Page 467 - Advanced engineering mathematics

13.4 Integration and Differentiation of Fourier Series 447

Proof Define
x 1

F(x) = f (t)dt − a 0 x
2
−L
for −L ≤ x ≤ L. Then F is continuous on [−L, L] and
1
F(−L) = F(L) = a 0 L.
2
Furthermore,
1

F (x) = f (x) − a 0
2
at every point on (−L, L) at which f is continuous. Therefore F (x) is piecewise continuous on

[−L, L], and the Fourier series of F(x) converges to F(x) on this interval. This series is
1 nπx nπx
∞
F(x) = A 0 + A n cos + B n sin .
2 L L
n=1
Here we obtain, using integration by parts,
1 L nπx
A n = F(t)cos dt
L −L L

1 L nπx L 1 L L nπx

= F(t) sin − sin F (t)dt
L nπ L L −L nπ L
−L
1 L 1 nπx
=− f (t) − a 0 sin dt
nπ −L 2 L
1 L nπx 1 L nπx
=− f (t)sin dt + a 0 sin dt
nπ −L L 2nπ −L L
L
=− b n .
nπ
Similarly,
1 L nπx L
B n = F(t)sin dt = a n ,
L −L L nπ

where the a n s and b n s are the Fourier coefficients of f (x) on [−L, L]. Therefore the Fourier

series of F(x) has the form
∞
1 L 1 nπx nπx
F(x) = A 0 + −b n cos + a n sin
2 π n L L
n=1
for −L ≤ x ≤ L. To determine A 0 , write
∞
L L
F(L) = a 0 − b n cos(nπ)
2 π
n=1
∞
1 L 1
n
= A 0 − b n (−1) .
2 π n
n=1
This gives us
∞
2L 1
n
A 0 = La 0 + b n (−1) .
π n
n=1
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October 14, 2010 14:57 THM/NEIL Page-447 27410_13_ch13_p425-464