Ejemplo 14
Calcular {{\mathfrak{F}}^{-1}}\left[ \frac{1}{\left( 4+{{\omega }^{2}} \right)\left( 9+{{\omega }^{2}} \right)} \right]
Entonces:
{{\mathfrak{F}}^{-1}}\left[ \frac{1}{\left( 4+{{\omega }^{2}} \right)}\frac{1}{\left( 9+{{\omega }^{2}} \right)} \right]
Si se define F\left( \omega \right)=\frac{1}{\left( 4+{{\omega }^{2}} \right)}~~ , ~G\left( \omega \right)=\frac{1}{\left( 9+{{\omega }^{2}} \right)} , f\left( t \right)=\frac{1}{4}{{e}^{-2\left| t \right|}}~ , ~g\left( t \right)=\frac{1}{6}{{e}^{-3\left| t \right|}}
AsÃ: {{\mathfrak{F}}^{-1}}\left[ \frac{1}{\left( 4+{{\omega }^{2}} \right)\left( 9+{{\omega }^{2}} \right)} \right]=f*g=\frac{1}{4}{{e}^{-2\left| t \right|}}*\frac{1}{6}{{e}^{-3\left| t \right|}}=\frac{1}{24}\underset{-\infty }{\overset{\infty }{\mathop \int }}\,{{e}^{-2\left| t-\tau \right|}}{{e}^{-3\left| \tau \right|}}d\tau
Para t > 0
\frac{1}{24}\left( f*g \right)=\underset{-\infty }{\overset{0}{\mathop \int }}\,{{e}^{-2\left| t-\tau \right|}}{{e}^{-3\left| \tau \right|}}d\tau +\underset{0}{\overset{t}{\mathop \int }}\,{{e}^{-2\left| t-\tau \right|}}{{e}^{-3\left| \tau \right|}}d\tau +\underset{t}{\overset{\infty }{\mathop \int }}\,{{e}^{-2\left| t-\tau \right|}}{{e}^{-3\left| \tau \right|}}d\tau
~~~~~~~~~~~\underset{-\infty }{\overset{0}{\mathop \int }}\,{{e}^{-2\left( t-\tau \right)}}{{e}^{-3\tau }}d\tau +\underset{0}{\overset{t}{\mathop \int }}\,{{e}^{-2\left( t-\tau \right)}}{{e}^{-3\tau }}d\tau +\underset{t}{\overset{\infty }{\mathop \int }}\,{{e}^{-2\left( t-\tau \right)}}{{e}^{-3\tau }}d\tau
~\frac{6}{5}{{e}^{-2t}}-~\frac{4}{5}{{e}^{-3t}}
Para t < 0
\frac{1}{24}\left( f*g \right)=\underset{-\infty }{\overset{t}{\mathop \int }}\,{{e}^{-2\left| t-\tau \right|}}{{e}^{-3\left| \tau \right|}}d\tau +\underset{t}{\overset{0}{\mathop \int }}\,{{e}^{-2\left| t-\tau \right|}}{{e}^{-3\left| \tau \right|}}d\tau +\underset{0}{\overset{\infty }{\mathop \int }}\,{{e}^{-2\left| t-\tau \right|}}{{e}^{-3\left| \tau \right|}}d\tau
~~~~~~~~~~~\underset{-\infty }{\overset{0}{\mathop \int }}\,{{e}^{-2\left( t-\tau \right)}}{{e}^{3\tau }}d\tau +\underset{0}{\overset{t}{\mathop \int }}\,{{e}^{2\left( t-\tau \right)}}{{e}^{3\tau }}d\tau +\underset{t}{\overset{\infty }{\mathop \int }}\,{{e}^{2\left( t-\tau \right)}}{{e}^{-3\tau }}d\tau
~\frac{6}{5}{{e}^{2t}}-~\frac{4}{5}{{e}^{3t}}
Para t = 0
\frac{1}{24}\left( f*g \right)=\underset{-\infty }{\overset{\infty }{\mathop \int }}\,{{e}^{-2\left| t-\tau \right|}}{{e}^{-3\left| \tau \right|}}d\tau =\frac{2}{5}
Entonces,
{{\mathfrak{F}}^{-1}}\left[ \frac{1}{\left( 4+{{\omega }^{2}} \right)\left( 9+{{\omega }^{2}} \right)} \right]=\frac{1}{24}\left( \frac{6}{5}{{e}^{-2\left| t \right|}}-~\frac{4}{5}{{e}^{-3\left| t \right|}} \right)=\frac{1}{20}{{e}^{-2\left| t \right|}}-~\frac{1}{30}{{e}^{-3\left| t \right|}}
