1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

98 Chapter 1 Fourier Series and Integrals

The addition formula for sines gives the equality

sin

((

N+^1

2

)

y

)

=cos(Ny)sin

( 1

2

y

)

+sin(Ny)cos

( 1

2

y

)

.

Substituting it in Eq. (13) and using simple properties of integrals, we ob-

tain

SN(x)−f(x)=

1

π

∫π

−π

(

f(x+y)−f(x)

) 1

2 cos(Ny)dy

+^1

π

∫π

−π

(

f(x+y)−f(x)

)cos(^12 y)

2sin(^12 y)

sin(Ny)dy. (14)

ThefirstintegralinEq.(14)canberecognizedastheFouriercosinecoeffi-

cient of the function

ψ(y)=^1

2

(

f(x+y)−f(x)

)

. (15)

Sincefis a sectionally smooth function, so isψ, and the first integral has

limit 0 asNincreases, by Lemma 3.

The second integral in Eq. (14) can also be recognized, as the Fourier sine

coefficient of the function

φ(y)=

f(x+y)−f(x)

2sin(^12 y) cos

( 1

2 y

)

. (16)

To proceed as before, we must show thatφ(y)is at least sectionally contin-

uous,−π≤y≤π. The only difficulty is to show that the apparent division

by 0 aty=0doesnotcauseφ(y)to have a bad discontinuity there.

First, iffis continuous and differentiable nearx,thenf(x+y)−f(x)is

continuous and differentiable neary=0. Then L’Hôpital’s rule gives

limy→ 0

f(x+y)−f(x)

2sin(^12 y) =limy→^0

f′(x+y)

cos(^12 y) =f

′(x). (17)

Under these conditions, the functionφ(y)of Eq. (16) has a removable dis-

continuity aty=0 and thus is sectionally continuous.

Second, iffis continuous atxbut has a corner there, thenf(x+y)−f(x)

is continuous with a corner aty=0. In this case, L’Hôpital’s rule applies

with the one-sided limits, which show

y→lim 0 +f(x+y)−f(x)

2sin(^12 y)

=ylim→ 0 +f

′(x+y)

cos(^12 y)

=f′(x+), (18)

y→lim 0 −f(x+y)−f(x)

2sin(^12 y)

=ylim→ 0 −f

′(x+y)

cos(^12 y)

=f′(x−). (19)