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Calculus and Beyond Homework Help
Manipulating Fourier transforms
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[QUOTE="sa1988, post: 5480921, member: 476715"] Ah, I see now, it can't be shifted over because it's a function of p so it needs to stay within the integral over dp. Furthermore I've just been looking through some old lecture notes and it turns out I was very much on the right track (woohoo!). The question is essentially proving the convolution theorem. So I'd got to the point: ##F_D(x_D) = \widetilde{f}(p)\int_{-\infty}^{\infty} \! e^{-ix_D p}\widetilde{w}(p) \, \mathrm{d}p## which [I]actually[/I] neatens up to ##F_D(x_D) = \int_{-\infty}^{\infty} \! e^{-ix_D p}\widetilde{f}(p)\widetilde{w}(p) \, \mathrm{d}p## which can be made into a standard Fourier transform by adding factors of ##\sqrt{2\pi}##: ##\frac{1}{\sqrt{2\pi}}F_D(x_D) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \! e^{-ix_D p}\widetilde{f}(p)\widetilde{w}(p) \, \mathrm{d}p## then taking the inverse: ##\frac{1}{\sqrt{2\pi}}\widetilde{F}_D(p) = \widetilde{f}(p)\widetilde{w}(p)## which clearly leads to the desired result: ##\widetilde{F}_D(p) = \sqrt{2\pi}\widetilde{f}(p)\widetilde{w}(p)## [/QUOTE]
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Manipulating Fourier transforms
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