# What is the Fourier transform of $1/|x|$?

I looked it up in several tables and calculated it in Mathematica and Matlab. Some tables say that the answer is simply $$\frac{1}{|\omega|}$$ and in other table it is $$-\sqrt{\frac{2}{\pi}}\ln|\omega|$$ and in Mathematica and Matlab (mupad) it is $$-\sqrt{\frac{2}{\pi}}\ln|\omega|-2\gamma$$ where $\gamma$ is the Euler–Mascheroni constant. Why are there so many answers? Are they all equivalent in some way or two (or all) of them are wrong?

• The correct answer is wolframalpha.com/input/?i=fourier+transform+%7C1%2Fx%7C . The first answer is clearly wrong. Adding a constant C to the FT of a function $f(x)$ adds $C\delta_0$ to the inverse transform - so the second, third and W|A answers differ but a multiple of $\delta_0$ which can be obscurred by the fact that $\frac 1 {|x|}$ diverges at $0$. You can obtain the correct constant by directly computing FT at $1$. – A.S. Nov 10 '15 at 21:10
• $$\int_\mathbb{R} \frac{1}{|x|} e^{-i\omega x}dx = \int_0^{+\infty} \frac{1}{x} e^{-i\omega x}dx - \int_{-\infty}^{0} \frac{1}{x} e^{-i\omega x}dx = \\ =\int_0^{+\infty} \frac{1}{x} e^{-i\omega x}dx - \int_0^{+\infty} \frac{1}{x} e^{i\omega x}dx = \\ =\int_0^{+\infty} \frac{1}{x} \left(e^{-i\omega x}- e^{i\omega x}\right)dx =-2i\int_0^{+\infty} \frac{\sin(\omega x)}{x} dx =\\ =-2i\omega\int_0^{+\infty} \frac{\sin(\omega x)}{\omega x} dx = \\ =-\frac{2i\omega}{2}\int_{-\infty}^{+\infty} \frac{\sin(\omega x)}{\omega x} dx = -i\omega$$ – the_candyman Nov 10 '15 at 21:15
• It seems that $-\int_{-\infty}^0\frac{1}{x}e^{-i\omega x}dx=+\int_0^{+\infty}\frac{1}{x}e^{i\omega x}$? @the_candyman – Molec Nov 10 '15 at 21:34
• @Molec yeah you are right, I forgot to change the sign when reversing the integration interval! – the_candyman Nov 10 '15 at 21:41
• $1/|x|$ is not locally integrable, hence not even a tempered distribution. So how would you define the Fourier transform? – PhoemueX Nov 11 '15 at 7:55

## 2 Answers

Although it is too late, let me give an explanation in hope for future reference.

Let me first compute the distributional Fourier transform (FT) of the function $f(x) = \log|x|$. Since $f$ is locally integrable and has slow growth, $f$ defines a tempered distribution on $\mathbb{R}$. So its FT is also well-defined.

Before we proceed, let us fix the convention for FT. Here, we will use the version

$$\mathcal{F}[f](\xi) = \hat{f}(\xi) = \int_{\mathbb{R}} e^{-2\pi i \xi x} f(x) \, dx.$$

Now with a bit of computation, you can check that for any Schwartz function $\varphi \in \mathcal{S}(\mathbb{R})$,

\begin{align*} \langle \hat{f}, \varphi \rangle = \langle \hat{\varphi}, f \rangle &= \int_{\mathbb{R}} \hat{\varphi}(x) \log|x| \, dx \\ &= -(\gamma + \log(2\pi))\varphi(0) - \int_{\mathbb{R}} \frac{\varphi(x) - \varphi(0)\mathbf{1}_{[-1,1]}(x)}{|x|} \, dx. \tag{1} \end{align*}

This means that $\hat{f}$ does not reduce to a linear combination of locally integrable functions and Dirac delta. But if we restrict the domain of $\hat{f}$ onto the space $X=\{\varphi \in \mathcal{S}(\mathbb{R}) : \varphi(0)= 0\}$, then the FT has the following nice representation:

$$\hat{f}(x) = -(\gamma + \log(2\pi))\delta_0(x) - \frac{1}{|x|} \quad \text{on } X. \tag{2}$$

Precisely in this sense, we may loosely say that the FT of $\frac{1}{|x|}$ is $- \log|\xi| - \gamma - \log(2\pi)$ in light of the inverse FT.

But wait! Since $\text{(21)}$ is being considered on the space $X$ where $\varphi(0) = 0$, taking Dirac delta gives no difference. So we can equally say that

$$\hat{f}(x) = -\frac{1}{|x|} \quad \text{on } X$$

and hence we can also say that the FT of $\frac{1}{|x|}$ is simply $-\log |\xi|$. There is no contradiction on this, since we are considering this distribution on the subspace

$$\mathcal{F}[X] = \left\{\varphi \in \mathcal{S}(\mathbb{R}) : \int_{\mathbb{R}} \varphi(x) \, dx = 0 \right\}$$

and thus adding constants makes no difference.

