How to find the integral of $\int_{-\infty}^{\infty}e^{ix}\frac{\arctan{x}}{1+x^2}dx$ How to find the integral of $$\int_{-\infty}^{\infty}e^{ix}\frac{\arctan{x}}{1+x^2}dx$$
Is it possible to find a closed form solution or only a power series representation is possible for this? I have tried the integration by parts with the substitution of $z=\arctan x$. But then $\int_{-\infty}^{\infty}e^{i\tan{z}} dz$ is not obvious for me to find.
 A: We will use the convention $\mathcal{F}(f) (\omega) = \int_\mathbb{R} f(t) \mathrm{e}^{\mathrm{i} \omega t} \, \mathrm{d} t$ for Fourier transforms. Following Ninad Munshi's idea from the comments, we will employ the convolution theorem (for two functions in $L^2(\mathbb{R})$, which is fine according to this question). It reads $\mathcal{F} (f g) = \frac{\mathcal{F}(f) * \mathcal{F}(g)}{2 \pi}$ with our definition of the Fourier transform.

Define $\phi \colon \mathbb{R} \to \mathbb{C},$
$$ \phi(\omega) = \mathcal{F} \left(t \mapsto \frac{\arctan(t)}{1+t^2}\right) (\omega) = \int \limits_{-\infty}^\infty \frac{\arctan{t}}{1+t^2} \mathrm{e}^{\mathrm{i} \omega t} \, \mathrm{d} t = \int \limits_{-\infty}^\infty \frac{t}{1+t^2} \frac{\arctan{t}}{t} \mathrm{e}^{\mathrm{i} \omega t} \, \mathrm{d} t \, .$$
Your integral is $\phi(1)$. The result
$ \mathcal{F} \left(t \mapsto \frac{t}{1+t^2}\right) (\omega) = \mathrm{i} \pi \operatorname{sgn}(\omega) \mathrm{e}^{-\lvert \omega \rvert}$ is well-known (it follows from the residue theorem) and $ \mathcal{F} \left(t \mapsto \frac{\arctan(t)}{t}\right) (\omega) = \pi \operatorname{E}_1 (\lvert \omega \rvert)$ is discussed here ($\operatorname{E}_1$ is an exponential integral). Therefore, for $\omega > 0$ ($\phi$ is purely imaginary and odd by symmetry) the convolution theorem yields
\begin{align}
\phi (\omega) &= \frac{1}{2\pi} \int \limits_{-\infty}^\infty \pi \operatorname{E}_1 (\lvert \nu \rvert) \mathrm{i} \pi \operatorname{sgn}(\omega - \nu) \mathrm{e}^{-\lvert \omega - \nu \rvert} \, \mathrm{d} \nu \\
&= \frac{\mathrm{i} \pi}{2} \left[\int \limits_{-\infty}^0 \operatorname{E}_1 (- \nu) \mathrm{e}^{\nu - \omega} \, \mathrm{d} \nu + \int \limits_0^\omega \operatorname{E}_1 (\nu) \mathrm{e}^{\nu - \omega} \, \mathrm{d} \nu - \int \limits_\omega^\infty \operatorname{E}_1 (\nu) \mathrm{e}^{\omega - \nu} \, \mathrm{d} \nu\right] \\
&= \frac{\mathrm{i} \pi}{2} \left[\mathrm{e}^{- \omega} \int \limits_0^\infty \operatorname{E}_1 (\nu) \mathrm{e}^{- \nu} \, \mathrm{d} \nu + \mathrm{e}^{- \omega} \int \limits_0^\omega \operatorname{E}_1 (\nu) \mathrm{e}^{\nu} \, \mathrm{d} \nu - \mathrm{e}^{\omega} \int \limits_\omega^\infty \operatorname{E}_1 (\nu) \mathrm{e}^{- \nu} \, \mathrm{d} \nu\right] \\
&\!\stackrel{\text{IBP}}{=} \frac{\mathrm{i} \pi}{2} \left[\mathrm{e}^{- \omega} \int \limits_0^\infty \frac{\mathrm{e}^{-\nu} - \mathrm{e}^{-2 \nu}}{\nu} \, \mathrm{d} \nu + \mathrm{e}^{- \omega} \left(\left(\mathrm{e}^{\omega} - 1\right) \operatorname{E}_1 (\omega) + \int \limits_0^\omega \frac{1 - \mathrm{e}^{-\nu}}{\nu} \, \mathrm{d} \nu \right) \right. \\
&\phantom{\!\stackrel{\text{IBP}}{=} \frac{\mathrm{i} \pi}{2} \left[\vphantom{\int \limits_0^\infty}\right.} + \left. \mathrm{e}^{\omega} \left(\left(1 - \mathrm{e}^{-\omega}\right) \operatorname{E}_1 (\omega) - \int \limits_\omega^\infty \frac{\mathrm{e}^{-\nu} - \mathrm{e}^{-2\nu}}{\nu} \, \mathrm{d} \nu \right)\right] \\
&= \frac{\mathrm{i} \pi}{2} \left[\mathrm{e}^{- \omega} \log(2) + \mathrm{e}^{-\omega}\left[\operatorname{Ein} (\omega) - \operatorname{E}_1 (\omega)\right] + \mathrm{e}^{\omega} \left[\operatorname{E}_1 (\omega) - \operatorname{E}_1 (\omega) + \operatorname{E}_1 (2 \omega)\right] \right] \\
&= \frac{\mathrm{i} \pi}{2} \left[\mathrm{e}^{\omega} \operatorname{E}_1 (2 \omega) + \mathrm{e}^{-\omega} \left[\gamma + \log(2 \omega)\right]\right] \, .
\end{align}
Here, $\gamma$ is the Euler-Mascheroni constant and $\operatorname{Ein}$ is another exponential integral connected to $\operatorname{E}_1$ by a simple relationship. In particular, your integral is
$$ \phi (1) = \frac{\mathrm{i} \pi}{2} \left[\mathrm{e} \operatorname{E}_1 (2) + \frac{\gamma + \log(2)}{\mathrm{e}}\right] \, . $$
