15—Fourier Analysis 461
∫∞
−∞
dω
2 π
e−iω(t−t
′)
−mω^2 −ibω+k
=− 2 πi
∑
ω±
Res
C 3
C 4
(15)
The denominator in Eq. ( 14 ) is−m(ω−ω+)(ω−ω−). Use this form to compute the residues. Leave the 1 / 2 π
aside for the moment and you have
e−iω(t−t
′)
−mω^2 −ibω+k
=
e−iω(t−t
′)
−m(ω−ω+)(ω−ω−)
The residues of this atω±are the coefficients of these first order poles.
atω+:
e−iω+(t−t
′)
−m(ω+−ω−)
and atω−:
e−iω−(t−t
′)
−m(ω−−ω+)
The explicit values ofω±are
ω+=
−ib+
√
−b^2 + 4km
2 m
and ω−=
−ib−
√
−b^2 + 4km
2 m
Let ω′=
√
−b^2 + 4km
2 m
and γ=
b
2 m
The difference that appears in the preceding equation is then
ω+−ω−= (ω′−iγ)−(−ω′−iγ) = 2ω′