Leading behaviour of Fourier transform for $\omega\to 0$ For his examination, today my son was asked to find the leading behaviour of $\hat f(\omega)$ for $\omega\to 0$, where
$$
\hat{f}(\omega)=\int_{-\infty}^\infty{e^{-i\omega x}\over 1+|x|}\,dx
$$ 
is the Fourier transform of $f(x)=1/(1+|x|)$.
The integral cannot be expressed in terms of elementary functions, but computing it with Mathematica I could find that $\hat f(\omega)\sim -2\ln |\omega|,$ as $\omega\to 0$. 
Of course there must be some obvious trick, allowing one to find the same result without actually computing the integral, but I haven't found it yet. Any ideas?
 A: Another solution: You need only to look at $\int_0^{+\infty}\frac{\cos(\omega x)}{1+x}dx$ for $\omega>0$. By the change of variable $\omega x=u$, this is $I=\int_0^{+\infty}\frac{\cos(u)}{u+\omega}du$. Write
 $$I=\int_0^{1}\frac{\cos(u)-1}{u+\omega}du+\int_0^1\frac{du}{u+\omega}+\int_1^{+\infty}(\frac{\cos(u)}{u+\omega}-\frac{\cos(u)}{u})du+\int_1^{+\infty}\frac{\cos(u)}{u}du=A+B+C+D$$
Now if $\omega\to 0$, $A\to \int_0^{1}\frac{\cos(u)-1}{u}du$ as this integral is convergent, $B$ is easily computable, $C$ is seen to converge to $0$, and $D$ is convergent, and it is easy to finish. Of course this does not gives the informations given in @Jack D'Aurizio's answer.   
A: By parity and the Laplace transform
$$ \widehat{f}(\omega) = 2\int_{0}^{+\infty}\frac{\cos(\omega x)}{1+x}\,dx = \int_{0}^{+\infty}\frac{2s\, e^{-s}}{s^2+\omega^2}\,ds\tag{1}$$
and by integration by parts
$$ \widehat{f}(\omega) = \int_{0}^{+\infty}\left(\log(s^2+\omega^2)-\log(\omega^2)\right)e^{-s}\,ds \tag{2} $$
from which:
$$ \widehat{f}(\omega) = -\log(\omega^2)-2\gamma+\int_{0}^{+\infty}\log\left(1+\frac{\omega^2}{s^2}\right)e^{-s}\,ds \tag{3} $$
and:
$$ \widehat{f}(\omega) = -2\log\left|\omega\right|-2\gamma + o(1)\qquad\text{as }\omega\to 0 \tag{4}$$
by the dominated convergence theorem, where $\gamma$ is Euler-Mascheroni constant.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\left.\hat{\mrm{f}}\pars{\omega}\right\vert_{\ \omega\ \not=\ 0} & \equiv
\int_{-\infty}^{\infty}{\expo{-\ic\omega x} \over 1 + \verts{x}}\,\dd x =
2\int_{0}^{\infty}{\cos\pars{\verts{\omega}x} \over 1 + x}\,\dd x =
2\int_{0}^{\infty}{\cos\pars{x} \over x + \verts{\omega}}\,\dd x
\\[5mm] & =
2\cos\pars{\verts{\omega}}
\int_{\verts{\omega}}^{\infty}{\cos\pars{x} \over x}\,\dd x +
2\sin\pars{\verts{\omega}}
\int_{\verts{\omega}}^{\infty}{\sin\pars{x} \over x}\,\dd x
\\[5mm] & =
-2\cos\pars{\verts{\omega}}\,\mrm{Ci}\pars{\verts{\omega}} +
2\sin\pars{\verts{\omega}}\bracks{{\pi \over 2} - \mrm{Si}\pars{\verts{\omega}}}
\end{align}

$\ds{\mrm{Ci}}$ is the
  Cosine Integral Function.
  Similarly, $\ds{\mrm{Si}}$ is the
  Sine Integral Function.

Expansions of those functions are given by
$\ds{\pars{~\gamma\ \mbox{is the}\ Euler\!-\!Mascheroni\ Constant~}}$:
$$
\left\{\begin{array}{rcl}
\ds{\mrm{Ci}\pars{z}} & \ds{=} &
\ds{\gamma + \ln\pars{z} +
\sum_{n = 1}^{\infty}\pars{-1}^{n}\,{z^{2n} \over \pars{2n}!\pars{2n}}}
\\[3mm]
\ds{\mrm{Si}\pars{z}} & \ds{=} &
\ds{\sum_{n = 0}^{\infty}\pars{-1}^{n}\,{z^{2n + 1} \over
\pars{2n + 1}!\pars{2n + 1}}}
\end{array}\right.
$$
such that to "the lowest order"
\begin{align}
\left.\hat{\mrm{f}}\pars{\omega}\right\vert_{\ \omega\ \not=\ 0} & \equiv
\int_{-\infty}^{\infty}{\expo{-\ic\omega x} \over 1 + \verts{x}}\,\dd x
\,\,\,\stackrel{\mrm{as}\ \omega\ \to\ 0}{\sim}
-2\gamma + 2\ln\pars{\verts{\omega}} + \pi\verts{\omega}
\end{align}
