Distributional Fourier transform of $x \arctan(x)$

Question

I want to find the Fourier transform $\hat{f}$ of $f(x) = x \arctan(x)$ in the sense of distributions. It is known that for any $\varphi \in \mathscr{S\mathbb{(R}})$, $\langle \hat f,\varphi\rangle = \langle f,\hat \varphi\rangle$. So we get \begin{align*} \langle \hat f,\varphi\rangle &=\int_\mathbb{R} x \arctan(x) \, \hat\varphi(x) \,\mathrm dx\\ &=\int_\mathbb{R}x \arctan(x) \int_\mathbb{R}\varphi(k)\,e^{-ikx}\,\mathrm dk \,\mathrm dx\\ &= \int_\mathbb{R}\left(\int_{\mathbb{R}}x \arctan(x)\,e^{-ikx}\,\mathrm dx \right)\varphi(k) \,\mathrm dk \end{align*}

We should rewrite $x \arctan(x)$ in exponential form in order to solve integral? Thanks for any help!

Answer 1

Let's define the Fourier transform of $f(x)\in L^1$ as

$$\mathscr{F}\{f\}(k)=\int_{-\infty}^\infty f(x)e^{ikx}\,dx$$

First, note that with $X(x)=x$ and $g(x)=\arctan(x)$ we have

$$\begin{align} \mathscr{F}\{Xg\}(k)&=-i\frac{d}{dk}\mathscr{F}\{g\}(k)\\\\ &=-i\frac{d}{dk}\left(\text{PV}\left(\frac{i\pi}{k}e^{-|k|}\right)\right)\\\\ \end{align}$$

where the derivative is taken in the sense of distributions and wehre we used THIS ANSWER to arrive at the Fourier transform of the arctangent.