Mathematical Physics Vol 1

Chapter 6. Trigonometric Fourier series. Fourier integral

326

Bu introducing the substitution

1 n

1 n

nx = y ⇒ x =

y dx =

dy

we obtain

1 π n 2 Z

n π

a n =

y cos ydy .

− n π

Partial integration

y = u , du = dy , cos ydy = dv , v = sin y

yields

y sin y   n π − n π

1 n 2 π 1 n 2 π

sin ydy =

− Z

n π

a n =

− n π

n π sin n π − n π sin ( − n π )+ cos y   n π − n π = ( cos n π − cos ( − n π ))= 1 n 2 π

=

1 n 2 π

( cos n π − cos n π )= 0 .

=

For b n , we obtain, in a similar way

1 π Z

1 n 2 π Z

π

n π

b n =

x sin nxdx = y cos y   n π − n π

y sin ydy =

− π 1 n 2 π − 1 n 2 π −

− n π

cos ydy =

+ Z

n π

=

− n π

n π cos n π − n π cos ( − n π )+ sin y   n π − n π = ( − 2 n π cos n π + sin n π + sin n π )= 1 n 2 π

=

1 n 2 π

2 n

( − 2 n π ( − 1 ) n )=

( − 1 ) n + 1 .

=

∞ ∑ n = 1

( − 1 ) n + 1 n

f ( x )= 2

sin nx .

(6.48)

R Note. The function f ( x )= x is odd and its expansion contains only coefficients next to the odd sine function.

Problem 219 Expand the function f ( x )= | x | into a Fourier series in the interval ( − π , π ) .

Solution

x 2   π 0

π Z

π Z 0

1 π

2 π

2 π

1 2

= π ,

a 0 =

| x | d x =

xdx =

− π