How to compute $ P.V. \int_{-\infty}^{\infty} \frac{e^{iwx}}{|w|} \,dw $? I came across the integral:
$$
\mbox{P.V.}\int_{-\infty}^{\infty} {\mathrm{e}^{\mathrm{i}\omega x} \over
\left\lvert\omega\right\rvert}
\,\mathrm{d}\omega
$$
in my research, and can’t figure it out after several hours of trial and error.
I didn’t know how to use Contour integration techniques on it because the pole is an absolute value, so I tried generating a power series for the exponential kernel, but ended up with a log function that blew up.
May you please explain how I can evaluate such an integral $?$. Wolfram has the answer, but I care more about the thought process than anything.
Many thanks in advance $!$.
 A: 1. The integral in question does not exist in the Cauchy principal value sense
$$ \lim_{\substack{\epsilon \to 0^+ \\ R \to \infty}} \int_{\epsilon < |x| < R} \frac{e^{i\omega x}}{\left| x \right|} \, \mathrm{d}x $$
precisely because of the logarithmic divergence mentioned in @Felix Marin's answer.
2. If we interpret the integral as the Fourier transform of a suitable distribution (a.k.a. generalized function), however, then it can be shown that
$$ \mathcal{F}\left[\operatorname{pf}\frac{1}{\left|x\right|}\right](\xi) = -2(\gamma + \log\left|\xi\right|). \tag{*} $$
Here, $\operatorname{pf}\frac{1}{\left|x\right|}$ is the distributional derivative of $\operatorname{sgn}(x)\log\left|x\right|$ and $\mathcal{F}$ is the Fourier transform of tempered distributions which extends
$$ \hat{f}(\xi) = \mathcal{F}\left[f\right](\xi) = \int_{-\infty}^{\infty} f(x)e^{i\omega x} \, \mathrm{d}x $$
for Schwarz functions $f \in \mathcal{S}(\mathbb{R})$. (More precisely, for a tempered distribution $T$ over $\mathbb{R}$, its Fourier transform $\hat{T}$ is defined via the relation $\langle \hat{T}, \varphi \rangle = \langle T, \hat{\varphi} \rangle$ for all $\varphi \in \mathcal{S}(\mathbb{R})$.)

*

*We first note that
$$ \operatorname{pf}\frac{1}{\left|x\right|} = \bigl[ \left|x\right| (\log\left|x\right|-1) \bigr]'' $$
in distributional sense. Since $\left|x\right|(\log\left|x\right|-1)$ is a continuous function with polynomial growth, it follows from the structure theorem that $\operatorname{pf}\frac{1}{\left|x\right|}$ is a tempered distribution, hence its Fourier transform is well-defiend.


*We first demonstrate a less rigorous argument. Using the idea of regularization, we may heuristically argue that
\begin{align*}
\mathcal{F}\left[\operatorname{pf}\frac{1}{\left|x\right|}\right](\xi)
&= \lim_{\epsilon \to 0^+} \left< \operatorname{pf}\frac{1}{\left|x\right|}, e^{-\epsilon x^2 + i\xi x} \right> \\
&= - \lim_{\epsilon \to 0^+} \left< \operatorname{sgn}(x)\log\left|x\right|, \bigl( e^{-\epsilon x^2 + i\xi x} \bigr)' \right> \\
&= - \lim_{\epsilon \to 0^+} \int_{-\infty}^{\infty} \operatorname{sgn}(x)\log\left|x\right| (-2\epsilon x + i\xi) e^{-\epsilon x^2 + i\xi x} \, \mathrm{d}x.
