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Chapter 1 Fourier Series and Integrals 61

∫π

−π

sin(nx)dx= 0

∫π

−π

cos(nx)dx=

{ 0 , n= 0

2 π, n= 0

∫π

−π

sin(nx)cos(mx)dx= 0

∫π

−π

sin(nx)sin(mx)dx=

{ 0 , n=m

π, n=m

∫π

−π

cos(nx)cos(mx)dx=

{ 0 , n=m

π, n=m=0.

Table 1 Orthogonality relations

Each of the terms in the integrated series is zero, so the right-hand side of this
equation reduces to 2π·a 0 ,giving

a 0 =^1

2 π

∫π

−π

f(x)dx.

(In words,a 0 is the mean value off(x)over one period.)
Now multiplying each side of Eq. (1) by sin(mx),wheremis a fixed integer,
and integrating from−πtoπ,wefind

∫π

−π

f(x)sin(mx)dx=

∫π

−π

a 0 sin(mx)dx+

∑∞

n= 1

∫π

−π

ancos(nx)sin(mx)dx

+

∑∞

n= 1

∫π

−π

bnsin(nx)sin(mx)dx.

All terms containinga 0 orandisappear, according to the orthogonality rela-
tions. Furthermore, of all those containing abn, the only one that is not zero is
the one in whichn=m.(Noticethatnis a summation index and runs through
all the integers 1, 2 ,....Wechosemto be a fixed integer, son=monce.) We
now have the formula

bm=

1

π

∫π

−π

f(x)sin(mx)dx.

By multiplying both sides of Eq. (1) by cos(mx)(mis a fixed integer) and
integrating, we also find

am=

1

π

∫π

−π

f(x)cos(mx)dx.