Thursday, August 18, 2011

Transform of a Product


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}