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## The Fourier Integral Theorem

The mathematically more precise statement of this theorem is as follows:

Theorem 23.1 (Fourier's Integral Theorem)
Given:(i)
is function piecewise continuous on every bounded closed interval of the -axis.
(ii)
At each point the function has both a left derivative and a right derivative ,
(iii)
, i.e. is absolutely integrable along the -axis:

Conclusion:

 (239)

1. This result can be restated as a Fourier transform pair,

 (240)

 (241)

whenever is continuous.
2. By interchanging integration order and letting one has

 (242)

This equation holds for all continuous functions . Thus is another delta convergent sequence:

 (243)

It is of course understood that one first do the integration over before taking to the indicated limit.
3. Either one of the two equations, Eq.(2.42) or (2.43), is a generalized completeness relation for the set of wave train'' functions,

However, these functions are not normalizable, i.e., they . Instead, as Eq.(2.43) implies, they are said to be  -function normalized''.

Proof of the Fourier integral theorem:

The proof of the Fourier integral theorem presupposes that the Fourier amplitude is well-defined for each . That this is indeed the case follows from the finiteness of :

The last inequality is an expression of the fact that . Thus is well-defined indeed.

The proof of the Fourier integral theorem runs parallel to the Fourier series theorem on page . We shall show that

where

The evaluation of the integrals is done by shifting the integration variable. For the second integral one obtains

Using the fact that

and the fact that

is piecewise continuous everywhere, including at , where

is the right hand derivative of at , one finds that

with the help the Riemann-Lebesgue lemma.

A similar analysis yields

The sum of the last two equations yields

This validates Fourier's integral theorem.

Next: The Fourier Transform as Up: The Fourier Integral Previous: Transition from Fourier Series   Contents   Index
Ulrich Gerlach 2010-12-09