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

730 Chapter 23 Fourier series
f(t)
f(t)
–a 0
a
t
Figure23.9
A typicalodd function, f (t).
–a 0
Figure23.10
Atypical even function, f (t).
a
t
If f (t)iseven andg(t) isodd, we find
P(−t) = f(−t)g(−t)
= f(t)(−g(t))
= −f(t)g(t)
= −P(t)
so the product is an odd function. These rules are obviously analogous to the rules for
multiplying positive and negative numbers.
Theresults ofExample 23.9 are summarizedthus:
(even) × (even) = even
(odd) ×(odd) =even
(even) × (odd) =odd
23.3.1 Integralpropertiesofevenandoddfunctions
Consider a typical odd function, f (t), such as that shown in Figure 23.9. Suppose we
wish to evaluate ∫ a
f (t)dt where the interval of integration [−a,a] is symmetrical
−a
abouttheverticalaxis.RecallfromChapter13thatadefiniteintegralcanberegardedas
theareaboundedbythegraphoftheintegrandandthehorizontalaxis.Areasabovethe
horizontalaxisarepositivewhilethosebelowarenegative.Weseethatbecausepositive
andnegativecontributionscancel,theintegralofanoddfunctionoveranintervalwhich
issymmetricalaboutthe vertical axiswill bezero.
Example23.10 Evaluate ∫ π
−π tcosnωtdt.
Solution The functiont is odd. The function cosnωt is even and hence the producttcosnωt is
odd. The interval [−π, π] is symmetrical about the vertical axis and hence the required
integral iszero.
Considernowatypicalevenfunction, f (t),suchasthatshowninFigure23.10.Suppose
we wish to evaluate ∫ a
f (t)dt. Clearly the area bounded by the graph and thet axis in
−a