15—Fourier Analysis 454

1. Ifx >+athen bothx+aandx−aare positive, which implies that both exponentials vanish rapidly as
   k→+i∞. Push the contourC 2 in this direction and the integrand vanishes exponentially, making the integral
   zero.


2. If−a < x <+a, then onlyx+ais positive. The integral of the first term is then zero by exactly the preceding
   reasoning, but the other term has an exponential that vanishes ask→ −i∞instead, implying that I have to
   push the contour down toward−i∞.

=i

∫

C 3

dk

2 π

1

k

eik(x−a)=

∫

C 4

= +i

1

2 π

(−1)2πiRes

k=0

eik(x−a)

k

=−i

1

2 π

. 2 πi= 1

C 3

C 4

The extra(−1)factor comes because the contour is clockwise.

3. In the third domain,x <−a, both exponentials have the forme−ik, requiring you to push the contour toward
   −i∞. The integrand now has both exponentials, so it is analytic at zero and there is zero residue. The integral
   vanishes and the whole analysis takes you back to the original function, Eq. ( 6 ).
   Another example of a Fourier transform, one that shows up often in quantum mechanics

f(x) =e−x

(^2) /σ 2
, so g(k) =

∫∞

−∞

dxe−ikxe−x

(^2) /σ 2

∫∞

−∞

dxe−ikx−x

(^2) /σ 2
The trick to doing this integral is to complete the square inside the exponent.
−ikx−x^2 /σ^2 =

− 1

σ^2

[

x^2 +σ^2 ikx−σ^4 k^2 /4 +σ^4 k^2 / 4

]

=

− 1

σ^2

[

(x+ikσ^2 /2)^2 +σ^4 k^2 / 4

]

The integral offis now

g(k) =e−σ

(^2) k (^2) / 4

∫∞

−∞

dx′e−x

′ (^2) /σ 2
where x′=x+ikσ/ 2