# Muhammed's question via email about an Inverse Fourier Transform (2)

#### Prove It

##### Well-known member
MHB Math Helper
Find the Inverse Fourier Transform of \displaystyle \begin{align*} \frac{4}{ \left( 1 + \mathrm{i}\,\omega + 2\mathrm{i} \right) ^2 } + \frac{6\mathrm{i}\,\omega}{ \left( 9 + \omega ^2 \right) ^2 } \end{align*}
Here we will use the following transforms: \displaystyle \begin{align*} \mathcal{F}^{-1} \left\{ \frac{n!}{ \left( a + \mathrm{i}\,\omega \right) ^{n+1} } \right\} = t^n\,\mathrm{e}^{-a\,t}\,\mathrm{H}(t) \end{align*} and \displaystyle \begin{align*} \mathcal{F}^{-1} \left\{ \frac{-4\mathrm{i}\,a\,\omega}{ \left( a^2 + \omega ^2 \right) ^2 } = t\,\mathrm{e}^{-a \, \left| t \right| } \right\} \end{align*}. With some rearranging we get...

\displaystyle \begin{align*} \frac{4}{ \left( 1 + \mathrm{i}\,\omega + 2\mathrm{i} \right) ^2 } + \frac{6\mathrm{i}\,\omega}{ \left( 9 + \omega ^2 \right) ^2 } &= 4 \left\{ \frac{1!}{ \left[ \left( 1 + 2\mathrm{i} \right) + \mathrm{i}\,\omega \right] ^2 } \right\} - \frac{1}{2} \left[ \frac{-4 \cdot 3\mathrm{i}\,\omega }{ \left( 3^2 + \omega ^2 \right) ^2 } \right] \end{align*}

Both terms are now in forms where the Inverse Fourier Transform can be found.