Muhammad's question via email about an Inverse Fourier Transform

[Prove It]

Find the Inverse Fourier Transform of $\displaystyle \begin{align*} \frac{2\mathrm{i}\,\omega}{\omega ^2 + 10\omega + 29} \end{align*}$.

Completing the square gives

$\displaystyle \begin{align*} \frac{2\mathrm{i}\,\omega}{\omega ^2 + 10\omega + 29} &= \frac{2\mathrm{i}\,\omega}{ \omega ^2 + 10\omega + 5^2 - 5^2 + 29} \\ &= \frac{2\mathrm{i}\,\omega}{ \left( \omega + 5 \right) ^2 + 4 } \\ &= \frac{2\mathrm{i}\,\omega}{ \left( \omega + 5 \right) ^2 + 2^2 } \\ &= \frac{ 2\mathrm{i} \left( \omega + 5 \right) - 10\mathrm{i} }{ \left( \omega + 5 \right) ^2 + 2^2 } \\ &= \frac{2\mathrm{i} \left( \omega + 5 \right) }{ \left( \omega + 5 \right) ^2 + 2^2} - \frac{10\mathrm{i}}{ \left( \omega + 5 \right) ^2 + 2^2 } \end{align*}$

Now applying the first shift theorem: $\displaystyle \begin{align*} \mathcal{F} ^{-1} \left\{ F \left( \omega - u \right) \right\} = \mathrm{e}^{\mathrm{i}\,u\,t} \mathcal{F}^{-1} \left\{ F \left( \omega \right) \right\} \end{align*}$ we find

$\displaystyle \begin{align*} \mathcal{F}^{-1} \left\{ \frac{2\mathrm{i}\,\left( \omega + 5 \right) }{ \left( \omega + 5 \right) ^2 + 2^2 } - \frac{10\mathrm{i}}{ \left( \omega + 5 \right) ^2 + 2^2 } \right\} &= \mathrm{e}^{-5\mathrm{i}\,t} \mathcal{F}^{-1} \left\{ \frac{2\mathrm{i}\,\omega}{\omega ^2 + 2^2 } - \frac{10\mathrm{i}}{ \omega ^2 + 2^2 } \right\} \end{align*}$

Now the rest is applying the following rules:

$\displaystyle \begin{align*} \mathcal{F}^{-1} \left\{ \frac{-2\mathrm{i}\,\omega}{ \omega ^2 + a^2 } \right\} = \textrm{sgn}\,{(t)}\,\mathrm{e}^{-a\, \left| t \right| } \end{align*}$

and

$\displaystyle \begin{align*} \mathcal{F}^{-1} \left\{ \frac{2a}{\omega ^2 + a^2 } \right\} = \mathrm{e}^{-a\,\left| t \right| } \end{align*}$

 

[jvicens]

Where $\textrm{sgn}\,{(t)}$ is the sign function.

Therefore, the inverse Fourier transform of the given function is:

$\displaystyle \begin{align*} \mathcal{F}^{-1} \left\{ \frac{2\mathrm{i}\,\omega}{\omega ^2 + 10\omega + 29} \right\} &= \mathrm{e}^{-5\mathrm{i}\,t} \left[ \textrm{sgn}\,{(t)}\,\mathrm{e}^{-2\left| t \right| } - \mathrm{e}^{-2\left| t \right| } \right] \\ &= \mathrm{e}^{-5\mathrm{i}\,t} \mathrm{e}^{-2\left| t \right| } \left[ \textrm{sgn}\,{(t)} - 1 \right] \end{align*}$

Thus, the inverse Fourier transform of $\displaystyle \begin{align*} \frac{2\mathrm{i}\,\omega}{\omega ^2 + 10\omega + 29} \end{align*}$ is $\displaystyle \begin{align*} \mathrm{e}^{-5\mathrm{i}\,t} \mathrm{e}^{-2\left| t \right| } \left[ \textrm{sgn}\,{(t)} - 1 \right] \end{align*}$.