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Complex Analysis

Russell W. Howell, John H. Mathews

Section 11.1 Fourier Series

In this chapter we show how Fourier series, the Fourier transform, and the Laplace transform are related to the study of complex analysis. We develop the Fourier series representation of a real-valued function \(U(t)\) of the real variable \(t\text{.}\) Complex Fourier series and Fourier transforms are then discussed. Finally, we develop the Laplace transform and the complex variable technique for finding its inverse. This chapter focuses on applying these ideas to solving problems involving real-valued functions, so many of the theorems throughout are stated without proof.
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Let \(U(t)\) be a real-valued function that is periodic with period \(2\pi\text{,}\) that is,
\begin{equation*} U(t+2\pi)=U(t) \; \text{ for all } \; t\text{.} \end{equation*}
One such function is \(s= U(t) = \sin(t-\frac{\pi}{2}) + 0.7\cos(2t-\pi -\frac{1}{4}) +1.7\text{,}\) and its graph is obtained by repeating the portion of the graph in any interval of length \(2\pi\text{,}\) as shown in Figure 11.1.1.
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Figure 11.1.1. A function \(U\) with period \(2\pi\)
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Familiar examples of real functions that have period \(2\pi\) are \(\sin nt\) and \(\cos nt\text{,}\) where \(n\) is an integer. This raises the question whether any periodic function can be represented by a sum of terms involving \(a_n\cos nt\) and \(b_n\sin nt\text{,}\) where \(a_n\) and \(b_n\) are real constants. As we shall soon see, the answer to this question is often yes.
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Definition 11.1.2. Piecewise Continuous.

The function \(U\) is piecewise continuous on the closed interval \([a,b]\) if there exists values \(t_0,t_1,\ldots,t_n\) with \(a = t_0\lt t_1\lt \cdots\lt t_n=b\) such that \(U\) is continuous in each of the open intervals \(t_{k-1}\lt t\lt t_k \;\; (\text{for } \, k=1,2,...n)\text{,}\) and has left- and right-hand limits at the values \(t_k\) for \(k=0,1,\ldots, n\text{.}\)
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We use the symbols \(U(a^-)\) and \(U(a^+)\) for the left- and right-hand limits, respectively, of a function \(U(t)\) as \(t\) approaches the point \(a\text{.}\) The graph of a piecewise continuous function is illustrated in Figure 11.1.3.
\begin{equation*} U(t)= \begin{cases} \frac{2}{3}\left(t-\frac{1}{2}\right)^2+\frac{1}{4} \amp \text{when} \quad 1 \le t \lt 2,\\ \frac{5}{2}-(t-2)^2 \amp \text{when} \quad 2 \lt t \lt 3,\\ 1+\frac{t-3}{4} \amp \text{when} \quad 3 \lt t \lt 4,\\ \frac{6}{5}-(t-5)^3 \amp \text{when} \quad 4 \lt t \le 6 \end{cases}\text{.} \end{equation*}
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Figure 11.1.3. A piecewise continuous function \(U\) over the interval \([1,6]\)
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The left- and right-hand limits at \(t_0=2\text{,}\) \(t_1=3\text{,}\) and \(t_2=4\) are easy to determine:
\begin{align*} \amp \text{At} \quad t=2, \quad U(2^-) = \frac{7}{4} \quad \text{and} \quad U(2^+) = \frac{5}{2};\\ \amp \text{At} \quad t=3, \quad U(3^-) = \frac{3}{2} \quad \text{and} \quad U(3^+) = \; 1;\\ \amp \text{At} \quad t=4, \quad U(4^-) = \frac{5}{4} \quad \text{and} \quad U(4^+) = \frac{11}{5}\text{.} \end{align*}
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Definition 11.1.4. Fourier Series.

If \(U(t)\) is periodic with period \(2\pi\) and is piecewise continuous on \([-\pi ,\pi]\) then the Fourier series \(S(t)\) for \(U(t)\) is
\begin{equation} S(t) = \frac{a_0}{2} + \sum_{n=1}^{\infty}(a_n\cos nt + b_n\sin jt)\text{,}\tag{11.1.1} \end{equation}
where the coefficients \(a_n\) and \(b_n\) are given by the so-called Euler’s Formulas:
\begin{equation} a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}U(t)\cos nt\,dt \quad \text{for} \quad n=0,1, \ldots\text{,}\tag{11.1.2} \end{equation}
and
\begin{equation} b_n = \frac{1}{\pi}\int_{-\pi}^{\pi}U(t) \sin nt\,dt \quad \text{for} \quad n=1,2, \ldots \,\text{.}\tag{11.1.3} \end{equation}
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The factor \(\frac{1}{2}\) in the constant term \(\frac{a_0}{2}\) on the right side of Equation (11.1.1) has been introduced for convenience so that \(a_0\) could be obtained from the general formula in Equation (11.1.2) by setting \(j=0\text{.}\) The reasons for this will be explained shortly. The next result discusses convergence of the Fourier series.
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Example 11.1.6.

