Fourier Transform of e^(ip0x)F(x) to F(p)

[kathrynlaa]

Homework Statement

f(p) is the Fourier transform of f(x). Show that the Fourier Transform of e^ip_ox f(x) is f(p- p₀).

Homework Equations

I'm using these versions of the Fourier transform:
f(x)=1/√(2π)∫e^ixpf(p)dx
f(p)=1/√(2π)∫e^-ixpf(x)dx

The Attempt at a Solution

I have:
f(p)=1/√(2π)∫e^ix(p_o-p)f(x)dx
which is the same as:
f(p)=1/√(2π)∫e^-ix(p-p_o)f(x)dx
but I don't know where to go from here. I think I need to make a substitution using the original transform as I don't need to solve the integral. My other idea is that I have nearly proved it so just need to state the theory as to why this proves it; however, I don't know what that theory would be.
Any help would be appreciated!

 

[jbunniii]

You are using $\underline{f}(p)$ to refer to two different integrals which are not equal.

To avoid confusion, I suggest defining a new function. Let $g(x) = e^{ip_0x}f(x)$. Then the goal is to show that $\underline{g}(p) = \underline{f}(p - p_0)$. You have already established that
$$\underline{g}(p) = \frac{1}{\sqrt{2\pi}}\int e^{-ix(p - p_0)}f(x) dx$$
Can you express this in terms of $\underline{f}$?

 

[kathrynlaa]

I understand why I should of renamed the functions but I really can't figure out the next step.
I realize this is the final aim:
g(p) = 1/√(2π)∫e^{ix(p - p₀}f(x) dx = f(p-p₀)
But I don't know how f(x) is related to f(p-p₀) and the best I seem able to do is get g(p) in terms of g(x), which is useless.
Sorry I'm being dense!

 

[jbunniii]

You have
$$\underline{f}(p) = \frac{1}{\sqrt{2\pi}} e^{-ixp} f(x) dx$$
so what is $\underline{f}(p - p_0)$? Just replace $p$ with $p - p_0$ in the equation above. What can you conclude?

Maybe to make it clearer, recognize that I can use any variable instead of $p$ and the equation will still be true. For example,
$$\underline{f}(y) = \frac{1}{\sqrt{2\pi}} e^{-ixy} f(x) dx$$
What does this give you when $y = p - p_0$?