- #1

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## Homework Statement

## Homework Equations

## The Attempt at a Solution

I think I'm ok with the first part. I start with:

##\widetilde{f}(p) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \! e^{-ipx}f(x) \, \mathrm{d}x##

Then moving on to the transform for ##e^{ip_0 x}f(x)## I get:

##\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \! e^{-ipx}e^{ip_0 x}f(x) \, \mathrm{d}x##

## = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \! e^{-i(p-p_0)x}f(x) \, \mathrm{d}x##

## =\widetilde{f}(p-p_0)##

However I'm really not sure what to do for part b).

Here's my attempt so far:

Starting with:

##F_D(x_D) = \int_{-\infty}^{\infty} \! f(x)\Big[w(x_D - x)\Big] \, \mathrm{d}x##

The ##w(x_D - x)## part is similar to the previous section so I plug that into get:

##F_D(x_D) = \int_{-\infty}^{\infty} \! f(x) \, \Big[\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \! e^{-i(x_D -x)p}\widetilde{w}(p) \, \mathrm{d}p\Big] \mathrm{d}x##

then rearrange a little to get:

##F_D(x_D) = \Big(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \! e^{ixp}f(x) \, \mathrm{d}x\Big) \int_{-\infty}^{\infty} \! e^{-ix_D p}\widetilde{w}(p) \, \mathrm{d}p##

The part in brackets is the transform giving ##\widetilde{f}(p)## so,

##F_D(x_D) = \widetilde{f}(p) \int_{-\infty}^{\infty} \! e^{-ix_D p}\widetilde{w}(p) \, \mathrm{d}p##

which can be neatened a little into:

##F_D(x_D) = \widetilde{f}(p)\widetilde{w}(p) \int_{-\infty}^{\infty} \! e^{-ix_D p} \, \mathrm{d}p##

so now I have ##F_D(x_D)## written in a way that's already starting to look like the desired outcome. However it only gets messy now as I do the Fourier transform:

##\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \! e^{ix_D p}\Big[F_D(x_D)\Big] \, \mathrm{d}x_D = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \! e^{ix_D p}\Big[\widetilde{f}(p)\widetilde{w}(p) \int_{-\infty}^{\infty} \! e^{-ix_D p} \, \mathrm{d}p\,\Big] \mathrm{d}x_D##

The exponents cancel to leave

##\frac{1}{\sqrt{2\pi}} \widetilde{f}(p)\widetilde{w}(p) \int_{-\infty}^{\infty} \! \int_{-\infty}^{\infty} \! \, \mathrm{d}p\, \mathrm{d}x_D##

which I really don't think is a very good result!

Any input would be much appreciated. Thanks.

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