Question: I'm sorry, I don't know what else to specify on the title..

the domain is all real numbers which $x \ne 1$

$\lim\limits\_{x \to \infty}$$f(x)=2$

$\lim\limits\_{x \to -\infty}$$f(x)=-2$

$\lim\limits\_{x \to \ 1+}$$f(x)=-\infty$

$\lim\limits\_{x \to \ 1-}$$f(x)=-\infty$

$f(x)$ should not be a piece-wise defined function.

Could you please help? I could only draw what the graph would look like, I have no clue what the function could be.

Answer: 
1. Since $(-ix)^\alpha\hat f = \widehat{D^\alpha f},$  

$$
\int\_{\mathbb{R^n}} x^\alpha \mathcal F\phi dx
= i^{|\alpha|}\int\_{\mathbb{R^n}}\mathcal F(D^\alpha\phi) dx
= (2\pi)^{n/2}i^{|\alpha|}\mathcal F(\mathcal F(D^\alpha\phi))(0)
= (2\pi)^{n/2}i^{|\alpha|}D^\alpha\phi(0).
$$
2. First note
$$
\int\_{\mathbb{R^n}} e^{i \langle a, x \rangle} \mathcal F\phi(x) dx
= (2\pi)^{n/2}\phi(a).
$$
Now since $2i\sin(t) = e^{it} -e^{-it}$ we get 
$$\int\sin(\langle a, x \rangle) \mathcal F\phi (x) dx = (2\pi)^{n/2}\frac{ \phi(a) - \phi (-a)}{2i}.$$
3. Yes, the fourier transform of a distribution induced by an $L^1$ function $f$ is that induced by the fourier transform of $f$, and
$$ \mathcal F(\mathbb1\_{[-1,1]})(x) = \frac1{\sqrt{2\pi}} \frac{2\sin(x)}x.$$

---

(convention for Fourier Transform) here, we use the Fourier/Inverse Fourier transform pair mentioned in comments
$$ \mathcal Ff(\xi) = (2\pi)^{-n/2}\int f(x) e^{-i\xi\cdot x} dx, \quad \mathcal F^{-1}f(\xi) = (2\pi)^{-n/2}\int f(x) e^{i\xi \cdot x} dx = \mathcal Ff(-\xi)$$
Wolfram Mathworld has helpfully laid out the different conventions of the Fourier Transform [here](https://mathworld.wolfram.com/FourierTransform.html).
For 2, try writing $$\sin(\langle a,x \rangle)=\Im\left(e^{i\langle a,x \rangle}\right)=\Im\left(\prod\_{k=1}^n e^{ia\_kx\_k}\right)$$ and use Fubini's Theorem