15—Fourier Analysis 376In the last line I interchanged the order of integration, and in the preceding line I had to be sure to use
another symbolt′in the second integral, nott. Now do theωintegral.
∫∞
−∞dω
2 π
e−iωt
−mω^2 −ibω+k
eiωt
′
=∫∞
−∞dω
2 π
e−iω(t−t
′)−mω^2 −ibω+k
(15.15)
To do this, use contour integration. The singularities of the integrand are at the roots of the denomi-
nator,−mω^2 −ibω+k= 0. They are
ω=
−ib±
√
−b^2 + 4km
2 m
=ω±
C 1
C 2
Both of these poles are in the lower half complex plane. The contour integralC 1 is along the real axis,
and now I have to decide where to push the contour in order to use the residue theorem. This will be
governed by the exponential,e−iω(t−t
′)
.First take the caset < t′, thene−iω(t−t
′)is of the forme+iω, so in the complexω-plane its
behavior in the±idirections is as a decaying exponential toward+i(∝e−|ω|). It is a rising exponential
toward−i(∝e+|ω|). This means that pushing the contourC 1 up towardC 2 and beyond will make
this integral go to zero. I’ve crossed no singularities, so that means that Eq. (15.15)iszero fort < t′.
Next, the case thatt > t′. Nowe−iω(t−t
′)is of the forme−iω, so its behavior is reversed from
that of the preceding paragraph. It dies off rapidly toward−i∞and rises in the opposite direction.
That means that I must push the contour in the opposite direction, down toC 3 and toC 4. Because
of the decaying exponential, the large arc of the contour that is pushed down to−i∞gives zero for its
integral; the two lines that parallel thei-axis cancel each other; only the two residues remain.
∫∞
−∞dω
2 π
e−iω(t−t
′)−mω^2 −ibω+k
=− 2 πi
∑
ω±ResC 3
C 4
(15.16)
The denominator in Eq. (15.15) is−m(ω−ω+)(ω−ω−). Use this form to compute the residues.
Leave the 1 / 2 πaside 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
′)