate the two terms and move constants outside.

Recognize the two remaining integrals as Tf and Tg on their common output domain.

Let K(s,t)=e^{-st} on t≥0. Compute T(2e^{-t}+3e^{-2t})(s) for real s>−1.

  1. Combine exponents: the two integrands are 2e^{-(s+1)t} and 3e^{-(s+2)t}.
  2. For a>0, ∫₀∞e^{-at}dt=1/a; s>−1 makes both a=s+1 and a=s+2 positive.
  3. Apply linearity to obtain 2/(s+1)+3/(s+2).

Result: Tf(s)=2/(s+1)+3/(s+2) for real s>−1.