Shift theorem
Let $\mathfrak{F}$ denote the Fourier transform. The shift theorem states \begin{equation} \mathfrak{F}\{f(x-a)\}(u)=e^{-2i\pi ua}F(u) \end{equation} where $F(u)=\mathfrak{F}\{f(x)\}(u)$.
The same theorem applies to the inverse Fourier transform as follows: \begin{equation} \mathfrak{F}^{-1}\{F(u-a)\}(x)=e^{2i\pi ax}f(x) \end{equation}
Proof
By letting $x’=x-a$, \begin{equation} \begin{split} \mathfrak{F}\{f(x-a)\}(u) &=\int_{-\infty}^\infty f(x-a)e^{-2i\pi ux}\,dx\\ &=\int_{-\infty}^\infty f(x’)e^{-2i\pi u(x’+a)}\,dx’\\ &=e^{-2i\pi ua}\int_{-\infty}^\infty f(x’)e^{-2i\pi ux’}\,dx’\\ &=e^{-2i\pi ua}F(u). \end{split} \end{equation}
By letting $u’=u-a$, \begin{equation} \begin{split} \mathfrak{F}^{-1}\{F(u-a)\}(x) &=\int_{-\infty}^\infty F(u-a)e^{2i\pi ux}\,du\\ &=\int_{-\infty}^\infty F(u’)e^{2i\pi (u’+a)x}\,du’\\ &=e^{2i\pi ax}\int_{-\infty}^\infty F(u’)e^{2i\pi u’x}\,du’\\ &=e^{2i\pi ax}f(x). \end{split} \end{equation}