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The Fourier transforms we use are \begin{align} \tilde{f}(\mathbf{p})&=\int d^3x\,f(\mathbf{x}) e^{-i\mathbf{p}\cdot\mathbf{x}}\\[5pt] f(\mathbf{x})&=\int \frac{d^3p}{(2\pi)^3}\,\tilde{f}(\mathbf{p}) e^{i\mathbf{p}\cdot\mathbf{x}} \end{align} I want to calculate the transfom of $1/p^3$ \begin{align} I=\int \frac{d^3p}{(2\pi)^3}\,\frac{1}{p^3} e^{i\mathbf{p}\cdot\mathbf{x}} \end{align}

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    $\begingroup$ Where $p=|\mathbf{p}|,$ I suppose. $\endgroup$
    – md2perpe
    Jun 17, 2020 at 13:29
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    $\begingroup$ @ md2perpe, yes $\endgroup$ Jun 17, 2020 at 13:36
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    $\begingroup$ I think it's important to clarify what do you mean by $1/\vert p\vert^3$; the function is not locally integrable and it doesn't have a direct interpretation as a distribution. You can define the distribution $1/t_+$ in $\mathbb{R}$ using integration by parts and the identity $\partial\log t = 1/t$ for $t>0$, so $\int_0^\infty \frac{1}{t}\phi(t)\,dt := -\int_0^\infty \log t\partial_t\phi\,dt$. We can carry this definition to your case as $\int\frac{1}{p^3}\phi := \int\frac{\log p}{p^2}\partial_r\phi$. By your notation I suspect you're a physicists studying QFT. $\endgroup$
    – user90189
    Jun 17, 2020 at 14:41
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    $\begingroup$ @user90189. Formally taking the Laplacian of the expression gives $$\begin{align} -\int \frac{d^3p}{(2\pi)^3}\,\frac{1}{p} e^{i\mathbf{p}\cdot\mathbf{x}} \end{align}$$ which is locally integrable and thus defines a distribution. That would at least take us a bit to a solution. $\endgroup$
    – md2perpe
    Jun 17, 2020 at 14:55
  • $\begingroup$ @md2perpe Nice! Formally the Fourier transform would be, up to a constant, $-\Delta^{-1}\frac{1}{\vert x\vert^2} = \frac{1}{\vert x\vert}*\frac{1}{\vert x\vert^2}$. We cannot define it directly as the Schwartz distribution $\langle \frac{1}{\vert x\vert^2},-\Delta^{-1}\phi\rangle$ because $-\Delta^{-1}\phi$ doesn't decay rapidly. Some modification is needed. $\endgroup$
    – user90189
    Jun 17, 2020 at 16:24

1 Answer 1

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In short, in $\mathbb{R}^d$, if I define the Fourier transform as $\mathcal{F}(f)(x) = \int_{\mathbb{R}^d} f(y) \,e^{-2iπxy}\,\mathrm{d}y$ the result is $$ \boxed{\mathcal{F}\!\left(\frac{1}{\omega_d\,|x|^d}\right) = \frac{\psi(d/2)-\gamma}{2} - \ln(|πx|)} $$ where $\omega_d = \frac{2\,\pi^{d/2}}{\Gamma(d/2)}$ is the size of the unit sphere (so $\omega_3 = 4\pi$), $\gamma$ is Euler–Mascheroni constant and $\psi$ is the digamma function. Since $\psi(3/2) = 2-\gamma-\ln(4)$, we deduce that in dimension $3$ $$ \mathcal{F}\!\left(\frac{1}{4\pi\,|x|^3}\right) = 1-\gamma -\ln(|2πx|) $$ which with your Fourier transform convention gives $$ \boxed{\tilde{\mathcal{F}}\!\left(\frac{1}{|x|^3}\right) = \frac{1-\gamma -\ln(|x|)}{2\pi^2}.} $$


Now the details. So first, what is the meaning of $\frac{1}{|x|^d}$? One can define the following distribution $$ v_d := \mathrm{pf}\!\left(\frac{1}{|x|^d}\right) := \mathrm{div}\!\left(\frac{x\ln(|x|)}{|x|^d}\right) $$ where the derivative is taken in the sense of distributions. One easily verifies that the distribution $v_d$ coincides on $\Bbb R^d\setminus\{0\}$ with the function $$ v_d(x) = \frac{1}{|x|^d} \text{ for any } x≠ 0. $$ One also notices that it is a tempered distribution as the derivative of a function in $L^1 + L^\infty$.


