For $a\in\mathbb{C}$, let us consider the following differential equation in $\mathcal{S}'(\mathbb{R})$, the set of tempered distributions on $\mathbb{R}$: $$xT''+2T'+(a-x)T=0.$$ For $T\in\mathcal{S}'(\mathbb{R})$, $xT''+2T'+(a-x)\in\mathcal{S}'(\mathbb{R})$ and since the fourier transform $\mathcal{F}:\mathcal{S}'(\mathbb{R})\to \mathcal{S}'(\mathbb{R})$ is an isomorphism, one gets: $$xT''+2T'+(a-x)T=0\Leftrightarrow\widehat{T}'+\frac{ia}{1+\xi^2}\widehat{T}=0.$$ Let $\Delta_a\subseteq\mathcal{S}'(\mathbb{R})$ be the set of the solutions of $xT''+2T'+(a-x)T=0$, one derives: $$\dim_{\mathbb{R}}\Delta_a=1.$$ I am asked to proof that:
$T\in\Delta_a$ is $\mathcal{C}^{\infty}(\mathbb{R}^*)$ and for $k\in\mathbb{N}$, $|x|^kT(x)\underset{{|x|\to +\infty}}{\longrightarrow}0$.
At that moment, I know that: $$\widehat{T}(\xi)=C\exp(-ia\arctan(\xi)).$$ I thought using Paley-Wiener theorem, since I am fairly new to Fourier-analysis, I am a bit distraught. Any help will be appreciated, preferably with some hints.