Advanced book on Mathematics Olympiad

Real Analysis 553

bm=

1

π

∫ 2 π

0

xsinmxdx=−

xcosmx

mπ

∣∣

∣

2 π

0

+

1

mπ

∫ 2 π

0

cosmxdx=−

2

m

, form≥ 1.

Therefore,

x=π−

2

1

sinx−

2

2

sin 2x−

2

3

sin 3x−···.

Divide this by 2 to obtain the identity from the statement. Substitutingx=π 2 , we obtain
the Leibniz series

π

4

= 1 −

1

3

+

1

5

−

1

7

+···.

In the series

π−x

2

=

∑∞

n= 1

sinnx

n

,

replacexby 2x, and then divide by 2 to obtain

π

4

−

x

2

=

∑∞

k= 1

sin 2kx

2 k

,x∈( 0 ,π).

Subtracting this from the original formula, we obtain

π

4

=

∑∞

k= 1

sin( 2 k− 1 )x

2 k− 1

,x∈( 0 ,π).

494.One computes
∫ 1

0

f(x)dx= 0 ,

∫ 1

0

f(x)cos 2πnxdx= 0 , for alln≥ 1 ,

∫ 1

0

f(x)sin 2πnxdx=

1

2 πk

, for alln≥ 1.

Recall that for a general Fourier expansion

f(x)=a 0 +

∑∞

n= 1

(

ancos

2 π

T

nx+bnsin

2 π

T

nx

)

,

one has Parseval’s identity