\end{align*}
Then
\begin{align*}
&\int_{-\infty}^{\infty} \operatorname{sgn}(x)\log\left|x\right| (-2\epsilon x + i\xi) e^{-\epsilon x^2 + i\xi x} \, \mathrm{d}x \\
&= 2 \int_{0}^{\infty} \log x (-2\epsilon x \cos(\xi x) - \xi \sin(\xi x)) e^{-\epsilon x^2} \, \mathrm{d}x \\
&= 2 \int_{0}^{1} \log x \bigl( e^{-\epsilon x^2} \cos(\xi x) - 1 \bigr)' \, \mathrm{d}x + 2 \int_{1}^{\infty} \log x \bigl( e^{-\epsilon x^2} \cos(\xi x) \bigr)' \, \mathrm{d}x \\
&= 2 \int_{0}^{1} \frac{1 - e^{-\epsilon x^2} \cos(\xi x)}{x} \, \mathrm{d}x - 2 \int_{0}^{1} \frac{e^{-\epsilon x^2} \cos(\xi x)}{x} \, \mathrm{d}x.
\end{align*}
Letting $\epsilon \to 0^+$ gives
\begin{align*}
\mathcal{F}\left[\operatorname{pf}\frac{1}{\left|x\right|}\right](\xi)
&= - 2 \int_{0}^{1} \frac{1 - \cos(\xi x)}{x} \, \mathrm{d}x + 2 \int_{0}^{1} \frac{\cos(\xi x)}{x} \, \mathrm{d}x \\
&= -2(\gamma + \log\left|\xi\right|)
\end{align*}
as desired.


*Here is a more rigorous proof. Let $\varphi \in \mathcal{S}(\mathbb{R})$. Then by the definition and the properties of the pairing $\langle \cdot, \cdot \rangle$,
\begin{align*}
\left< \mathcal{F}\left[\operatorname{pf}\frac{1}{\left|x\right|}\right], \varphi \right> 
&= \left< \operatorname{pf}\frac{1}{\left|x\right|}, \hat{\varphi} \right>
= - \left< \operatorname{sgn}(x) \log\left|x\right|, \hat{\varphi}' \right> \\
&= - \int_{-\infty}^{\infty} \hat{\varphi}'(x) \operatorname{sgn}(x) \log\left|x\right| \, \mathrm{d}x,
\end{align*}
where the last integral is the usual Lebesgue integral. Then
\begin{align*}
&= - \int_{0}^{\infty} (\hat{\varphi}'(x) - \hat{\varphi}'(-x)) \log x \, \mathrm{d}x \\
&= - \int_{0}^{1} (\hat{\varphi}(x) + \hat{\varphi}(-x) - 2\hat{\varphi}(0))' \log x \, \mathrm{d}x \\
&\qquad  - \int_{1}^{\infty} (\hat{\varphi}(x) + \hat{\varphi}(-x))' \log x \, \mathrm{d}x \\
&= \int_{0}^{\infty} \frac{\hat{\varphi}(x) + \hat{\varphi}(-x) - 2\hat{\varphi}(0) \mathbf{1}_{[0,1]}(x)}{x} \, \mathrm{d}x \\
&= \lim_{\substack{\epsilon \to 0^+ \\ R\to\infty}} \int_{\epsilon}^{R} \frac{\hat{\varphi}(x) + \hat{\varphi}(-x) - 2\hat{\varphi}(0) \mathbf{1}_{[0,1]}(x)}{x} \, \mathrm{d}x. \tag{1}
\end{align*}
Assuming $0 < \epsilon < 1 < R$ without losing the generality, the last integral can be computed by the Fubini's theorem:
\begin{align*}
&\int_{\epsilon}^{R} \frac{\hat{\varphi}(x) + \hat{\varphi}(-x) - 2\hat{\varphi}(0) \mathbf{1}_{[0,1]}(x)}{x} \, \mathrm{d}x \\
&= \int_{-\infty}^{\infty} \varphi(\xi) \biggl( \int_{\epsilon}^{R} \frac{e^{ix\xi} + e^{-ix\xi} - 2 \cdot \mathbf{1}_{[0,1]}(x)}{x} \, \mathrm{d}x \biggr) \, \mathrm{d}\xi \\
&= 2 \int_{-\infty}^{\infty} \varphi(\xi) \left( \operatorname{Ci}(R\left|\xi\right|) - \operatorname{Ci}(\epsilon\left|\xi\right|) - \log(1/\epsilon) \right) \, \mathrm{d}\xi,
