082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

482 Chapter 16 Further topics in integration
Inasimilarmanner, itiseasytoshow
∫ π
1cos(nt)dt = 0
−π
Also,usingthe trigonometric identities inSection 3.6
∫ π
=
−πcos(nt)sin(mt)dt 1 ∫ π
sin(n +m)t −sin(n −m)tdt
2 −π
We have seen that ∫ π
−πsinnt dt = 0 for any n ∈ N. Noting that (n +m) ∈ N and
(n −m) ∈ N,weseethat
∫ π
−π
sin(n +m)t −sin(n −m)tdt =0
Itis left as anexercise for the reader toshow that
∫ π
−π
∫ π
−π
sinntsinmtdt =0 n≠m
cosntcosmtdt =0
n≠m
The functions thus form an orthogonal set across [−π,π].
TheresultofExample 16.3 can beextended:
{1,sint,cost,sin2t,cos2t,...}isanorthogonalsetoveranyintervaloflength2π.
More generally:
{
1,sin
( 2πt
T
)
,cos
( 2πt
T
)
,sin
( 4πt
T
)
,cos
( 4πt
T
) }
,...
isanorthogonalsetoveranyintervaloflengthT.Inparticular,thesetisorthogonal
[
over [0,T] and − T ]
2 ,T .
2
These results areused extensively inFourier analysis.
Example16.4 Find
(a)
(b)
∫ π
−π
∫ π
−π
sin 2 (nt)dt n∈ Z n ≠0
cos 2 (nt)dt n∈ Z n ≠0