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##### Well-known member
MHB Math Helper
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*}

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