So I am given this simple example, where $T \in \mathcal{S}(\mathbb{R})$:
\begin{equation} T=(\mu +\lambda x+\beta x^2)H(x) \end{equation} where $H(x)\in \mathcal{S}(\mathbb{R})$ (also notated as the Heaviside $\theta$). Also I use the following definition for the Fourier transform: \begin{equation} \hat{f}(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx, k\in \mathbb{R} \end{equation}
Now I would like to derive the Fourier transform of the above,so I make use of the properties of the Fourier transform for tempered distributions. For example, since I know the Fourier transform of $H(x)$ to be equal to: \begin{equation} \hat{H}(k)=-\frac{i}{k\sqrt{2\pi}}+\sqrt{\frac{\pi}{2}}\delta(k) \end{equation} I can use the relation:
\begin{equation} \mathcal{F}[-ixH(x)]=\hat{H}'(k) \Leftrightarrow \mathcal{F}[xH(x)]=i\hat{H}'(k) \end{equation} and derive the Fourier transform for $\lambda(xH(x))$. I guess I can do that also for the expression $\beta (x^2H(x))$, right? In the end I found my result to be: \begin{equation} \hat{T}(k)=\frac{1}{\sqrt{2\pi}}\left( \frac{2i\beta -\lambda k-\mu ik^2}{k^3}\right)+\sqrt{\frac{\pi}{2}}\left( \delta(k)+i\delta'(k)-\delta''(k)\right) \end{equation} and Mathematica gives: \begin{equation} \hat{T}(k)=-\sqrt{\frac{2}{\pi}}\frac{i \beta }{k^3}+\sqrt{\frac{\pi}{2}} \mu \delta(k)-\frac{\lambda}{\sqrt{2\pi}k^2}+\frac{i \mu }{\sqrt{2\pi}k} \end{equation} So I have a couple of $\delta$ derivatives which should not be there.. Why am I so wrong?
Thank you!