Show that the function \(U(t) =\frac{t}{2}\) for \(t \in (-\pi ,\pi)\text{,}\) extended periodically by the equation \(U(t+2\pi)=U(t)\text{,}\) has the Fourier series expansion
\begin{equation*} U(t) = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin nt\text{.} \end{equation*}
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Solution.
Using Euler’s Formulas (11.1.2) and integration by parts, we obtain
\begin{align*} a_n \amp = \frac{1}{\pi}\int_{-\pi}^{\pi}\frac{t}{2}\cos nt\,dt\\ \amp = \left.\left(\frac{t\sin nt }{2\pi n} + \frac{\cos nt}{2\pi n^2}\right)\right|_{-\pi}^{\pi}\\ \amp = 0 \quad \text{for} \quad n=1,2,\ldots\text{,} \end{align*}
and then using Formula (11.1.3) we get
\begin{align*} b_n \amp = \frac{1}{\pi}\int_{-\pi}^{\pi}\frac{t}{2}\sin nt\,dt\\ \amp = \left.\left(\frac{ -t\cos nt}{2\pi n}+\frac{\sin nt}{2\pi n^2}\right)\right|_{-\pi}^{\pi}\\ \amp =\frac{-\cos n\pi}{n}\\ \amp =\frac{(-1)^{n+1}}{n} \quad \text{for} \quad n=1,2,\ldots\text{.} \end{align*}
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The coefficient \(a_0\) is computed by the calculation
\begin{equation*} a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}\frac{t}{2}\,dt = \frac{t^2}{4\pi}\bigg|_{-\pi}^{\pi}=0\text{.} \end{equation*}
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Substituting the coefficients \(a_j\) and \(b_j\) in Equation (11.1.1) produces the required solution. The graphs of \(U(t)\) and the first three partial sums \(S_1(t) = \sin t, \, S_2(t) = \sin t-\frac{1}{2}\sin 2t\text{,}\) and \(S_3(t) = \sin t-\frac{1}{2}\sin 2t + \frac{1}{3}\sin 3t\) are shown in Figure 11.1.7.
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Figure 11.1.7. \(U(t) =\frac{t}{2}\text{,}\) and approximations \(S_1(t), \, S_2(t)\text{,}\) and \(S_3(t)\)
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We now state some general properties of Fourier series that are useful for calculating the coefficients. The proofs are left for the reader.
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Example 11.1.13.

Show that the function \(U(t) = |t|\) for \(t \in (-\pi ,\pi)\text{,}\) extended periodically the equation \(U(t+2\pi) =U(t)\text{,}\) has the Fourier series representation
\begin{equation*} U(t) = |t| = \frac{\pi}{2}-\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}\cos[(2n-1)t]\text{.} \end{equation*}
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Solution.
The function \(U(t)\) is an even function, hence we can use Theorem 11.1.9 to conclude that \(b_n=0\) for all \(n\text{,}\) and
\begin{align*} a_n \amp = \frac{2}{\pi}\int_0^{\pi}t\cos nt\,dt\\ \amp = \left.\left(\frac{2t\sin nt}{\pi n} + \frac{2\cos nt}{\pi n^2}\right)\right|_0^{\pi}\\ \amp = \frac{2\cos n\pi -2}{\pi n^2}\\ \amp =\frac{2(-1)^n-2}{\pi n^2} \quad \text{for} \quad n=1,2, \ldots\text{.} \end{align*}
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The coefficient \(a_0\) is computed by the calculation:
\begin{equation*} a_0=\frac{2}{\pi}\int_0^{\pi}t\,dt=\left.\frac{t^2}{\pi}\right|_0^{\pi}=\pi\text{.} \end{equation*}
Using the \(a_n\) and Theorem 11.1.9 produces the required solution.
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Subsection 11.1.1 Proof of Euler’s Formulas

