Two questions on Fourier transform and inverse Fourier transform 
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*I want to find the Fourier transform of $$ \int_{-\infty}^{\infty} {\frac{\sin{x}}{x^{3}+x}\,\mathrm{d}x}$$ I have solved it using Residue and contour but I am stuck when I am using Fourier transform.


*What will be the inverse Fourier of the following?
$$\frac{1-iw}{1+iw}\frac{\sin{w}}{w}$$ I think it could be solved using convolution theorem but still not sure.
 A: About $(1)$
We make use of Plancherel-Parseval's Theorem:
$$\int_{-\infty}^{+\infty} f(x)\overline{g(x)}\text{d}x = \int_{-\infty}^{+\infty} \hat{f(s)} \overline{\hat{g(s)}}\text{d}x$$
Here we can choose for simplicity
$$f(x) = \dfrac{\sin(x)}{x}$$
$$g(x) = \dfrac{1}{x^2+1}$$
By the very standard calculus of Fourier Transform you should know that
$$\hat{f(s)} = \frac{1}{2} \sqrt{\frac{\pi }{2}} (\text{sgn}(1-s)+\text{sgn}(s+1))$$
and
$$\hat{g(s)} = \sqrt{\frac{\pi }{2}} e^{-| s| }$$
whence we get
\begin{equation*}
\begin{split}
\int_{-\infty}^{+\infty} & \left(\frac{1}{2} \sqrt{\frac{\pi }{2}} (\text{sgn}(1-s)+\text{sgn}(s+1))\right)\cdot\left(\sqrt{\frac{\pi }{2}} e^{-| s| }\right)\text{d} s \\\\
& = \dfrac{\pi}{4}\int_{-\infty}^{+\infty}(\text{sgn}(1-s)+\text{sgn}(s+1))e^{-| s| }\text{d} s \\\\
& = \frac{(e-1) \pi }{e}
\end{split}
\end{equation*}
Remark 1: constants may vary, according to the definition of Fourier Transform used.
If you have doubts about how to compute the Fourier Transforms above, write a comment!
Remark 2: Notice that you could have chosen a different $f$ and $g$, like $f = \sin(x)$ and consequently $g$, and the result would have been the same. Perhaps easy, since F.T. of the sine function gives you two Dirac Delta. You can try it for fun.
About $(2)$
We use the convolution property of the Fourier Transform, that is:
$$\mathcal{F}^{-1}(f\cdot g) = \mathcal{F}^{-1}(f) * \mathcal{F}^{-1}(g)$$
with
$$f = \dfrac{1 + i\omega}{1 + i\omega}$$
$$g = \dfrac{\sin(\omega)}{\omega}$$
We have respectively:
$$\mathcal{F}^{-1}(f) = 2 \sqrt{2 \pi } e^s \theta (-s)-\sqrt{2 \pi } \delta (s)$$
and the other one comes from above:
$$\mathcal{F}^{-1}\left(\dfrac{\sin(\omega)}{\omega}\right) = \frac{1}{2} \sqrt{\frac{\pi }{2}} (\text{sgn}(1-s)+\text{sgn}(s+1))$$
Whence the inverse Fourier Transform of their product is give by the convolution of the two transforms:
$$\left(2 \sqrt{2 \pi } e^s \theta (-s)-\sqrt{2 \pi } \delta (s)\right) * \left(\frac{1}{2} \sqrt{\frac{\pi }{2}} (\text{sgn}(1-s)+\text{sgn}(s+1))\right)$$
that is
$$\dfrac{\pi}{2}\left(2e^s \theta (-s)-\sqrt{2 \pi } \delta (s)\right) * \left(\text{sgn}(1-s)+\text{sgn}(s+1))\right)$$
Assuming you know how to manage convolutions, the final result will be
$$4 \left(e^{x+1}-1\right) \theta (-x-1)- 4 \left(e^{x-1}-1\right) \theta (1-x) + (\text{sgn}(1-x)+\text{sgn}(x+1))$$
