<s>[INST] Inverse Fourier transform with a $\delta$-function integrand I'm self-studying math and trying to find the inverse Fourier transform of $\frac{4+w^2}{1+w^2}(4\pi * (\delta(w-2)+\delta(w+2)))$
Based on wolframalpha, the result is $32/5\sqrt{2\pi}\cos(2t)$.
But I can't even find the fourier transform of $\frac{4+w^2}{1+w^2}$ because there doesn't seem to have a Fourier transform pair in my table for $w^2$ in the numerator.
 [/INST] You don't need to know, just apply the definition
\begin{eqnarray}
f(t) &=& \frac{1}{2\pi}\int_{-\infty}^{+\infty} \frac{4 + \omega^2}{1 + \omega^2}[4\pi\delta(\omega - 2) + 4\pi\delta(\omega + 2)] e^{i\omega t}{\rm d}\omega \\
&=& 2\int_{-\infty}^{+\infty} \frac{4 + \omega^2}{1 + \omega^2}4\pi\delta(\omega - 2) e^{i\omega t}{\rm d}\omega + 2\int_{-\infty}^{+\infty} \frac{4 + \omega^2}{1 + \omega^2} \delta(\omega + 2) e^{i\omega t}{\rm d}\omega \\
&=& 2 \frac{4 + 2^2}{1 + 2^2}e^{2it} + 2\frac{4+2^2}{1 + 2^2}e^{-2it} \\
&=&\frac{32}{5}\left(\frac{e^{2it} + e^{-2it}}{2}\right) = \frac{32}{5}\cos(2t)
\end{eqnarray}
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