The following intuitive proof will justify the Euler formulas given in Equations (11.1.2) and (11.1.3) To determine \(a_0\) we integrate both \(U(t)\) and the Fourier series representation in Equation (11.1.1) from \(-\pi\) to \(\pi\text{,}\) which results in
\begin{equation*} \int_{-\pi}^{\pi}U(t)\,dt = \int_{-\pi}^{\pi}\left[\frac{a_0}{2} + \sum_{n=1}^{\infty}(a_n\cos nt+b_n\sin nt)\right]dt\text{.} \end{equation*}
Next, we perform integration term by term, and we obtain
\begin{equation*} \int_{-\pi}^{\pi}U(t)\,dt = \frac{a_0}{2}\int_{-\pi}^{\pi}1\,dt + \sum_{n=1}^{\infty}a_n\int_{-\pi}^{\pi}\cos nt\,dt + \sum\limits_{n=1}^{\infty}b_n\int_{-\pi}^{\pi}\sin nt\,dt\text{.} \end{equation*}
The value of the first integral on the right side of this equation is \(2\pi\) and all the other integrals are zero. Hence we obtain the desired formula for \(a_0\text{:}\)
\begin{equation*} a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}U(t)\,dt\text{.} \end{equation*}
To determine \(a_m\text{,}\) we let \(m>1\) denote a fixed integer and multiply both \(U(t)\) and the Fourier series representation in Equation (11.1.1) by the term \(\cos mt\text{.}\) Then we integrate and obtain
\begin{align} \int_{-\pi}^{\pi}U(t)\cos mt\,dt \amp = \frac{a_0}{2}\int_{-\pi}^{\pi}\cos mt\,dt\notag\\ \amp + \sum_{n=1}^{\infty}a_j\int_{-\pi}^{\pi}\cos mt\cos nt\,dt\notag\\ \amp + \sum_{n=1}^{\infty}b_n\int_{-\pi}^{\pi}\cos mt\sin nt\,dt\text{.}\tag{11.1.5} \end{align}
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The value of the first term on the right side of Equation (11.1.5) is easily seen to be zero:
\begin{equation} \frac{a_0}{2}\int_{-\pi}^\pi\cos mt\,dt = \frac{a_0\sin mt}{2m}\bigg|_{-\pi}^\pi = 0\text{.}\tag{11.1.6} \end{equation}
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The value of the term involving \(\cos mt\cos jt\) is found by using the trigonometric identity:
\begin{equation*} \cos mt\cos nt = \frac{1}{2} \Big(\cos [(m+n) t] +\cos [(m-n) t]\Big)\text{.} \end{equation*}
Calculation reveals that if \(m \ne n\) (and \(m>0\)), then
\begin{align} a_n\int_{-\pi}^{\pi}\cos mt\cos nt\,dt \amp = \frac{a_m}{2} \left[\int_{-\pi}^{\pi}\cos[(m+n)t]\,dt\right.\notag\\ \amp + \left.\int_{-\pi}^{\pi}\cos [(m-n) t] dt\right]\tag{11.1.7}\\ \amp = 0\text{.}\notag \end{align}
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When \(m=n\text{,}\) the value of the integral becomes:
\begin{equation} a_m\int_{-\pi}^{\pi}\cos^2mt\,dt = \pi a_m\text{.}\tag{11.1.8} \end{equation}
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The value of the term on the right side of Equation (11.1.5) involving the integrand \(\cos mt\sin nt\) is found by using the trigonometric identity
\begin{equation*} \cos mt\sin nt=\frac{1}{2}\Big(\sin \big[(m+n)t\big] + \sin\big[(m-n)t\big]\Big)\text{,} \end{equation*}
and for all values of \(m\) and \(n\text{,}\) we obtain
\begin{align} b_n\int_{-\pi}^{\pi}\cos mt\sin nt\,dt \amp = \frac{b_m}{2}\left[\int_{-\pi}^\pi\sin[(m+n)t]\,dt\right.\notag\\ \amp + \left. \int_{-\pi}^\pi\sin[(m-n)t]\,dt\right]\tag{11.1.9}\\ \amp = 0\text{.}\notag \end{align}
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ATherefore, we can use the results of Equations (11.1.6)–(11.1.9) in Equation (11.1.5) to obtain
\begin{equation*} \int_{-\pi}^{\pi}U(t)\cos mt\,dt=\pi a_m \quad \text{fo } \quad m=0,1,2, \ldots\text{,} \end{equation*}
and Equation (11.1.2) is established. We leave it as an exercise to establish Euler’s Formula (11.1.3) for the coefficients \(\left\{ b_n\right\}\text{.}\) A complete discussion of the details of the proof of Theorem 11.1.5 can be found in some advanced texts. See for instance, John W. Dettman, Chapter 8, in Applied Complex Variables, The Macmillan Company, New York, 1965.
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Exercises Exercises

