33.6. INTEGRATING AND DIFFERENTIATING FOURIER SERIES 655
Then plugging in π to the Fourier series for G we get
0 = A0 +∞
∑n=1
An (−1)n , A0 =−∞
∑n=1
An (−1)n (*)
Next consider An,n > 0.
LAn =∫ L
−L
∫ x
−L( f (t)−a0)dt cos
(nπxL
)dx
=n−1
∑k=0
∫ xk+1
xk
(∫ xk
−L( f (t)−a0)dt +
∫ x
xk
( f (t)−a0)dt)
cos(nπx
L
)dx
=n−1
∑k=0
∫ xk+1
xk
∫ xk
−L( f (t)−a0)dt cos
(nπxL
)dx
+n−1
∑k=0
(L
nπsin( nπx
L
)∫ xxk( f (t)−a0)dt|xk+1
xk
−∫ xk+1
xkL
nπsin( nπx
L
)( f (x)−a0)dx
)
=n−1
∑k=0
∫ xk+1
xk
G(xk)cos(nπx
L
)dx+
n−1
∑k=0
Lnπ
sin(nπxk+1
L
)(G(xk+1)−G(xk))
−n−1
∑k=0
∫ xk+1
xk
Lnπ
sin(nπx
L
)( f (x)−a0)dx
Now do an integration on the first sum. This yields
Lnπ
n−1
∑k=0
G(xk)(
sin(nπxk+1
L
)− sin
(nπxk
L
))+
Lnπ
n−1
∑k=0
sin(nπxk+1
L
)(G(xk+1)−G(xk))
−n−1
∑k=0
∫ xk+1
xk
Lnπ
sin(nπx
L
)( f (x)−a0)dx
The sums simplify and the result one obtains is
Lnπ
n−1
∑k=0
G(xk+1)sin(nπxk+1
L
)−G(xk)sin
(nπxk
L
)
−n−1
∑k=0
∫ xk+1
xk
Lnπ
sin(nπx
L
)( f (x)−a0)dx
The series telescopes and the result is 0 because G(L) = G(−L) = 0. Thus the result of itall is
LAn = −n−1
∑k=0
∫ xk+1
xk
Lnπ
sin(nπx
L
)( f (x)−a0)dx
=∫ L
−L− L
nπsin(nπx
L
)( f (x)−a0)dx