Then Fourier s integral theorem states that the Fourier integral of a function f is 1 f x in .NET framework

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Then Fourier s integral theorem states that the Fourier integral of a function f is 1 f x
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fA cos x B sin xg d
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1 8 > A 1 > f x cos x dx <  1 1 > > B 1 : f x sin x dx  1
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A and B with 1 < < 1 are generalizations of the Fourier coe cients an and bn . The right-hand side of (1) is also called a Fourier integral expansion of f . (Since Fourier integrals are improper integrals, a review of 12 is a prerequisite to the study of this chapter.) The result (1) holds if x is a point of continuity of f x . If x is a point of discontinuity, we must replace f x by f x 0 f x 0 as in the case of Fourier series. Note that the above conditions are su cient but not 2 necessary. 363
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FOURIER INTEGRALS
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[CHAP. 14
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1 In the generalization of Fourier coe cients to Fourier integrals, a0 may be neglected, since whenever f x dx exists, 1   L  1  as L!1 ja0 j  f x dx ! 0  L L
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EQUIVALENT FORMS OF FOURIER S INTEGRAL THEOREM Fourier s integral theorem can also be written in the forms 1 1 1 f x f u cos x u du d  0 u 1 1 1 1 e i x d f u ei u du f x 2 1 1 1 1 1 f u ei u x du d 2 1 1
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where it is understood that if f x is not continuous at x the left side must be replaced by f x 0 f x 0 . 2 These results can be simpli ed somewhat if f x is either an odd or an even function, and we have 1 2 1 f x cos x d f u cos u du if f x is even 5  0 0 1 1 2 sin x d f u sin u du if f x is odd 6 f x  0 0 An entity of importance in evaluating integrals and solving di erential and integral equations is introduced in the next paragraph. It is abstracted from the Fourier integral form of a function, as can be observed by putting (4) in the form & ' 1 1 i x 1 1 i u p f x p e e f u du d 2 1 2 1 and observing the parenthetic expression.
FOURIER TRANSFORMS From (4) it follows that 1 F p 2 then 1 f x p 2 1 f u ei u du
F e i x d
The function F is called the Fourier transform of f x and is sometimes written F ff f x g. The function f x is the inverse Fourier transform of F and is written f x f 1 fF g. p Note: The constants preceding the integral signs in (7) and (8) were here taken as equal to 1= 2. However, they can be any constants di erent from zero so long as their product is 1=2. The above is called the symmetric form. The literature is not uniform as to whether the negative exponent appears in (7) or in (8).
CHAP. 14]
FOURIER INTEGRALS Determine the Fourier transform of f if f x e x for x > 0 and e2x when x < 0. & 0 ' 1 1 1 1 F p ei x f x dx p ei x e2x dx ei x e x dx 2 2 1 2 1 0 ( )   & ' 1 ei 2 x!0 ei 1 x!1 1 1 1   p p 2 i 2 x! 1 i 1 x!0 2 2 i 1 i
EXAMPLE.
If f x is an even function, equation (5) yields r 1 8 > > F 2 > c f u cos u du <  0 r 1 > > > f x 2 : F cos x d  0 c and we call Fc and f x Fourier cosine transforms of each other. If f x is an odd function, equation (6) yields r 1 8 > > F 2 > s f u sin u du <  0 r 1 > > > f x 2 : F sin x d  0 s and we call Fs and f x Fourier sine transforms of each other.
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Note: The Fourier transforms Fc and Fs are (up to a constant) of the same form as A and B . Since f is even for Fc and odd for Fs , the domains can be shown to be 0 < < 1. When the product of Fourier transforms is considered, a new concept called convolution comes into being, and in conjunction with it, a new pair (function and its Fourier transform) arises. In particular, if F and G are the Fourier transforms of f and g, respectively, and the convolution of f and g is de ned to be 1 1 f u g x u du 11 f g p  1 then 1 F G p  1 f g p  1 ei u f g du
1 1 1
12 13
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