X 43 1 further reading can be found in the many

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Unformatted text preview: nition. If f : R → R is a piecewise continuous function which vanishes outside some finite interval, its Fourier transform is ˆ f (ξ ) = ∞ f (t)e−iξt dt. (3.22) −∞ The integral in this formula is said to be improper because it is taken from −∞ to ∞; it is best to regard it as a limit, ∞ −∞ f (t)e−iξt dt = lim L→∞ 79 πL −πL f (t)e−iξt dt. To explain (3.22), we suppose that f vanishes outside the interval [−πL, πL]. We can extend the restriction of f to this interval to a function which is periodic of period 2πL. Then ˆ f (k/L) = ∞ f (t)e−ikt/L dt = −∞ πL ˜ f (t)e−ikt/L dt −πL represents 2πL times the Fourier coefficient of this extension of frequency k/L; indeed, it follows from (3.21) that we can write f (t) = . . . + c−2 e−2it/L + c−1 e−it/L + c0 + c1 eit/L + c2 e2it/L + . . . , for t ∈ [−πL, πL], where ck = 1ˆ f (k/L), 2πL or alternatively, f (t) = . . . + 1ˆ 1ˆ f (−2/L)e−2it/L + f (−1/L)e−it/L + 2πL 2πL 1ˆ 1ˆ 1ˆ f (0) + f (...
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This document was uploaded on 01/12/2014.

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