Proving the Fourier Transform Property: e^(ip0x)f(x) = f'(p - p0)

[12x4]

Homework Statement

f'(p) is the Fourier transform of f(x). Show that the Fourier transform of e^(ip₀x)f(x) is f'(p - p0). (using f'(p) for transform)

Homework Equations

f(x) = 1/√(2pi) ∫e^(ipx) f'(p) dp (intergral from -∞ to ∞)

f'(p) = 1/√(2pi) ∫e^(-ipx) f(x) dx (also from -∞ to ∞)

The Attempt at a Solution

So I substituted e^(ip₀x) f(x) into the f'(p) formula and after rearranging came up with:

f'(p) = 1/√(2pi) ∫ e^(ix(p - p₀))f(x) dx

If I'm completely honest I'm not really sure what I am doing here. I can see that this looks similar to the formula for f'(p) but I was just jiggling things around to see what happened. Its worth mentioning as well that this is a 5 mark question on a past exam paper so I assume that I have to do a lot more than what I have already done. Any advice would be hugely appreciated. Cheers

 

[Noctisdark]

What about setting P = p-p0 ?
It won't mean anything to the integral because p is constant to x, it's just to make our lives easier !
In case what i said isn't clear, here's what i mean
∫e^(ix(p-p0))f(x) dx = ∫e^(ixP)f(x) dx = f'(P) = f(p-p0)

 

[12x4]

Thanks Noctisdark, think i was just doubting myself but have found that it follows on from there quite nicely. Seems like an easy 5 marks on a paper!