\end{align*}
where $\operatorname{Ci}(\cdot)$ is the cosine integral. Then by using the asymptotic formulas for $\operatorname{Ci}(\cdot)$,
$$ \operatorname{Ci}(R\left|\xi\right|) = (\log(R\left|\xi\right|)) \mathbf{1}_{\{\left|\xi\right|< 1/R\}} + \mathcal{O}(1) $$
$$ - \operatorname{Ci}(\epsilon\left|\xi\right|) - \log(1/\epsilon) = -(\log\left|\xi\right|) \mathbf{1}_{\{\left|\xi\right| < 1/\epsilon\}} + (\log \epsilon) \mathbf{1}_{\{\left|\xi\right| > 1/\epsilon\}} + \mathcal{O}(1) $$
uniformly in $\epsilon$, $R$, and $\xi$. Since $\varphi$ is rapidly decreasing, it is not hard to show that
\begin{align*}
&2 \int_{-\infty}^{\infty} \varphi(\xi) \left( \operatorname{Ci}(R\left|\xi\right|) - \operatorname{Ci}(\epsilon\left|\xi\right|) - \log(1/\epsilon) \right) \, \mathrm{d}\xi \\
&= 2 \int_{-\infty}^{\infty} \varphi(\xi) \left( \operatorname{Ci}(R\left|\xi\right|) - \operatorname{Ci}(\epsilon\left|\xi\right|) - \log(1/\epsilon) \right)\mathbf{1}_{\{\epsilon < \left|\xi\right|<R\}} \, \mathrm{d}\xi + o(1)
\end{align*}
uniformly as $\epsilon \to 0^+$ and $R\to \infty$, and then by the dominated convergence theorem, the limit in $\text{(1)}$ is computed as
\begin{align*}
\text{(1)}
&= 2 \int_{-\infty}^{\infty} \varphi(\xi) \lim_{\substack{\epsilon \to 0^+ \\ R\to\infty}} \left( \operatorname{Ci}(R\left|\xi\right|) - \operatorname{Ci}(\epsilon\left|\xi\right|) - \log(1/\epsilon) \right)\mathbf{1}_{\{\epsilon < \left|\xi\right|<R\}} \, \mathrm{d}\xi \\
&= -2 \int_{-\infty}^{\infty} \varphi(\xi) (\gamma + \log\left|\xi\right|) \, \mathrm{d}\xi.
\end{align*}
By the definition of the Fourier transform, this implies $\text{(*)}$ as desired.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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{\on}[1]{\operatorname{#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}}}
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\begin{align}
&\bbox[5px,#ffd]{\mbox{P.V.}\int_{-\infty}^{\infty} {\expo{\ic\omega x} \over \verts{\omega}}\,\dd\omega}
\\[2mm] \stackrel{\mrm{DEF.}}{=}\,\,\,&
\lim_{\epsilon \to 0^{+}}\,\,\pars{%
\int_{-\infty}^{-\epsilon}{\expo{\ic\omega x} \over
-\omega}\,\dd\omega +
\int_{\epsilon}^{\infty}{\expo{\ic\omega x} \over
\omega}\,\dd\omega}
\\[5mm] = &\
\lim_{\epsilon \to 0^{+}}\,\,\pars{%
\int_{\epsilon}^{\infty}{\expo{-\ic\omega x} \over
\omega}\,\dd\omega +
\int_{\epsilon}^{\infty}{\expo{\ic\omega x} \over
\omega}\,\dd\omega}
\\[5mm] = &\
\lim_{\epsilon \to 0^{+}}\,\,\bracks{2%
\int_{\epsilon}^{\infty}{\cos\pars{\omega x} \over
\omega}\,\dd\omega}
\\[5mm] = &\
2\int_{0}^{\infty}{\cos\pars{\omega\verts{x}} \over
\omega}\,\dd\omega =
2\int_{0}^{\infty}{\cos\pars{\omega} \over
\omega}\,\dd\omega
\\[2mm] &\
\mbox{which}\ diverges\ \mbox{logarithmically.}
\end{align}
