Mathematical Methods for Physics and Engineering : A Comprehensive Guide

INTEGRAL TRANSFORMS

two- or three-dimensional functions of position. For example, in three dimensions

we can define the Fourier transform off(x, y, z)as

̃f(kx,ky,kz)=^1

(2π)^3 /^2

∫∫∫

f(x, y, z)e−ikxxe−ikyye−ikzzdx dy dz, (13.47)

and its inverse as

f(x, y, z)=

1

(2π)^3 /^2

∫∫∫

̃f(kx,ky,kz)eikxxeikyyeikzzdkxdkydkz. (13.48)

Denoting the vector with componentskx,ky,kzbykand that with components

x, y, zbyr, we can write the Fourier transform pair (13.47), (13.48) as

̃f(k)=^1

(2π)^3 /^2

∫

f(r)e−ik·rd^3 r, (13.49)

f(r)=

1

(2π)^3 /^2

∫

̃f(k)eik·rd^3 k. (13.50)

From these relations we may deduce that the three-dimensional Diracδ-function

canbewrittenas

δ(r)=

1

(2π)^3

∫

eik·rd^3 k. (13.51)

Similar relations to (13.49), (13.50) and (13.51) exist for spaces of other dimen-

sionalities.

In three-dimensional space a functionf(r)possesses spherical symmetry, so thatf(r)=

f(r). Find the Fourier transform off(r)as a one-dimensional integral.

Let us choose spherical polar coordinates in which the vectorkof the Fourier transform
lies along the polar axis (θ=0).Thiswecandosincef(r) is spherically symmetric. We
then have

d^3 r=r^2 sinθdrdθdφ and k·r=krcosθ,

wherek=|k|. The Fourier transform is then given by

̃f(k)=^1

(2π)^3 /^2

∫

f(r)e−ik·rd^3 r

=

1

(2π)^3 /^2

∫∞

0

dr

∫π

0

dθ

∫ 2 π

0

dφ f(r)r^2 sinθe−ikrcosθ

=

1

(2π)^3 /^2

∫∞

0

dr 2 πf(r)r^2

∫π

0

dθsinθe−ikrcosθ.

The integral overθmay be straightforwardly evaluated by noting that

d

dθ

(e−ikrcosθ)=ikrsinθe−ikrcosθ.

Therefore

̃f(k)=^1

(2π)^3 /^2

∫∞

0

dr 2 πf(r)r^2

[

e−ikrcosθ

ikr

]θ=π

θ=0

=

1

(2π)^3 /^2

∫∞

0

4 πr^2 f(r)

(

sinkr

kr

)

dr.