Leading behaviour of Fourier transform for $\omega\to 0$

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}