How to find the inverse Fourier Transform of $F(w)=\frac{1}{w^2-a^2}$ I tried to use partial fractions but still I cannot find the way to inverse transform this Fourier transform.
$$F(w)=\frac{1}{w^2-a^2}$$
 A: If $a\in\mathbb{R}$, $a\ne 0$, then we will interpret the inverse Fourier Transform as a Cauchy Principal Value.
Then, the integral of interest can be written 
$$\begin{align}
\text{PV}\int_{-\infty}^\infty \frac{e^{i\omega t}}{\omega^2-a^2}\,d\omega&=\frac1{2|a|}\text{PV}\left(\int_{-\infty}^\infty \frac{e^{i\omega t}}{\omega-|a|}\,d\omega-\int_{-\infty}^\infty \frac{e^{i\omega t}}{\omega+|a|}\,d\omega\right)\\\\
&=\frac1{2|a|}\lim_{\epsilon\to 0^+}\left(\int_{-\infty}^{|a|-\epsilon}\frac{e^{i\omega t}}{\omega-|a|}\,d\omega +\int_{|a|+\epsilon}^\infty \frac{e^{i\omega t}}{\omega-|a|}\,d\omega\right)\\\\
&-\frac1{2|a|}\lim_{\epsilon\to 0^+}\left(\int_{-\infty}^{-|a|-\epsilon}\frac{e^{i\omega t}}{\omega+|a|}\,d\omega +\int_{-|a|+\epsilon}^\infty \frac{e^{i\omega t}}{\omega+|a|}\,d\omega\right)\\\\
&=\frac{i\pi}{2|a|}\left(e^{i|a|t}-e^{-i|a|t}\right)\text{sgn}(t)\\\\
&=-\left(\frac{\pi \sin(at)}{a}\right)\,\text{sgn}(t)\tag1
\end{align}$$
For $a=0$, the Fourier Transform is the distribution given by the limit as $a\to 0$ in $(1)$, namely $-\pi t\text{sgn}(t)$.
A: From FT table, we have:
$$F[sin(ax)u(x)] = F_1 = \frac{j\pi}{2}[\delta(w+a)-\delta(w-a)] - \frac{a}{w^2-a^2}$$
$$F[sin(ax)] = F_2 = j\pi[\delta(w+a)-\delta(w-a)]$$
$\frac{F_2-2F_1}{2a}$ is what you ask in Fourier domain, which corresponds to 
$sin(ax)\frac{1-2u(x)}{2a}$, if I'm not mistaken.
