15—Fourier Analysis 371For a given value ofk, define the integral
gL(k) =
∫L
−Ldx′e−ikx
′f(x′)
If the functionfvanishes sufficiently fast asx′→∞, this integral will have a limit asL→∞. Call that
limitg(k). Look back at Eq. (15.3) and you see that for largeLthe last factor will be approximately
g(kn), where the approximation becomes exact asL→∞. Rewrite that expression as
f(x)≈
1
2 π
∑∞
n=−∞eiknx∆kng(kn) (15.4)
AsL→∞, you have∆kn→ 0 , and that turns Eq. (15.4) into an integral.
f(x) =
∫∞
−∞dk
2 π
eikxg(k), where g(k) =
∫∞
−∞dxe−ikxf(x) (15.5)
The functiongis called* the Fourier transform off, andfis the inverse Fourier transform ofg.
Examples
For an example, take the step function
f(x) =
{
1 (−a < x < a)
0 (elsewhere)theng(k) =
∫a−adxe−ikx 1
=
1
−ik
[
e−ika−e+ika
]
=
2 sinka
k
(15.6)
The first observation is of course that the dimensions check: Ifdxis a length then so is 1 /k.
After that, there is only one parameter that you can vary, and that’sa. Asaincreases, obviously the
width of the functionfincreases, but now look atg. The first place whereg(k) = 0is atka=π.
This value,π/adecreasesasaincreases. Asfgets broader,ggets narrower (and taller). This is a
general property of these Fourier transform pairs.
Can you invert this Fourier transform, evaluating the integral ofgto get back tof? Yes, using
the method of contour integration this is very easy. Without contour integration it would be extremely
difficult, and that is typically the case with these transforms; complex variable methods are essential
to get anywhere with them. The same statement holds with many other transforms (Laplace, Radon,
Mellin, Hilbert,etc.)
The inverse transform is
∫∞−∞dk
2 π
eikx
2 sinka
k
=
∫
C 1dk
2 π
eikx
eika−e−ika
ik
=−i
∫
C 2dk
2 π
1
k
[
eik(x+a)−eik(x−a)
]
C 1
C 2
* Another common notation is to definegwith an integral dx/
√
2 π. That will require a cor-
responding dk/
√
2 πin the inverse relation. It’s more symmetric that way, but I prefer the other
convention.