For Exercises 1–2 and 6–11, find the Fourier series representation.
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1.
\(U(t)= \begin{cases} \;\;\, 1 \quad \text{for} \qquad \; 0\lt t\lt \pi, \\ -1 \quad \text{for} \quad -\pi \lt t\lt 0. \qquad \end{cases}\) See Figure 11.1.14.
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Figure 11.1.14. Graph of \(U(t)\) for Exercise 1
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Solution.
\(U(t) =\frac{4}{\pi}\sum\limits_{n=1}^{\infty }\frac{1}{2n-1}\sin [(2n-1)t]\text{.}\)
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2.
\(V(t)= \begin{cases} \frac{\pi}{2}-t \quad \text{for} \qquad \; 0 \le t \le \pi, \\ \frac{\pi}{2}+t \quad \text{for} \quad -\pi \lt t\lt 0. \qquad \end{cases}\) See Figure 11.1.15.
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Figure 11.1.15. Graph of \(V(t)\) for Exercise 2
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3.
For Exercises 1 and 2, verify that \(U(t) =-V'(t)\) by termwise differentiation of the Fourier series representation for \(V(t)\text{.}\)
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Solution.
\begin{align*} V\,'(t)\amp =\frac{4}{\pi}\sum\limits_{n=1}^{\infty }\frac{1}{(2n-1)^2}\frac{d}{dt}\cos[(2n-1)t]\\ \amp= -\frac{4}{\pi} \sum\limits_{n=1}^{\infty}\frac{1}{2n-1}\sin[(2n-1)t]\\ \amp= -U(t). \end{align*}
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4.
For Exercise 1, set \(t=\frac{\pi}{2}\) and conclude that \(\frac{\pi}{4}=\sum\limits_{j=1}^{\infty}\frac{(-1)^{j-1}}{2j-1}\text{.}\)
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5.
For Exercise 2, set \(t=0\) and conclude that \(\frac{\pi^2}{8} = \sum\limits_{j=1}^{\infty}\frac{1}{(2j-1)^2}\text{.}\)
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Solution.
\(\frac{\pi }{2} = V(0) = \frac{4}{\pi}\sum\limits_{n=1}^{\infty} \frac{1}{(2n-1)^2}\cos(0)= \frac{4}{\pi }\sum\limits_{n=1}^{\infty}\frac{1}{(2n-1)^2} \text{.}\) Now solve for \(\sum\limits_{n=1}^{\infty}\frac{1}{(2n-1)^2}\text{.}\)
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6.
\(U(t)= \begin{cases} -1 \quad \text{for} \qquad \, \frac{\pi}{2} \lt t \lt \pi, \\ \;\;\, 1 \quad \text{for} \quad -\frac{\pi}{2}\lt t \lt \frac{\pi}{2}, \\ -1 \quad \text{for} \quad -\pi \lt t \lt -\frac{\pi}{2}. \qquad \end{cases}\) See Figure 11.1.16.
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Figure 11.1.16. Graph of \(U(t)\) for Exercise 6
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7.
\(U(t)= \begin{cases} \;\;\, \pi -t \quad \text{for} \qquad \frac{\pi}{2}\lt t\lt \pi, \\ \;\;\;\;\;\; t \quad \;\;\; \text{for} \quad -\frac{\pi}{2}\lt t\lt \frac{\pi}{2}, \\ -\pi -t \quad \text{for} \quad -\pi \lt t \lt -\frac{\pi}{2}. \qquad \end{cases}\) See Figure 11.1.17.
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Figure 11.1.17. Graph of \(U(t)\) for Exercise 7
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Solution.
\(U(t)=\frac{4}{\pi}\sum\limits_{n=1}^{\infty}\frac{(-1)^{n-1}}{(2n-1)^2}\sin[(2n-1)t]\text{.}\)
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9.
\(U(t)= \begin{cases} \;\;\, 1 \quad \text{for} \qquad \, \frac{\pi}{2} \lt t \lt \pi, \\ \;\;\, 0 \quad \text{for} \quad -\frac{\pi}{2}\lt t\lt \frac{\pi}{2}, \\ -1 \quad \text{for} \quad -\pi \, \lt t \lt \frac{-\pi}{2}. \qquad \end{cases}\) See Figure 11.1.19.
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Figure 11.1.19. Graph of \(U(t)\) for Exercise 9
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Solution.
\(U(t)=\frac{2}{\pi} \sum\limits_{n=1}^{\infty}\frac{1}{2n-1}\sin[(2n-1)t] - \frac{4}{\pi }\sum\limits_{n=1}^{\infty}\frac{1}{2(2n-1)}\sin[2(2n-1)t],\) where \(a_n=0\) for all \(n\text{,}\) and \(b_{4n}=0\) for all \(n\text{.}\)
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11.
\(U(t)\text{,}\) as defined in Figure 11.1.21.
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Figure 11.1.21. Graph of \(U(t)\) for Exercise 11
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Solution.
\(U(t)=\frac{2}{\pi} \sum\limits_{n=1}^{\infty }\frac{1}{2n-1}\sin[(2n-1)t] + \frac{4}{\pi}\sum\limits_{n=1}^{\infty }\frac{1}{2(2n-1)}\sin[2(2n-1)t],\) where \(a_n=0\) for all \(n\text{,}\) and \(b_{4n}=0\) for all \(n\text{.}\)
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