Understanding the Fourier Shift Theorem

Question

The Fourier transform of $f(x)$ is, by definition,

$$ \mathcal{F}[f(x)](u) = \int f(x)e^{-ixu}dx.$$

Now, the Fourier transform of $f(x-a)$ is, according to the Fourier shift theorem,

$$ \mathcal{F}[f(x-a)](u) = \int f(x-a)e^{-ixu}dx.$$

Why doesn't $(x-a)$ appear in the exponent too? That is, $$ \mathcal{F}[f(x-a)](u) = \int f(x-a)e^{-i(x-a)u}dx.$$

Does it follow from the definition of Fourier transform or is there another reason?

Answer 1

The shift is defined by $g_a(x) = f(x-a)$. Then you write

$$F[g_a](\xi) = \int_{\Bbb R}g_a(x)\exp(-ix\xi)dx = \int_{\Bbb R}f(x-a)\exp(-ix\xi)dx.$$

Conceptually, you first apply the shift and then apply the Fourier transform, but you can apply the shift only to the function, there is no sense in applying it to the exponent.

Answer 2

I laid out the same question a little differently here.

Essentially, it does appear in the exponent, but is constant over the integration, and is taken outside. The new integral is the definition of $F[f(x)]$.