Let $u_d = \frac{x\ln(|x|)}{|x|^d}$. Multiplying by a test function $\varphi\in C^\infty_c$, one gets for any $\lambda>0$ $$ \begin{align*} \langle v_d,\varphi\rangle &= -\int_{\mathbb{R}^d} u_d\cdot\nabla\varphi \\ &= -\int_{|x|\leq\lambda} u_d\cdot\nabla(\varphi(x)-\varphi(0)) - \int_{|x|>\lambda} u_d\cdot\nabla \varphi \end{align*} $$ which by integration by parts yields $$ \begin{align}\tag{1}\label{eq1} \langle v_d,\varphi\rangle &= \int_{|x|\leq\lambda} \frac{\varphi(x)-\varphi(0)}{|x|^d}\,\mathrm{d}x \\ &\quad+ \int_{|x|>\lambda} \frac{\varphi(x)}{|x|^d} \,\mathrm{d}x + \omega_d \ln(\lambda) \,\varphi(0). \end{align} $$ One can take $\lambda = 1$ to get $$ \begin{align*} \langle v_d,\varphi\rangle &= \int_{|x|\leq 1} \frac{\varphi(x)-\varphi(0)}{|x|^d}\,\mathrm{d}x + \int_{|x|> 1} \frac{\varphi(x)}{|x|^d} \,\mathrm{d}x. \end{align*} $$ But with the formula \eqref{eq1} with $\lambda\neq 1$, one can also compute easily the distribution $v_d(\lambda x)$ defined by $$ \langle v_d(\lambda x),\varphi(x)\rangle = \frac{1}{|\lambda|^d}\langle v_d(x),\varphi(x/\lambda)\rangle $$ using Equation \eqref{eq1} with $\varphi(x)$ replaced by $\varphi(x/\lambda)$. Doing the change of variable $x/\lambda \to x$ yields $$ v_d(\lambda\,\cdot) = \frac{1}{|\lambda|^d}\,v_d + \omega_d\frac{\ln(\lambda)}{|\lambda|^d}\,\delta_0. $$ Therefore, we can now use the scaling properties of the Fourier transform to get for any $r=1/\lambda>0$ $$ \begin{align*} (\mathcal{F}(v_d))(r\tilde{x}) &= r^{-d} (\mathcal{F}(v_d(y/r)))(\tilde{x}) \\&= \mathcal{F}(v_d-\omega_d\ln(r)\delta_0)(\tilde{x}) \\&= \mathcal{F}(v_d)(\tilde{x}) -\omega_d\ln(r). \end{align*} $$ Taking $\tilde{x} = \frac{x}{|x|}$ and $r=|x|$ gives $$ \boxed{\mathcal{F}(v_d)(x) = C_d -\omega_d\ln(|x|)} $$ where $C_d = \mathcal{F}(v_d)(\tilde{x})$ is a constant since the Fourier transform of a radial function is radial.


If you want to know the constant $C_d$, the usual trick is to multiply by a Gaussian and use the fact that we know the Fourier transform of a Gaussian. Here remark first that by the Fourier inversion theorem we have $$ \mathcal{F}(\ln(|x|)) = C_d\, \delta_0 - \frac{v_d}{\omega_d} $$ Therefore $$ \begin{align*} C_d - \langle\mathcal{F}(\ln(|x|)), e^{-\pi|x|^2}\rangle &= \frac{1}{\omega_d}\langle v_d, e^{-\pi|x|^2}\rangle \\ &= \int_0^1 \frac{e^{-\pi r^2}-1}{r}\,\mathrm{d} r + \int_1^\infty \frac{e^{-\pi r^2}}{r}\,\mathrm{d} r \\ &= \int_0^\pi \frac{e^{-t}-1}{2\,t}\,\mathrm{d} t + \int_\pi^\infty \frac{e^{-t}}{2\,t}\,\mathrm{d} t \\ &= \frac{-\ln(\pi)}{2} + \int_0^\pi \frac{\ln(t)\,e^{-t}}{2}\,\mathrm{d} t + \int_\pi^\infty \frac{\ln(t)\,e^{-t}}{2}\,\mathrm{d} t \\ &= \frac{-\gamma - \ln(\pi)}{2} \end{align*} $$ where $\gamma = -\Gamma'(1) = -\psi(1)$ and I did a polar change of variable and the change $t = πr^2$ and then integrations by parts. But since $\mathcal{F}(e^{-\pi|x|^2}) = e^{-\pi|x|^2}$, we can also compute $$ \begin{align*} \langle\mathcal{F}(\ln(|x|)), e^{-\pi|x|^2}\rangle &= \int_{\mathbb{R}^d} \ln(|x|)\, e^{-\pi|x|^2}\,\mathrm{d} x \\ &= \omega_d \int_0^\infty \ln(r)\,e^{-\pi r^2} r^{d-1}\,\mathrm{d} r \\ &= \frac{1}{2\,\Gamma(d/2)} \int_0^\infty (\ln(t)-\ln(\pi))\,e^{-t}\, t^{d/2-1}\,\mathrm{d} t \\ &= \frac{1}{2} \,\big(\psi(d/2) - \ln(\pi)\big). \end{align*} $$ with the same changes of variable. We deduce that $C_d = \frac{\psi(d/2)-\gamma}{2} - \ln(\pi)$. Tell me if you spot any errors!

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    $\begingroup$ Thanks for your reply. Though I can not understand your process right now, I think your result is right. By the way, can you calculate the Fourier transform of $\frac{1}{|\mathbf{x}|^\alpha}$ in general $d$ dimensions? $\endgroup$ Jun 18, 2020 at 8:59
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    $\begingroup$ Yes, this one is more classical when $a \in (0,d)$ , then $\mathcal{F}(1/(\omega_a |x|^a)) = 1/(\omega_{d-a} |x|^{d-a})$ $\endgroup$
    – LL 3.14
    Jun 18, 2020 at 16:50
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    $\begingroup$ You can find the result in a lot of textbooks, for example "Functional Analysis" from Lieb and Loss. Then you can take derivatives to get the other cases. An other special case is the case when $a=-2n$ is a nonnegative integer. Then you get the Dirac delta if $n=0$ and powers of Laplacian of the Dirac delta when $n$ is larger. $\endgroup$
    – LL 3.14
    Jun 18, 2020 at 16:59
  • $\begingroup$ @LL3.14 Hi LL! I hope that you're doing well. I always enjoy reading your posted solutions. I thought THIS solution I posted on the Fourier Transforms of $|x|^s$, $s\in \mathbb{R}$ and THIS ONE on the Fourier Transform of $\log(|x|)$ might be of interest. $\endgroup$
    – Mark Viola
    Aug 10, 2022 at 18:46
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    $\begingroup$ This is ingenuous. The end product is very polished. But I am very curious how you came up with the idea. I guess the process was a lot messier? $\endgroup$ Jan 5, 2023 at 3:56

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