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Mathematical Methods for Physicists: A concise introduction - Site Map

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CALCULATIONS OF FOURIER TRANSFORMS<br />

provided that u and @u=@x ! 0asx !1. In general, we can show that<br />

<br />

F<br />

@n u<br />

@x n ˆ…i† n F…u†<br />

if u;@u=@x; ...;@ n1 u=@x n1 !1as x !1.<br />

(c) By de®nition<br />

<br />

F<br />

@u Z 1<br />

@u<br />

ˆ<br />

@t 1 @t eix dx ˆ @ Z 1<br />

ue ix dx ˆ @<br />

@t 1<br />

@t F…u†:<br />

Example 4.12<br />

Solve the inhomogeneous di€erential equation<br />

!<br />

d 2<br />

dx 2 ‡ p d dx ‡ q f …x† ˆR…x†; 1x 1;<br />

where p and q are constants.<br />

Solution:<br />

We trans<strong>for</strong>m both sides<br />

( )<br />

F<br />

d2 f<br />

dx 2 ‡ p df<br />

dx ‡ qf ˆ‰…i† 2 ‡ p…i†‡qŠFff…x†<br />

g<br />

ˆ FfR…x†<br />

g:<br />

If we denote the Fourier trans<strong>for</strong>ms of f …x† and R…x† by g…† and G…†, respectively,<br />

we have<br />

and hence<br />

Fff…x†<br />

g ˆ g…†; FfR…x†<br />

g ˆ G…†;<br />

… 2 ‡ ip ‡ q†g…† ˆG…†; or g…† ˆG…†=… 2 ‡ ip ‡ q†<br />

f …x† ˆp<br />

1<br />

2<br />

ˆ p 1<br />

2<br />

Z 1<br />

1<br />

Z 1<br />

e ix<br />

1<br />

e ix g…†d<br />

G…†<br />

2 ‡ ip ‡ q d:<br />

We will not gain anything if we do not know how to evaluate this complex<br />

integral. This is not a dicult problem in the theory of functions of complex<br />

variables (see Chapter 7).<br />

191

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