Let \(F\) and \(G\) be the Fourier transforms of \(f\) and \(g\), i.e. \begin{eqnarray*} F(p)&=&\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{+\infty}f(x)e^{-ipx}dx\\ G(p)&=&\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{+\infty}g(y)e^{-ipy}dy\\ \end{eqnarray*}
Then
\[ \F^{-1}\left\{F(p)\cdot G(p)\right\}(z) = \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{+\infty}\left( \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{+\infty} f(x)e^{-ipx}dx \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{+\infty}g(y)e^{-ipy} \;dy\right) e^{ipz} dp\] |
Since
\[\frac{1}{2\pi}\int\limits_{-\infty}^{+\infty} e^{-ip(x+y-z)}dp= \delta(x+y-z)\] |
We finally get \begin{eqnarray*} \F^{-1}\left\{F(p)\cdot G(p)\right\}(z)& = & \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{+\infty} \int\limits_{-\infty}^{+\infty} f(x) g(y) \delta(x+y-z)dx dy\\ & = & \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{+\infty} f(x) g(z-x) dx\\ & = & \frac{1}{\sqrt{2\pi}}\left\{f \star g\right\}(z) \end{eqnarray*}
Example \begin{eqnarray*}F(p)&=&\sqrt{\frac{2}{\pi}}\frac{\alpha}{\alpha^2+p^2}\qquad f(x)=e^{-\alpha|x|}\\ G(p)&=&\frac{2aA}{\sqrt{2\pi}}\frac{\sin(pa)}{pa}\quad g(x)=\left\{\begin{array}{cl}A&|x| \lt a\\0&|x| \gt a\end{array}\right.\\\\ \rightarrow \F\left\{\sqrt{2\pi}F(p)G(p)\right\}&=&\int\limits_{-\infty}^{+\infty}e^{-\alpha|x-t|}Adt=A\overbrace{\int\limits_{-a}^{+a}e^{-\alpha|x-t|}dt}^{\mbox{\normalsize different cases}}\\ &=&h(x)=e^{-\alpha|x|}\frac{A}{\alpha}\left(e^{-\alpha a}-e^{\alpha a}\right)\;\text{for } |x| \gt a\\ &&\qquad \text{for } |x|\lt a \text{ more difficult!} \end{eqnarray*}
Fourier Transformation: Properties
© J. Carstensen (Math for MS)