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446 CHAPTER 13 Fourier Series

for −π< x <π. Differentiate this series term by term to get

∞
n+1
2(−1) cos(nx).
n=1
This series does not converge on (−π,π).

However, under fairly mild conditions, we can integrate a Fourier series term by term.

THEOREM 13.4 Integration of Fourier Series

Let f be piecewise continuous on [−L, L], with Fourier series
∞
1
a 0 + (a n cos(nπx/L) + b n sin(nπx/L)).
2
n=1
Then, for any x with −L ≤ x ≤ L,
1
x
f (t)dt = a 0 (x + L)
2
−L
∞
L 1
n
+ a n sin(nπx/L) − b n (cos(nπx/L) − (−1) ) .
π n
n=1
The expression on the right in this equation is exactly what we obtain by integrating the
Fourier series term by term, from −L to x. This means that we can integrate any piecewise
continuous function f from −L to x by integrating its Fourier series term by term. This is true
even if the function has jump discontinuities and its Fourier series does not converge to f (x) for
all x in [−L, L]. Notice, however, that the result of this integration is not a Fourier series.

EXAMPLE 13.12
From Example 13.11,
∞
2 n+1
f (x) = x = (−1) sin(nx)
n
n=1
on (−π,π). Term by term differentiation results in a series that does not converge on the interval.
However, we can integrate this series term by term:

1
x
2
2
tdt = (x − π )
2
−π
∞ x
2

= (−1) n+1 sin(nt)dt
n
n=1 −π
∞
2 1 1
n+1
= (−1) − cos(nx) + cos(nπ)
n n n
n=1
∞
2
n+1 n
= (−1) (cos(nx) − (−1) ).
n
n=1
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October 14, 2010 14:57 THM/NEIL Page-446 27410_13_ch13_p425-464