11.5 Integrating And Differentiating Fourier Series
First here is a definition of what it means for a function to be piecewise continuous.
Definition 11.5.1Let f be a function defined on
[a,b]
. It is called piecewisecontinuousif there is a partition of
[a,b]
,
{x0,⋅⋅⋅,xn}
such that on
[xk−1,xk]
there isa continuous function g_{k}such that f
(x)
= g_{k}
(x)
for all x ∈
(xk−1,xk)
.
You can typically integrate Fourier series term by term and things will work out according
to your expectations. More precisely, if the Fourier series of f is
∞
∑ a eikx
k=−∞ k
then it will be true for x ∈
[− π,π ]
that
∫ x ∑n ∫ x
F (x) ≡ f (t)dt = lnim→∞ ak eiktdt
−π k=−n −π
I shall show this is true for the case where f is an arbitrary 2π periodic function which is
piecewise continuous according to the above definition. However, with a better
theory of integration, it all works for much more general functions than these. It is
limited here to a simpler case because we don’t have a very sophisticated theory of
integration.
Note that it is not necessary to assume anything about the function f being the limit of
its Fourier series. Let
∫ x
G (x) ≡ F (x)− a0(x+ π) = (f (t)− a0)dt
−π
Then G equals 0 at −π and π because
∫
2πa = π f (t)dt
0 −π
Therefore, the periodic extension of G, still denoted as G, is continuous. Also
|∫ x |
|G (x) − G (x1)| ≤ || M dt||≤ M |x − x1|
|x1 |
where M is an upper bound for
|f (t)− a0|
. Thus the Dini condition of Corollary 11.4.3 holds.
Therefore for all x ∈ ℝ,
∑∞
G (x) = Akeikx (11.21)
k=−∞
(11.21)
where
1 ∫ π
Ak = --- G(x)eikxdx
2π −π
Now from 11.21 and the definition of the Fourier coefficients for f,
∑n
G (π ) = F (π )− a02π = 0 = A0 + lim Ak (− 1)k (11.22)
n→ ∞k=− n,k⁄=0
(11.22)
and so
∑n
A0 = − lim Ak (− 1)k
n→ ∞ k= −n,k⁄=0
Next consider A_{k} for k≠0.
1 ∫ π 1 ∫ π ∫ x
Ak ≡ --- G (x )e− ikxdx ≡ --- (f (t) − a0)dte−ikxdx
2π − π 2π −π −π
By assumption, f is the restriction of a continuous function on
[xi−1,xi]
, and so it follows
that
(t,x) → (f (t)− a )e−ikx
0
is continuous on
[xi−1,xi]
×
[− π,π]
. Therefore, from Problem 48 on Page 627, we can
interchange the order of integration on this rectangle, similarly on any other such rectangle
where the function is continuous. Actually, it is even easier than this. The interchange is
obvious because the function is the product of a function of t with a function of x. It is
nothing more than the observation that
( )
∫ b∫ d ∫ b ∫ d
a c f (x)g(y)dxdy = a g (y) c f (x)dx dy
( ∫ ) ( ∫ )
d b
= c f (x)dx a g(y)dy
with the same result in the other order.
Thus
-1-∫ π∫ x −ikx
Ak = 2π −π − π(f (t)− a0) dte dx
n−1∫ x ( ∫ x ∫ x )
= -1-∑ j+1 (f (t)− a0)e−ikxdt+ j(f (t)− a0)e−ikxdt dx
2π j=0 xj xj −π
A B
1 n−∑ 1∫ xj+1 −ikx∫ x 1 n∑−1∫ xj+1∫ xj −ikx
2π- e (f (t)− a0)dtdx + 2π (f (t)− a0)e dtdx
j=0 xj xj j=0 xj −π
( ∫ ∫ )
-1-n∑−1 e−ikx x xj+1 xj+1e−ikx
A = 2π − ik xj (f (t)− a0)dt|xj + xj ik (f (x) − a0)dx
j=0 ∫ ∫
-1-n∑−1 e−-ikxj+1- xj+1 -1- π e−ikx-
= 2π − ik xj (f (t)− a0)dt+ 2π −π ik f (x)dx
j=0
( ◜------+A◞◟0------◝)
| ∑n ∑n |
F (x)− a0(x+ π) = lim || ak eikx− ak(− 1)k||
n→∞ ( k= −n,k⁄=0 ik k=−n,k⁄=0 ik )
and so
∫ x ∫ x ∑n ak( ikx k)
F (x) = − πf (t)dt = −π a0dt + ln→im∞ ik e − (− 1)
∫ ∫ k=−n,k⁄=0
x ∑n x ikt
= − πa0dt+ lnim→∞ ak −π e dt
k=−n,k⁄=0
This proves the following theorem.
