Definitions, Theorems, Examples, Topics for Asmar 2nd edition
Chapter 7: 7.1 only

 7.1 The Fourier Integral Representation
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 FOURIER SERIES for 2p-PERIODIC FUNCTIONS

   Assume f(x) is 2p-periodic and piecewise smooth.

   f(x) = a_0 + series of terms (a_n cos(n Pi x/p) + b_n sin(n Pi x/p))

    FOURIER COEFFICIENT FORMULAS
    a_0 = (1/2p) integral of f(t) over [-p,p]
    a_n = (1/p) integral of f(t)cos(n Pi t/p) over [-p,p]
    b_n = (1/p) integral of f(t)sin(n Pi t/p) over [-p,p]

 APPROXIMATION for LARGE PERIOD p

    a_n == (1/p) integral of f(t)cos(n Pi t/p) over (-inf,inf)
        == (1/Pi) [same integral](Pi/p)
        ==  A(n Pi/p)(Pi/p)
        where
        A(omega) = (1/Pi) integral of f(t)cos(omega t) over (-inf,inf)

    b_n == (1/p) integral of f(t)sin(n Pi t/p) over (-inf,inf)
        == (1/Pi) [same integral](Pi/p)
        ==  B(n Pi/p)(Pi/p)
        where
        B(omega) = (1/Pi) integral of f(t)sin(omega t) over (-inf,inf)

 DEF. The integrals A(omega), B(omega) are called the FOURIER COSINE and
      SINE INTEGRALS of signal f(x), assumed to be defined on the whole
      real line. The integrals make sense with the additional
      restriction: integral of |f(x)| over (-inf,inf) is finite.

 SUMMATION to INTEGRAL

   The Fourier summation then becomes approximately a Riemann sum for
   the interval [0,inf). Formally, write w_n = n Pi/p, and then the
   summation looks like

        sum over n=1 to inf of terms G(w_n)(w_n - w_{n-1})
            where
                   G(w) =  A(w)cos(w x)+B(w)sin(w x)

   This summation is a Riemann sum for G(w) over 0 <= w < inf.
   Therefore, the summation limits to the integral of G over [0,inf).

 THEOREM 1. [Fourier Representation Theorem]
   Assume f(x) is piecewise smooth on every finite interval and the
   integral of |f(x)| over (-inf,inf) is finite. Define

        A(omega) = (1/Pi) integral of f(t)cos(omega t) over (-inf,inf)
        B(omega) = (1/Pi) integral of f(t)sin(omega t) over (-inf,inf)

   Then

       (f(x+)+f(x-))/2 = integral of A(w)cos(wx)+B(w)sin(wx)
                           for w=0 to infinity.

 QUESTION. What happened to a_0? Answer: a_0 == (1/2p)integral of f,
           which limits to zero as p goes to infinity.

 EXAMPLE 1. Find the Fourier integral representation for the unit pulse
            f(x) = 1 on -1<x<1, f(x) = 0 elsewhere.

 DETAILS. The integral of |f(x)| on (-inf,inf) is finite. And f is
 piecewise smooth on the line. Compute
        A(w) = (2/Pi)(sin(w)/w)
        B(w) = 0  [because f is even]
        h(x) = integral of A(w)cos(wx)+B(w)sin(wx)
             = integral of (2/Pi)sin(w)cos(wx)/w
             = (f(x+)+f(x-))/2
             = f(x) except at |x|=1, where h(x)=1/2

 DIRICHLET INTEGRAL
   Setting x=0 in the Fourier integral representation of the unit pulse
   gives the identity

                      integral of sin(w)/w = Pi/2

   known as the Dirichlet integral.

 EXAMPLE 2. Let f(x) = cos(x) pulse(x,-Pi/2,Pi/2). Find the Fourier
 integral representation.

 DETAILS. This modified pulse can be constructed from the unit pulse
 U=pulse(x,-1,1) by a change of variables. As in EXAMPLE 1, the function
 f is even, giving B(w)=0.

   A(w) = (2/Pi) integral of cos(x)cos(wx) over [0,Pi/2]
        = (2/Pi)cos(w Pi/2)/(1-w^2)  except for w=1
   A(1) = (2/Pi) integral of cos(x)^2 over [0,Pi/2]
        = (2/Pi)(Pi/4)
        = 1/2

 Then the Fourier representation theorem gives the identity

   (2/Pi) integral of cos(Pi w/2)cos(wx)/(1-w^2) = f(x)
   except for |x|=Pi/2 [The value at |x|=Pi/2 is zero]

 GIBBS PHENOMENON

  DEF. The partial Fourier integral is

          S_v(x) = Fourier integral with inf replaced by v

  DEF. The sine integral is

          Si(v) = integral of sin(t)/t over 0<=t<=v

       Known from the Dirichlet integral is Si(inf)=Pi/2.

 EXAMPLE 3. Gibbs overshoot for the unit pulse on [-1,1].

  LEMMA. S_v(x) = (1/Pi)(Si(v+xv)+Si(v-xv))
  PROOF. Trig identity plus a change of variable in the integrals.

  MAPLE PLOT
   # Gibbs Phenomenon for Fourier Integral Representation
   S:=(x,v)->(1/Pi)*(Si(v+v*x)+Si(v-v*x));
   # make an animation of Gibbs overshoot
   plots[animate]( plot, [S(x,v), x=-1.5..1.5],v=20..40);