Fourier transform of $\frac{\arctan(x)}{x}$ If I try to compute the fourier transform of 
$$
f(x)=\frac{\arctan(x)}{x}
$$
Maple return the function
$$
g(\omega) = \frac{1}{2} \pi e^{-|\omega|}
$$
However when I try to compute the inverse fourier transform of $g$ I get
$$
h(x) = \frac{1}{2(x^2 + 1)}.
$$
command I used are "fourier" and "invfourier". I was wondering if there's a way to derive by "hand" the fourier transform, the function is in $L^2$ so it exists.
Update: 
Below a screenshot of the commands I used.

 A: Write 
$$\frac{\arctan{x}}{x} = \int_0^1 \frac{du}{1+x^2 u^2} $$
so that the FT may be written as
$$\int_0^1 \frac{du}{u^2} \int_{-\infty}^{\infty} dx \frac{e^{i k x}}{\frac1{u^2}+x^2} $$
The inner integral is simply the FT of the Lorentzian function, or $\pi u e^{-|k|/u}$.  The FT is then
$$\pi \int_0^1 \frac{du}{u} e^{-|k|/u} = \pi \int_1^{\infty} \frac{du}{u} e^{-|k| u} = \pi \Gamma(0,|k|)$$
A: 
Herein, we will use the form of the Fourier Transform of a function $f$ given by $$\mathscr{F}\left(f\right)=\int_{-\infty}^\infty f(x)e^{i\omega x}\,dx$$Integrals are interpreted in terms of Cauchy Principal Values.


We can use the Convolution Theorem to find the Fourier Transform of $\frac{\arctan(x)}{x}$.  First, we note that the Fourier Transforms of $\frac1x$ and $\arctan(x)$ are given respectively by 
$$\mathscr{F}\left(\frac{1}{x}\right)=i\pi \,\text{sgn}(\omega) \tag 1$$
$$\begin{align}
\mathscr{F}\left(\arctan(x)\right)&=\frac{i}{\omega}\,\mathscr{F}\left(\frac{d\arctan(x)}{dx}\right)\\\\
&=\frac{i}{\omega}\,\mathscr{F}\left(\frac{1}{1+x^2}\right)\\\\
&=\frac{i\,\pi}{\omega} \,e^{-|\omega|} \tag 2
\end{align}$$
Then, using the Convolution Theorem, we find that for $\omega\ne 0$, the Fourier Transform of $\frac{\arctan(x)}{x}$ is given by
$$\begin{align}\mathscr{F}\left(\frac{\arctan(x)}{x}\right)&=\text{PV}\left(\int_{-\infty}^\infty i\pi \,\text{sgn}(\omega-\omega')\,\frac{i\pi}{\omega'} \,e^{-|\omega'|}\,d\omega'\right)\\\\
&=-\frac{\pi}{2}\,\lim_{\epsilon\to0}\left(\int_{-\infty}^{-\epsilon}\text{sgn}(\omega-\omega')\,\frac{e^{-|\omega'|}}{\omega'} \,d\omega'+\int_{\epsilon}^\infty \text{sgn}(\omega-\omega')\,\frac{e^{-|\omega'|}}{\omega'}\,d\omega'\right)\\\\
&=-\frac{\pi}{2}\,\left(\int_{-\infty}^{-|\omega|}\text{sgn}(\omega-\omega')\,\frac{e^{-|\omega'|}}{\omega'} \,d\omega'+\int_{|\omega|}^\infty \text{sgn}(\omega-\omega')\,\frac{e^{-|\omega'|}}{\omega'}\,d\omega'\right)\\\\
&=\frac{\pi}{2}\,\left(-\int_{-\infty}^{-|\omega|}\frac{e^{\omega'}}{\omega'} \,d\omega'+\int_{|\omega|}^\infty \,\frac{e^{-\omega'}}{\omega'}\,d\omega'\right)\\\\
&=\pi \int_{|\omega|}^\infty \frac{e^{-\omega'}}{\omega'}\,d\omega'\\\\
&=\pi \Gamma\left(0,|\omega|\right)
\end{align}$$
where $\Gamma(x,y)=\int_y^\infty u^{x-1}e^{-u}\,du$ is the upper Incomplete Gamma Function.