Theorem 11.5.2Let f be 2π periodic and piecewise continuous. Then
∫ x ∫ x ∑n ∫ x
f (t)dt = a0dt+ lim ak eiktdt
−π −π n→ ∞ k= −n,k⁄=0 −π
where a_{k}are the Fourier coefficients of f.
Example 11.5.3Let f
(x)
= x for x ∈ [−π,π) and extend f to make it 2π periodic. Thenthe Fourier coefficients of f are
(− 1)k i
a0 = 0,ak =---k--
Therefore,
1-
2π
∫_{−π}^{π}te^{−ikt} =
-i
k
cosπk
∫ x 1 2 1 2
tdt = 2x − 2π
−π
∑n k ∫ x
= lim (−-1)-i eiktdt
n→ ∞ k=−n,k⁄=0 k −π
∑n k ( k)
= lim (−-1)-i sinxk-+ i−-cosxk+-(− 1)
n→ ∞ k=−n,k⁄=0 k k k
For fun, let x = 0 and conclude
∑n k ( k)
− 1π2 = lim (−-1)-i i−-1+-(− 1)
2 n→ ∞k=− n,k⁄=0 k k
∑n k+1 ( k)
= lim (−-1)--- −-1+-(− 1)
n→ ∞k=− n,k⁄=0 k k
∑n k ∞∑
= lim 2 (− 1)-+2(− 1)-= ---− 4-2
n→ ∞ k=1 k k=1(2k− 1)
and so
∞
π2-= ∑ ----1---
8 k=1(2k− 1)2
Of course it is not reasonable to suppose that you can differentiate a Fourier series term
by term and get good results.
Consider the series for f
(x )
= 1 if x ∈ (0,π] and f
(x)
= −1 on
(− π,0)
with f
(0)
= 0. In
this case a_{0} = 0.
( ∫ ∫ )
1-- π −ikt 0 −ikt icosπk-−-1
ak = 2π 0 e dt− −π e dt = π k
so the Fourier series is
∑ ( k )
(−-1)--− 1 ieikx
k⁄=0 πk
What happens if you differentiate it term by term? It gives
∑ (− 1)k − 1
− --------eikx
k⁄=0 π
which fails to converge anywhere because the k^{th} term fails to converge to 0. This is in spite
of the fact that f has a derivative away from 0.
However, it is possible to prove some theorems which let you differentiate a Fourier series
term by term. Here is one such theorem.
Theorem 11.5.4Suppose for x ∈
[− π,π]
∫ x
f (x) = f′(t)dt+ f (− π)
− π
and f^{′}
(t)
is piecewise continuous. Denoting by f the periodic extension of the above, then if
∑∞ ikx
f (x) = ake
k=−∞
it follows the Fourier series of f^{′}is
∞∑ ikx
akike .
k=−∞
Proof: Since f^{′} is piecewise continuous, 2π periodic it follows from Theorem
11.5.2
(∫ )
∑∞ x ikt
f (x)− f (− π ) = bk − πe dt
k=−∞
where b_{k} is the k^{th} Fourier coefficient of f^{′}. Thus
∫
b = 1-- π f′(t)e−iktdt
k 2π −π
Breaking the integral into pieces if necessary, and integrating these by parts yields finally
1 [ ∫ π ]
bk = ---f (t)e−ikt|π−π + ik f (t)e−iktdt
2π ∫ π −π
= ik 1-- f (t)e−iktdt = ikak
2π −π
where a_{k} is the Fourier coefficient of f. Since f is periodic of period 2π, the boundary term
vanishes. It follows the Fourier series for f^{′} is
∑∞
ikakeikx
k=−∞
as claimed. ■
Note the conclusion of this theorem is only about the Fourier series of f^{′}. It does not say
the Fourier series of f^{′} converges pointwise to f^{′}. However, if f^{′} satisfies a Dini
condition, then this will also occur. For example, if f^{′} has a bounded derivative
at every point, then by the mean value theorem
|f′(x)− f′(y)|
≤ K
|x − y|
and
this is enough to show the Fourier series converges to f^{′}