Fourier transform of the regular tempered distributions induced by $x^\alpha$, $\sin(\langle a, x \rangle_{\mathbb R^n})$, and $\mathbb 1_{[-1,1]}$ 
Compute the Fourier transform of the following tempered distributions on $\mathbb R^n$, which, for $\phi \in \mathcal S(\mathbb R^n)$ are given by



*

*$T_1(\phi) := \int_{\mathbb R^n} x^{\alpha} \phi(x) dx$, $\alpha \in \mathbb N_0^n$.





*$T_2(\phi) := \int_{\mathbb R^n}\sin(\langle a, x \rangle_{\mathbb R^n}) \phi(x)dx$, $a \in
    \mathbb R^n$.





*$T_3(\phi) := \int_{\mathbb R} \mathbb 1_{[-1,1]}(x) \phi(x) dx$.


Here's what I know and what I've tried so far
$\mathcal F$ denotes the Fourier transform
$$
\mathcal{F} f (\xi)
:= \hat{f}(\xi)
:= (2 \pi)^{-n/2} \int_{\mathbb R^n} f(x) e^{-i \langle x, \xi \rangle} dx
$$
and $\mathcal{S}$ the Schwartz space.
We defined $(\mathcal{F} T)(\phi) = T(\mathcal{F}(\phi))$. I am also going to use the following rules for the Fourier transform: $D^{\alpha}(\mathcal{F} \phi)(\xi) = (-i)^{|\alpha|} \mathcal{F}(x^{\alpha} \phi)(\xi)$ and $\xi^{\alpha} (\mathcal{F}\phi)(\xi) = (-i)^{| \alpha |} \mathcal{F}(D^{\alpha} \phi)(\xi)$.

*

*Is the following calculation correct? \begin{align*}
            \mathcal{F}(T_1(\phi))
            & = T_1(\mathcal{F}(\phi))
            = \int_{\mathbb{R}^n} x^{\alpha} \hat{\phi}(x) dx
            = (-1)^{| \alpha |} \int_{\mathbb{R}^n} (\hat{\phi}(x))^{(\alpha)} dx \\
            & = (-1)^{|\alpha|} \int_{\mathbb{R}^n}  (-i)^{|\alpha|} \mathcal{F}(t^{\alpha} \phi)(x) dx
            = i^{|\alpha|} \int_{\mathbb{R}^n} \mathcal{F}(t^{\alpha} \phi)(x) dx \\
            & = i^{|\alpha|} \mathcal{F}^{-1}(\mathcal{F}(t^{\alpha} \phi))(0)
            = i^{| \alpha |} 0^{\alpha} \cdot \phi(0)
            = \begin{cases} i^{| \alpha |} \phi(0), & \alpha = 0, \\
            0, & \text{else.} \end{cases}
        \end{align*}


*I am pretty stuck with this one as I haven't found a way to further reduce this expression. Any hints?
\begin{align*}
        \mathcal{F}(T_2(\phi))
        = T_2(\hat{\phi})
        = \int_{\mathbb R^n} \sin(\langle a, x \rangle) \hat{\phi}(x) dx
    \end{align*}
The next step I will try is to shift the Fourier transform to the sin, as I have done in 3. But I am even struggling to evaluate the integral for $n = 1$, which is
$$
\int_{\mathbb R} \sin(a x) e^{-i x \xi} dx
= \frac{e^{-i \xi x}(a \cos(ax) + i \xi \sin(ax))}{\sqrt{2 \pi} (a - \xi)(a + \xi)}\bigg|_{x = - \infty}^{\infty}
$$


*Here I redristribute the Fourier transform inside the integral, which should be valid by Fubini. Is it valid in this case? \begin{align*}
        \mathcal{F}(T_3(\phi))
        & = T_3(\hat{\phi})
        = \int_{\mathbb{R}} \mathbb 1_{[-1,1]}(x) \hat{\phi}(x) dxf
        = \int_{\mathbb{R}} \widehat{\mathbb 1_{[-1,1]}}(x) \phi(x) dx \\
        & = \int_{\mathbb{R}} \sqrt{\frac{2}{\pi}} \frac{\sin(x)}{x} \phi(x) dx
    \end{align*}
Edit
With @guy3141's hint I proceeded as follows: With $\Im(x) = \frac{x - \bar{x}}{2i}$ we have
\begin{align}
\int_{\mathbb R^n} \sin(\langle a, x \rangle) e^{-i \langle x, \xi \rangle} dx
& = \frac{1}{2i} \int_{\mathbb R^n} \left(e^{i \langle a, x \rangle} - e^{-i \langle a, x \rangle}\right)e^{-i \langle x, \xi \rangle} dx \\
& = \frac{1}{2i} \int_{\mathbb R^n} e^{-i \langle \xi - a, x \rangle} - e^{-i \langle \xi + a, x \rangle} dx.
\end{align}
Now we can use Fubini's theorem (F) to evaluate both summands separately:
\begin{align}
\int_{\mathbb R^n} e^{- i \langle \xi \pm a, x \rangle} dx
& = \int_{\mathbb R^n} \prod_{k = 1}^{n} e^{-i x_k (\xi_k \pm a_k)} dx 
\overset{\text{(F)}}{=} \prod_{k = 1}^{n} \int_{\mathbb R} e^{-i x_k (\xi_k \pm a_k)} dx_k,
\end{align}
but I fear that I can't use Fubini's theorem as the integral on the RHS doesn't converge.
 A: *

*Since $(-ix)^\alpha\hat f = \widehat{D^\alpha f},$
$$
\int_{\mathbb{R^n}} x^\alpha \mathcal F\phi dx
= i^{|\alpha|}\int_{\mathbb{R^n}}\mathcal F(D^\alpha\phi) dx
= (2\pi)^{n/2}i^{|\alpha|}\mathcal F(\mathcal F(D^\alpha\phi))(0)
= (2\pi)^{n/2}i^{|\alpha|}D^\alpha\phi(0).
$$

*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 
$$\int\sin(\langle a, x \rangle) \mathcal  F\phi (x) dx = (2\pi)^{n/2}\frac{ \phi(a) - \phi (-a)}{2i}.$$

*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.
A: 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 
