Fourier transform of the regular tempered distributions induced by $x^\alpha$, $\sin(\langle a, x \rangle_{\mathbb R^n})$, and $\mathbb 1_{[-1,1]}$

Question

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

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

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

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

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 $\xi^{\alpha} (\mathcal{F}\phi)(\xi) = (-i)^{| \alpha |} \mathcal{F}(D^{\alpha} \phi)(\xi)$.

1. 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*}

2. 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 $$ \int_{\mathbb R} \sin(a x) e^{-i x \xi} dx = \frac{e^{-i \xi x}(a \cos(ax) + i \xi \sin(ax))}{\sqrt{2 \pi} (a - \xi)(a + \xi)}\bigg|_{x = - \infty}^{\infty} $$

3. 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.

Answer 1

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 $$ \mathcal F(\mathbb1_{[-1,1]})(x) = \frac1{\sqrt{2\pi}} \frac{2\sin(x)}x.$$

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(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.

Answer 2

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