I am not good at physics, but I guess we can give a physical interpretation of it. Notice that $\text{(1)}$ has a term that cancels out the singularity of $\frac{1}{|x|}$. So in a very loose sense, it has the form

$$(\log |x|)^{\wedge}(\xi) = - \frac{1}{|\xi|} + \infty \cdot \delta_0(\xi) \qquad \Leftrightarrow \quad \left( \frac{1}{|x|}\right)^{\wedge}(x) = \infty - \log |\xi|.$$

Thus in order that we can ignore the cancellation part, we should work on the situation where the addition of constant does not affect the meaning of the FT of $\frac{1}{|x|}$. Probably this is the case for potentials, where only the difference in values has actual physical meaning.

Addendum. (Derivation of $\text{(1)}$) The key trick is the following Frullani-type integral

$$\forall a > 0 \ : \quad \int_{0}^{\infty} \frac{e^{-2\pi s} - e^{-2\pi as}}{s} \, ds = \log a.$$

From this, we find that

\begin{align*} \int_{\mathbb{R}} \hat{\varphi}(x) \log|x| \, dx &= \int_{\mathbb{R}} \hat{\varphi}(x) \left( \int_{0}^{\infty} \frac{e^{-2\pi s} - e^{-2\pi |x|s}}{s} \, ds \right) \, dx \\ &= \int_{0}^{\infty} \frac{1}{s} \left( \int_{\mathbb{R}} \hat{\varphi}(x) (e^{-2\pi s} - e^{-2\pi |x|s}) \, dx \right) \, ds \tag{i} \\ &= \int_{0}^{\infty} \frac{1}{s} \left( e^{-2\pi s} \varphi(0) - \int_{\mathbb{R}} \hat{\varphi}(x) e^{-2\pi |x|s} \, dx \right) \, ds \tag{ii} \\ &= \int_{0}^{\infty} \left( \frac{e^{-2\pi s}}{s} \varphi(0) - \int_{\mathbb{R}} \frac{\varphi(x)}{\pi(s^2 + x^2)} \, dx \right) \, ds \tag{iii} \end{align*}

Here,

(i) We interchanged the order of integrations by Fubini's theorem. It is not hard to check that the integrability condition is indeed met.

(ii) We utilized Fourier inversion $\varphi(\xi) = \int_{\mathbb{R}} \hat{\varphi}(x) e^{2\pi i \xi x} \, dx$.

(iii) We utilized the fact that $\langle \hat{\varphi}, \psi \rangle = \langle \varphi, \hat{\psi} \rangle$ for any integrable functions $\varphi, \psi$, together with the fact that the FT of $e^{-2\pi s |\cdot|}$ is $\frac{1}{\pi(s^2 + (\cdot)^2)}$.

Now we add and subtract the term

$$\int_{\mathbb{R}} \frac{\varphi(0) \mathbf{1}_{[-1,1]}(x)}{\pi(s^2 + x^2)} \, dx = \frac{2\arctan(1/s)}{\pi s} \, \varphi(0)$$

inside the outer integral. Then

\begin{align*} \int_{\mathbb{R}} \hat{\varphi}(x) \log|x| \, dx &= \varphi(0) \int_{0}^{\infty} \left( \frac{e^{-2\pi s}}{s} - \frac{2\arctan(1/s)}{\pi s} \right) \, ds \\ &\qquad + \int_{0}^{\infty} \int_{\mathbb{R}} \left( \frac{\varphi(x) - \varphi(0) \mathbf{1}_{[-1,1]}(x)}{\pi(s^2 + x^2)} \right) \, dx \, ds \end{align*}

Now the first term is just a constant multiple of $\varphi(0)$, and this constant turns out to have the value

$$\int_{0}^{\infty} \left( \frac{e^{-2\pi s}}{s} - \frac{2\arctan(1/s)}{\pi s} \right) \, ds = -\gamma - \log(2\pi).$$

For the second term, interchanging the order of integrations using Fubini's theorem gives

$$\int_{\mathbb{R}} \frac{\varphi(x) - \varphi(0) \mathbf{1}_{[-1,1]}(x)}{|x|} \, dx$$

Therefore the claim follows.

This can be computed rigorously in terms of distributions, see here. The transform of a distribution is defined as $$(\mathcal F[f], \phi) = (f, \mathcal F[\phi]).$$ If $f$ is a regular distribution, induced by an ordinary function $f_o$, and the order of integration in the resulting double integral can be changed, we get $$(\mathcal F[f], \phi) = \int dx f_o(x) \mathcal F[\phi](x) = \\ \int dx f_o(x) \int d\xi \phi(\xi) e^{i x \xi} = \\ \int d\xi \left( \int dx f_o(x) e^{i \xi x} \right) \phi(\xi) = \\ \int d\xi F_o(\xi) \phi(\xi),$$ which means that $\mathcal F[f]$ is a regular distribution induced by $F_o$. Further, $F_o = (f, e^{i \xi x})$ (this can be thought of as just a shorthand notation for the integral, since $e^{i \xi x}$ is not in the space of test functions), which explains why it's valid to compute $\mathcal F[f]$ by directly applying $f$ to $e^{i \xi x}$: $$\mathcal F\!\left[ |x|^{-1} \right] = 2\int_0^{1} dx \frac {\cos \xi x - 1} x + 2\int_1^\infty dx \frac {\cos \xi x} x = -2 \ln |\xi| - 2 \gamma.$$