# 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 $$!$$.

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|

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|

By the definition of the Fourier transform, this implies $$\text{(*)}$$ as desired.

• $+1$. Thanks. Nice job. We need it the mathematician opinion. Nov 16 '20 at 3:53

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

• firstly, thank you for your effort! It hasn’t gone unnoticed. I think you’re right until the second last integral, where you introduce the absolute value of x. But you’re on the right track. The real question at this point might be how to carefully compute the cos(wx)/w integral. Wolfram predicts that it is proportional to log(abs(x)). See link below
– user564215
Nov 16 '20 at 2:44
• wolframalpha.com/input/…
– user564215
Nov 16 '20 at 2:45
• ${\tt Mathematica}$ evaluates it as $+\infty$. Nov 16 '20 at 3:09
• ${\tt Mathematica}$ evaluates it As $+\infty$ whenever you set $\verb*\mathrm{PrincipalValue}*$ option to $\verb*True*$. However, if you ask for the ${\tt Mathematica}$ $\verb*FourierTransform*$ function it yields (besides some multiplicative constant) $-\left[\log\left(\left\vert x\right\vert\right) + \gamma\right]$. It's difficult to know what ${\tt Mathematica}$ is doing internally. I guess it would be nice you set this difference as a question in the SE ${\tt Mathematica}$ grooup. Nov 16 '20 at 3:39
• The classical transform diverges, but there is a distributional meaning of the Fourier transform of $1/|w|$. Nov 17 '20 at 23:02