Radial Functions and The Fourier Transform: 1 Area of A Sphere

Radial functions and the Fourier transform
Notes for Math 583A, Fall 2008

December 6, 2008
1 Area of a sphere
The volume in n dimensions is
vol = dn x = dx1 · · · dxn = rn−1 dr dn−1 ω. (1)
Here r = |x| is the radius, and ω = x/r it a radial unit vector. Also dn−1 ω
denotes the angular integral. For instance, when n = 2 it is dθ for 0 ≤ θ ≤ 2π,
while for n = 3 it is sin(θ) dθ dφ for 0 ≤ θ ≤ π and 0 ≤ φ ≤ 2π.
The radial component of the volume gives the area of the sphere. The radial
directional derivative along the unit vector ω = x/r may be denoted
1 ∂ ∂ ∂
ωd = (x1 + · · · + xn )= . (2)
r ∂x1 ∂xn ∂r
The corresponding spherical area is
ωdcvol = rn−1 dn−1 ω. (3)
Thus when n = 2 it is (1/r)(x dy−y dx) = r dθ, while for n = 3 it is (1/r)(x dy dz+
y dz dx + x dx dy) = r2 sin(θ) dθ dφ.
The divergence theorem for the ball Br of radius r is thus
Z Z
n
div v d x = v · ω rn−1 dn−1 ω. (4)
Br Sr
Notice that if one takes v = x, then div x = n, while x · ω = r. This shows that
n times the volume of the ball is rn times the surface areaR∞of the sphere.
Recall that the Gamma function is defined by Γ(z) = 0 tz e−t dt t . It is easy
to show that Γ(z + 1) = zΓ(z). Since Γ(1) = 1, it follows that Γ(n) = (n − 1)!.
1
The result Γ( 21 ) = π 2 follows reduction to a Gaussian integral. It follows that
1
Γ( 23 ) = 12 π 2 .
Theorem 1 The area of the unit sphere Sn−1 ⊆ Rn is

n
2π 2
ωn−1 = . (5)
Γ( n2 )
1
Thus in 3 dimensions the area of the sphere is ω2 = 4π, while in 2 dimensions
the circumference of the circle is ω1 = 2π. In 1 dimension the two points get
count ω0 = 2.
To prove this theorem, consider the Gaussian integral
Z
n x2
(2π)− 2 e− 2 dn x = 1. (6)
Rn
In polar coordinates this is

Z ∞
n r2
ωn−1 (2π)− 2 e− 2 rn−1 dr = 1. (7)
0
Let u = r2 /2. Then this is

Z ∞
n n−2 n−2
ωn−1 (2π)− 2 2 2 e−u u 2 du = 1. (8)
0
That is
n n
ωn−1 π − 2 2−1 Γ( ) = 1. (9)
2
This gives the result.
2 Fourier transform of a power

Theorem 2 Let 1 < a < n. The Fourier transform of 1/|x|a is Ca /|k|n−a ,
where n−a
n−a
n 2 2 Γ(
2 )
Ca = (2π) 2 a . (10)
2 2 Γ( a2 )
This is not too difficult. It is clear from scaling that the Fourier transform
of 1/|x|a is C/|k|n−a . It remains to evaluate the constant C.
Take the inner product with the Gaussian. This gives
Z Z
n x2 1 n x2 1
(2π)− 2 e− 2 a
d x = (2π)−n e− 2 C n−a dn k. (11)
Rn |x| Rn |k|
Writing this in polar coordinates gives

Z ∞ 2
Z ∞
r2
−n − r2 n−1−α −n
(2π) 2 e r dr = C(2π) e− 2 r1−α dr. (12)
0 0
This in turn gives

n n−a−2 n−a a−2 a
(2π)− 2 2 2 Γ( ) = C(2π)−n 2 2 Γ( ). (13)
2 2
2
3 The Hankel transform
Define the Bessel function
Z π
tν
Jν (t) = ω2ν e−it cos(θ) sin(θ)2ν dθ. (14)
(2π)ν+1 0
This makes sense for all real numbers ν ≥ 0, but we shall be interested mainly
in the cases when ν is an integer or ν is a half-integer. In the case when ν is a
half-integer the exponent 2ν is odd, and so it is possible to evaluate the integral
in terms of elementary functions. Thus, for example,
1 Z π 1
t2 −it sin(θ) t2 sin(t)
J 12 (t) = 1 2π e sin(θ) dθ = 1 2 . (15)
(2π) 2 0 (2π) 2 t
This is not possible when ν is an integer. Thus for ν = 0 we have the relatively
mysterious expression Z
1 π it cos(θ)
J0 (t) = e dθ. (16)
π 0
Fix a value of ν. If we consider a function g(r), its Hankel transform is the
function ĝν (s) given by
Z ∞
ĝν (s) = Jν (sr)g(r)r dr. (17)
0
We shall see that the Hankel transform is related to the Fourier transform.
4 The radial Fourier transform

The first result is that the radial Fourier transform is given by a Hankel trans-
form. Suppose f is a function on Rn . Its Fourier transform is
Z
fˆ(k) = e−ik·x f (x) dn x. (18)
Let r = |x| and s = |k|. Write f (x) = F (r) and fˆ(k) = Fn (s).
Theorem 3 The radial Fourier transform in n dimensions is given in terms of

the Hankel transform by
Z ∞
n−2 n n−2
s 2 F̂n (s) = (2π) 2 J n−2 (sr)r 2 F (r)r dr. (19)
2
0
Here is the proof of the theorem. Introduce polar coordinates with the z
axis along k, so that k · x = sr cos(θ). Suppose that the function is radial, that
is, f (x) = F (r).
Z ∞Z π
ˆ
f (k) = F̂n (s) = e−isr cos(θ) F (r)ωn−2 sin(θ)n−2 dθrn−1 dr. (20)
0 0
3
Use n−2 Z π
t 2
J n−2 (t) = n ωn−2 e−it cos(θ) sin(θ)n−2 dθ. (21)
2 (2π) 2 0
For the case n = 3 the Bessel function has order 1/2 and has the above
expression in terms of elementary functions. So
Z ∞
sin(sr)
F̂3 (s) = 4π F (r)r2 dr. (22)
0 sr
For n = 2 the Bessel function has order 0. We get

Z ∞
F̂2 (s) = 2π J0 (sr)F (r)r dr. (23)
0

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Radial functions and the Fourier transform

Notes for Math 583A, Fall 2008

vol = dn x = dx1 · · · dxn = rn−1 dr dn−1 ω. (1)

ωdcvol = rn−1 dn−1 ω. (3)

Theorem 1 The area of the unit sphere Sn−1 ⊆ Rn is

In polar coordinates this is

Let u = r2 /2. Then this is

2 Fourier transform of a power

Writing this in polar coordinates gives

This in turn gives

4 The radial Fourier transform

Theorem 3 The radial Fourier transform in n dimensions is given in terms of

For n = 2 the Bessel function has order 0. We get

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