Mathematical Methods for Physics and Engineering : A Comprehensive Guide

INTEGRAL TRANSFORMS

given by

̃h(k)=√^1

2 π

∫∞

−∞

dz e−ikz

{∫∞

−∞

f(x)g(z−x)dx

}

=

1

√

2 π

∫∞

−∞

dx f(x)

{∫∞

−∞

g(z−x)e−ikzdz

}

.

If we letu=z−xin the second integral we have

̃h(k)=√^1

2 π

∫∞

−∞

dx f(x)

{∫∞

−∞

g(u)e−ik(u+x)du

}

=

1

√

2 π

∫∞

−∞

f(x)e−ikxdx

∫∞

−∞

g(u)e−ikudu

=

1

√

2 π

×

√

2 π ̃f(k)×

√

2 π ̃g(k)=

√

2 π ̃f(k) ̃g(k). (13.38)

Hence the Fourier transform of a convolutionf∗gis equal to the product of the

separate Fourier transforms multiplied by

√

2 π; this result is called theconvolution

theorem.

It may be proved similarly that the converse is also true, namely that the

Fourier transform of the productf(x)g(x) is given by

F[f(x)g(x)]=

1

√

2 π

̃f(k)∗ ̃g(k). (13.39)

Find the Fourier transform of the function in figure 13.3 representing two wide slits by

considering the Fourier transforms of(i)twoδ-functions, atx=±a,(ii)a rectangular

function of height 1 and width 2 bcentred onx=0.

(i) The Fourier transform of the twoδ-functions is given by

̃f(q)=√^1

2 π

∫∞

−∞

δ(x−a)e−iqxdx+

1

√

2 π

∫∞

−∞

δ(x+a)e−iqxdx

=

1

√

2 π

(

e−iqa+eiqa

)

=

2cosqa

√

2 π

.

(ii) The Fourier transform of the broad slit is

̃g(q)=

1

√

2 π

∫b

−b

e−iqxdx=

1

√

2 π

[

e−iqx

−iq

]b

−b

=

− 1

iq

√

2 π

(e−iqb−eiqb)=

2sinqb

q

√

2 π

.

We have already seen that the convolution of these functions is the required function
representing two wide slits (see figure 13.6). So, using the convolution theorem, the Fourier
transform of the convolution is

√

2 πtimes the product of the individual transforms, i.e.
4cosqasinqb/(q

√

2 π). This is, of course, the same result as that obtained in the example
in subsection 13.1.2.