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Introduction to Linear Algebra with Mathematica
Glossary
Preface
Integration of Fourier Series
Theorem 1:
If 2π-periodic function f(๐ฅ) is an absolutely integrable over (โฯ, π) and specified by the giving Fourier series
\[
f(x) \,\sim\, \frac{a_0}{2} + \sum_{n\ge 1} a_n \cos (nx) + b_n \sin (nx) ,
\]
then the integral of ๐(๐ฅ) has the Fourier series expansion as
\begin{equation} \label{RqInt.1}
\int_0^x f(x)\,{\text d}x = \sum_{n\ge 1} \frac{b_n}{n} - \sum_{n\ge 1} \frac{b_n}{n}\,\cos (nx) + \sum_{n\ge 1} \, \frac{a_n - (-1)^{n} a_0 }{n} \, \sin (nx) , \quad -\pi < x < \pi .
\end{equation}
Let us define the antiderivative
\[
F(x) = \int_0^x \left[ f(x) - \frac{a_0}{2} \right] {\text d}x
\]
This function is continuous and has an absolutely integrable derivative. In addition, it is 2π-periodic, as evident from
\begin{align*}
F(x+ 2\pi ) &= \int_0^x \left[ f(x) - \frac{a_0}{2} \right] {\text d}x + \int_x^{x+2\pi} \left[ f(x) - \frac{a_0}{2} \right] {\text d}x
\\
&= F(x) + \int_{-\pi}^{\pi} \left[ f(x) - \frac{a_0}{2} \right] {\text d}x
\\
&= F(x) + \int_{-\pi}^{\pi} f(x)\,{\text d}x - a_0 \pi = F(x) .
\end{align*}
It should be mentioned that in the second step of the above expression, the function ๐(๐ฅ) is treated as a function of period 2π, which ensured by the assumption of periodicity of ๐(๐ฅ). Hence, ๐น(๐ฅ) can be expanded into trigonometric series
\[
F(x) = \frac{A_0}{2} + \sum_{n\ge 1} A_n \cos (nx) + B_n \sin (nx) ,
\]
where for n ≥ 1,
\begin{align*}
A_n &= \frac{1}{\pi}\,\int_{-\pi}^{\pi} F(x) \,\cos (nx)\,{\text d} x
\\
&= \frac{1}{\pi}\,F(x) \,\frac{\sin (nx)}{n} \,\Big\vert_{x=-\pi}^{x=\pi} - \frac{1}{n\pi}\,\int_{-\pi}^{\pi} \left[ f(x) - \frac{a_0}{2} \right] \sin (nx)\,{\text d}x = - \frac{b_n}{n} ,
\end{align*}
and similarly,
\[
B_n = \frac{a_n}{n} .
\]
Therefore, we have
\[
\int_0^x f(x)\,{\text d}x = \frac{a_0}{2}\,x + \frac{A_0}{2} + \sum_{n\ge 1} \frac{-b_n \cos nx + a_n \sin nx}{n} .
\]
Setting ๐ฅ = 0 in this equation, we get
\[
\frac{A_0}{2} = \sum_{n\ge 1} \,\frac{b_n}{n} .
\]
Corollary 1:
If 2π-periodic function f(๐ฅ) is an absolutely integrable over (โฯ, π) and its Fourier coefficient 𝑎₀ = 0, then
\[
\int_0^x f(x)\,{\text d}x = \sum_{n\ge 1} \,\frac{b_n}{n} + \sum_{n\ge 1}\,\frac{-b_n\,\cos nx + a_n \sin nx}{n} ,
\]
that is, the Fourier coefficients of the integral can be obtained from the term-by-term integration of the series for f(๐ฅ).
Note:
If ๐(๐ฅ) is absolutely integrable over (−π, π) and
\[
f(x) = \frac{a_0}{2} + \sum_{n\ge 1} a_n \cos (nx) + b_n \sin (nx) ,
\]
then the integral \( \quad \int_a^b \,f(x)\,{\text d}x \quad \) can be obtained directly from the term-by-term integration of the Fourier series for ๐(๐ฅ), that is,
\begin{equation} \label{EqInt.2}
\int_a^b \,f(x)\,{\text d}x = \frac{a_0}{2} \left( b-a \right) + \sum_{n=1}^{\infty} \frac{a_n \sin nx - b_n \cos nx}{n}\,\Big\vert_{x=a}^{x=b} ,
\end{equation}
where 𝑎 and b represent any two points in the interval [−π, π]. ▣
As a matter of fact, even when 𝑎 or b (or both) falls outside of the interval [−π, π], Eq.\eqref{EqInt.2} is still valid because ๐น(๐ฅ) is a function of period 2π.
