Define $F$ such that
$$
f(x) = \frac{1}{\sqrt {2 \pi}} \int{F(k) e^{ikx} {dk}}
$$
and therefore
$$
F(k) = \frac{1}{\sqrt {2 \pi}} \int {f(x) e^{-ikx}}
$$
Then we have
$$
\int{f(x) g(x) {dx} } = \int{F(k)G(-k) {dk} }
$$
Since $ G^\star(-k) \iff g^*(x)$, we have
$$
\int{f(x) g^\star(x) {dx} } = \int {F(k) G^*(k) {dk}}
$$
Since $g(x) e^{-i \alpha x} \iff G(k + \alpha)$, a similar manipulation gives
\begin{align}
\int{ f(x) [g(x) e^{-ikx}] {dx} } &= \int{ F(k') G( - k'+ k) ) {dk'}} \\
&= \int{ F(k') G( k - k') ) {dk'}}
\end{align}
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