Equation (4-25) gives a Fourier transform of a particular function. If equation (4-21) is correct, then multiplying this amplitude A(k) times eikx and integrating over all ft from -∞ to +∞ should return that particular function. Actually, because A(k) is an even function of k, multiplying it by the imaginary part of eikx (i.e., i sin kx) would produce an integrand that is an odd function of ft. which would integrate to 0. Therefore, it is only necessary to multiply A (k) by the real part, cos kx.
(a). Define f(x) to be the integral of this A(k) times cos kx, with these provisions: For simplicity, define C to be 1 and w to be 2. Although the integral from -∞ to +∞ can be done exactly (or looked up in a table of integrals), it isn’t a very easy one. To see what is really happening, it is better to use a numerical integration routine on a computer. So, define f(x) as the numerical integral, with limits on k from -200 to 200. (You may need to tweak the limits just a bit to avoid k = 0, where the function is really finite but may not appear so to your computer.) Afterward, plot f(x) as a function of x from —3 to +3. What do you see, and why?
(b). Change w to 3, then plot f(x) and discuss the result.
(c). Restrict the range of k in the numerical integration to —5.0 to +50. What happens? Can you explain qualitatively why?
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