Fourier transform of $f(x+h)$?

Question

Show that $f(x+h)\to \hat{f}(w)e^{2\pi i h w}$

Let $g(y) = f(x+h)$, then $\hat{g}(w) = \int_{-\infty}^\infty g(y) e^{-2\pi i y w} dy = \int_{-\infty}^\infty f(x+h) e^{-2\pi i (x+h) w} dx$, then I simply have no idea how to continue?

Answer 1

You're confusing yourself with $x$ and $y$. You want to find the transform of $g(x) = f(x+h)$, not $g(y) = f(x+h)$.

Putting $g(x) = f(x+h)$, We see \begin{align*} \hat g(x) = \int_\mathbb R g(x) e^{-2\pi i x w} dx = \int_{\mathbb R} f(x+h) e^{-2\pi i xw} dx &= \int_\mathbb R f(x) e^{-2\pi i(x-h)w} dx \\ &= e^{2\pi i h w}\int_{\mathbb R}f(x) e^{-2\pi i x w} dx \\ &= e^{2\pi i h w} \hat f(w)\end{align*} where the last step of the first line, I made the transformation $x \mapsto x-h$.