Advanced book on Mathematics Olympiad

3.2 Continuity, Derivatives, and Integrals 165

This expansion is unique, and

a 0 =

1

2 π

∫T

0

f(x)dx,

an=

1

π

∫T

0

f(x)cos

2 nπ

T

xdx,

bn=

1

π

∫T

0

f(x)sin

2 nπ

T

xdx.

Of course, we can requirefto be defined only on an interval of lengthT, and then
extend it periodically, but if the values offat the endpoints of the interval differ, then
the convergence of the series is guaranteed only in the interior of the interval.
Let us discuss a problem from the Soviet Union University Student Contest.

Example.Compute the sum

∑∞

n= 1

cosn

1 +n^2

.

Solution.The sum looks like a Fourier series evaluated at 1. For this reason we concen-
trate on the general series

∑∞

n= 0

1

n^2 + 1

cosnx.

The coefficientsn (^21) + 1 should remind us of the integration formulas
∫
excosnxdx=

1

n^2 + 1

ex(cosnx+nsinnx),

∫

exsinnxdx=

n

n^2 + 1

ex(sinnx+ncosnx).

These give rise to the Fourier series expansion

ex=

1

2 π

(e^2 π− 1 )+

1

π

(e^2 π− 1 )

∑∞

n= 1

1

n^2 + 1

cosnx+

1

π

(e^2 π− 1 )

∑∞

n= 1

n

n^2 + 1

sinnx,

which holds true forx∈( 0 , 2 π). Similarly, fore−xandx∈( 0 , 2 π), we have

e−x=

1

2 π

( 1 −e−^2 π)+

1

π

( 1 −e−^2 π)

∑∞

n= 1

1

n^2 + 1

cosnx

−

1

π

( 1 −e^2 π)

∑∞

n= 1

n

n^2 + 1

sinnx.