C+
C−
ω=−iFigure 12.2: Evaluatingθ(t) via contour integration.Fort > 0 , one instead closes the path usingC−in the lower half-plane, and
finds that the integral can be evaluated in terms of the residue of the pole at
ω=−i(with the minus sign coming from orientation of the curve), giving
θ(t) =− 2 πi(
i
2 π)
= 1
By similar arguments one can show thatθ(−t) has (as a distribution) Fourier
transform
lim
→ 0 +−
i
√
2 π1
ω−iand the integral representation
θ(−t) = lim
→ 0 +−
i
2 π∫+∞
−∞1
ω−i
e−iωtdωTaking 1/
√
2 πtimes the sum of the Fourier transforms forθ(t) andθ(−t) gives
the distribution
lim
→ 0 +i
2 π(
1
ω+i−
1
ω−i)
= lim
→ 0 +i
2 π− 2 i
ω^2 +^2= lim
→ 0 +1
πω^2 +^2
=δ(ω) (12.11)