1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

12.l • FOURIER SERIES 519

The value of the first integral on the right side of this equation is 27T, and all the
other integrals are zero. Thus,

<l() = ; 11" _.,. u (t) dt.

To determine a,,., we let m (m > 1) denote a fixed integer and multiply both U (t)

and the Fourier series representation in Equation (12-1) by the term cos mt. We

then integrate to obtain

J:,, U(t)cosmt dt = a

2

° [ cosmt dt (12-5)

• ~ = ai 1" .,.cos mt cos jt dt + ~ = bj 1" ,, cos mt sin jt dt.

The value of the first term on the right side of Equation (12-5) is easily seen to

be zero:

ao 1" d _ aosinmt

1

.,, _ 0

2

cos mt t -

2

_.,. -.

_.,, m

(12-6)

We find the value of the term involving cosmtcosjt by using the trigonometric

identity:

cos mt cosjt = ~{cos ((m + j) t) +cos [(m -j) t)}.

Calculation reveals that if m =/= j and m > 0, then

ai L: cosmtcosjt dt =a;' {1-: cos((m+ j)t]dt

• L: cos [(m -j) t] dt} = o.

When m = j, the value of the integral becomes

am L: cos

2

mt dt =?Tam.

(12-7)

(12-8)

We find the value of the term on the right side of Equation (12-5) involving the

integrand cos mt sin jt by using the trigonometric identity

cosmtsinjt = ~ {sin[(m+j)t) +sin[(m-j)t]}.

Then, for all values of m and n, we have

bj L: cosmtsinjtdt= b; {1-: sin ((m+j)t]dt+ 1-: sin[(m- j)t)dt} =O.

(12-9)