Example 4: ■
End of Example 4
Term-by-term integration of infinite series
Lemma 3:
Let f(x) be a periodic integrable function of a real variable. The antiderivative F of f defined by
\[
F(x) = \int_0^x f(t)\,{\text d} t
\]
is T-periodic if and only if
\[
\int_0^T f(x)\,{\text d}x = 0.
\]
Suppose that the integral condition is satisfied. We need to show that the antiderivative is periodic,, F(x + T) = F(x). We have
This yields the main statement about term-by-term integration of Fourier series.
\[
F(x+T) - F(x) = \int_0^{x+T} f(t)\,{\text d}t - \int_0^{x} f(t)\,{\text d}t = \int_x^{x+T} f(t)\,{\text d}t = \int_0^{T} f(t)\,{\text d}t = 0.
\]
Conversely, if the antiderivative is T-periodic, then
\[
F(T) - F(0) = \int_0^{T} f(t)\,{\text d}t
\]
Theorem 15:
Let f(x) be a periodic with period 2π integrable function satisfying condition
\[
\int_0^{2\pi} f(x)\,{\text d}x = 0 \qquad \Longrightarrow \qquad a_0 (f) = 0 .
\]
Consider the antiderivative of f defined by \( F(x) = \int_0^x f(t)\,{\text d} t . \) Then the Fourierseries of F is obtained from that of f by termwise integration. That is, if
\[
f(x) \sim \sum_{k\ge 1} a_k \cos kx + b_k \sin nx ,
\]
then
\[
F(x) = \int_0^x f(t)\,{\text d}t \sim A_0 + \sum_{k\ge 1} \frac{-b_k}{k}\, \cos kx + \frac{a_k}{k}\, \sin nx ,
\]
where \( \displaystyle A_0 = \sum_{k\ge 1} \frac{b_k}{k} . \)
The antiderivative F is continuous and it is also 2ฯ-periodic function according to the previous lemma. Since Fโฒ = f is piecewise smooth, then we can apply the previous Theorem aboutdifferentiation of Fourier series. Consider the Fourier series of F:
\[
F(x) = A_0 + \sum_{k\ge 1} A_k \cos (kx) + k\,B_k \sin (kx) .
\]
Then its derivative becomes
\[
F'(x) = f(x) = \sum_{k\ge 1} k\. B_k \cos (kx) - k\, \frac{a_k}{k}\,\sin (kx) .
\]
Hence,
\[
k\, B_k = a_k , \qquad -k\,A_k = b_k , \qquad k=1,2,3,\kdots .
\]
The coefficient A0 can be found by using the initial condition F(0) = 0 and equating it to the value of the Fourier series at the origin.
Corollary 2:
If function f(x) has a Fourier series \( S[f] = \frac{1}{2}\,a_0 + \sum_{k\ge 1} a_k \cos kx + b_k \sin kx , \) then
\[
\sum_{k\ge 1} \frac{b_k}{k} = \frac{1}{2\pi} \int_0^{2\pi} f(t)\left( \pi - t \right) {\text d}t .
\]
The formula follows from the identity
\[
S^{\ast} {f} = \sum_{\k \ge 1} \left( a_k \sin kx - b_k \cos kx \right) = \frac{1}{\pi} \int_0^{2\pi} f(t)\, \sum_{k\ge 1} \frac{\sin k\left( x-t \right)}{k} \, {\text d}t .
\]
Example 22: Let us consider the function f(x) = |x| that has Fourier series
\[
|x| = \frac{\pi}{2} - \frac{4}{\pi} \sum_{k\ge 1} \frac{1}{(2k-1)^2}\,\cos (2k-1)x , \qquad |x| < \pi .
\]
Integrate[Abs[x], {x, -Pi, Pi}]/Pi
Pi
Integrate[Abs[x]*Cos[(2*k-1)*x], {x, -Pi, Pi}]/Pi
(1/((-1 + 2 k)^2 \[Pi]))(-1 - 2 Cos[k \[Pi]]^2 - Cos[2 k \[Pi]] +
2 \[Pi] Cos[k \[Pi]] Sin[k \[Pi]] -
4 k \[Pi] Cos[k \[Pi]] Sin[k \[Pi]] + \[Pi] Sin[2 k \[Pi]] -
2 k \[Pi] Sin[2 k \[Pi]])
Integrating both sides, we get
\[
\int_0^x |t|\.{\text d}t = \frac{\pi}{2}\,x - \frac{4}{\pi} \sum_{k\ge 1} \frac{1}{(2k-1)^3}\,\sin (2k-1)x , \qquad |x| < \pi .
\]
■
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