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Integration in Differential Equations

  1. Integration over a simplex
    Examples: [Wat1, p.258, §12.5; p.75, §4.51].

  2. Infinite integrals and improper integrals [Wat1, §4.4-§4.5]

  3. Evaluating the values of a function through a recursion formula
    Example. [Wat1, p.282, l.12] ® [Wat1, p.282, l.-1].

  4. Calculating d(òg(t)h(t) f(k(t),s)ds)/dt [Cod, p.37, Problem 1].
    Solution. Let F(u,v,w) = òuv f(w,s)ds,
    u=g(t), v=h(t), w=k(t). Then
    d(òg(t)h(t) f(k(t),s)ds)/dt
    = (F/u)(du/dt)+(F/v)(dv/dt)+(F/w)(dw/dt).

  5. For a solution of a differential equation or an integral, we should express it in closed form first, and then discuss its properties. This way may help systematize its properties and thereby organize and facilitate related discussions.
    After introducing the integral given in [Hob, p.185, l.-2], Hobson discusses its properties first [Hob, p.186, (a), (b), (c) and (d)]. In contrast, Watson expresses the integral in closed form first [Wat1, p.257, l.-2]. Once the integral is expressed in closed form, the proofs of [Hob, p.186, (a), (b), (c) and (d)] become trivial.

  6. Contours for integral transforms
    1. A line segment
      1. Laplace transforms [Guo, p.82, l.-9-l.-4]
        1. Method: elementary calculus
        2. Domain restriction for z: [Guo, p.82, l.-4]
          Domain restriction for l: [Guo, p.302, l.-10-l.-9; p.352, l.-12-l.-11]
        3. Example 1: [Guo, p.302, l.1-l.-1]
          Example 2: [Guo, p.352, l.-12-p.353, l.6].
    2. Figure of eight [Wat, p.161, Fig.1]
      1. Laplace transforms
        1. Method: analytic functions
        2. Domain restriction for z: none
        3. Example: [Wat, p.162, l.4-p.163, l.6].
        4. Remark. For practical calculations, we can reduce case B to case A [Wat1, p.365, l.13-p.366, l.11]. Thus, case A can be considered a special case of case B.
    3. Pochhammer's contour [Guo, p.83, Fig.3] (method: analytic functions; domain restriction for z: none)
      1. Laplace transforms
        1. Example: [Guo, p.303, l.7-l.-6]
      2. Euler transforms
        1. Example 1: [Guo, p.152, l.-11-p.153, l.9]
        2. Example 2: [Guo, p.247, l.-8-p.248, l.11]
      Remark. [Guo, p.303, (6); p.153, (7); p.247, (8)] are proved using the primitive formula [Guo, p.105, (1)].
    4. The contour starts from infinity, then goes towards a figure. After circling around the figure, it goes back to infinity [Guo, p.83, Fig.4; p.355, Fig.26].
      1. Laplace transforms
        1. Example 1 (coefficients are linear polynomials of z [Guo, p.81, (1)]): [Guo, p.84, l.5-l.-11]
          Domain restriction for a: [Guo, p.84, l.6]
        2. Example 2 (coefficients are linear polynomials of z [Guo, p.85, (19); Wat, l.19, (1)]): [Guo, p.355, l.1-p.356, l.-1]
          Domain restriction for z, t and n: [Guo, p.356, l.10]
          Remark. [Guo, p.356, (10); p.357, (13)] are proved using the primitive formula [Guo, p.102, (5)].

  7. Prove ò0¥ cos (x2) dx = ò0¥ sin (x2) dx = 2-1(p/2)1/2 [Inc1, p.173, l.2].
    Proof. ò0¥ e-ax2 dx = 2-1(p/a)1/2 [Reif, p.609, (A.5.1)].
    Let a = i.
    (-i)1/2 = (e-pi/2)1/2 = e-pi/4.

  8. Let a, b, m, n be integers with 0 and m, n ³ 0. Then
    (2p)-1ò[-p,p] (cos ax) cosn (bx) dx is 2-n C(n, m) if |a| = |b(2m-n)| and 0 otherwise.
    Proof. We may assume b=1.
    By the binomial theorem, cosn x = 2-n (eix + e-ix) n is 2-n Sk = 0(n-1)/2 C(n, k) cos ((n - 2k)x) if n is odd and
    2-n C(n, n/2) + 21-n Sk = 0(n/2) -1 C(n, k) cos ((n - 2k)x) if n is even.

  9. Using the method of contraction for continued fractions to increase the rate of convergence [Perr, p.200, l.-7-p.201, l.2].
    Example. [Perr, p.298, (13)] ® [Perr, p.298, (16)]

  10. For Re n³-1/2, [Wat, p.47, (1)] can be proved with a routine procedure, but for -1/2<n<1/2, the proof requires the use of integration by parts.