1
$\begingroup$

I encountered the following integral in dimensional regularization $$ I=\int d^d k \,e^{i\vec{k}\cdot \vec{x}}\frac{1}{\vec{k}^2}\frac{1}{(\vec{q}-\vec{k})^2}, $$ say that we already Wick rotated the integral. This looks like something which would be possible to evaluate but I'm not managing to do so. I tried Schwinger parametrization but got nowhere so far. I know it's possible to evaluate the integral at $x=0$ but I was wondering if anyone know a closed expression which is also a function of $x$. This is essentially the Fourier transform of the bubble diagram.

UPDATE

Here's where I got so far. Using a Feynman parameter $I$ can be written as

$$ I=\int {d^d k} \,e^{i\vec{k}\cdot \vec{x}}\frac{1}{\vec{k}^2}\frac{1}{(\vec{q}-\vec{k})^2}= \int_0^1 dt \int {d^d k} \,e^{i\vec{k}\cdot \vec{x}}\frac{1}{(\vec{k}^2-2t \vec{k}\cdot {q +t \vec{q}^2} )^2}\\= \int_0^1 dt \, e^{it \vec{x}\cdot \vec{q}} \int d^d k e^{i\vec{k}\cdot \vec{x}} \frac{1}{(\vec{k}+ \Delta)^2}, \,\,\, \Delta=t(1-t)\vec{q}^2. $$ Where I have shifted integration variable $k \to k-x q$ in the last step.

Using the suggested result the loop integral should be (not being careful about numerical factors) $$ \int d^d k\, e^{i\vec{k}\cdot \vec{x}} \frac{1}{(\vec{k}+ \Delta)^2}= |\vec{x}|^{2-d/2} (t(1-t)\vec{q}^2)^{d/4-1} K_{d/2-2}( \sqrt{t(1-t)} |\vec{q}||\vec{x}|), $$ The last $t$ integral looks then pretty nasty then $$\int_0^1 dt \, e^{it \vec{x}\cdot \vec{q}} (t(1-t))^{d/4-1} K_{d/2-2}( \sqrt{t(1-t)}|\vec{q}||\vec{x}|), $$

maybe it can be done using residues and the integral representation on $K$? This looks weird since in the end I want to take $d=3$ but what's $K_{-1/2}$ then?

$\endgroup$
6
  • 2
    $\begingroup$ Have you tried combining denominators a la Feynman? $\endgroup$ Nov 16, 2020 at 18:30
  • $\begingroup$ That would amount to solving the FT of $\int d^d k \,e^{i\vec{k}\cdot \vec{x}}\frac{1}{({k^2}+\Delta)^2}$ for some $\Delta$, is this known? $\endgroup$ Nov 16, 2020 at 18:42
  • $\begingroup$ Possibly. What is $\Delta$? Does it depend on $\vec x$? If it doesn't, then $I$ only depends on $\vec x^2$ instead of $\vec x$, so you can choose a convenient orientation, say $\vec x\propto \vec e_z$. Smooth sailing from there. $\endgroup$ Nov 16, 2020 at 18:50
  • $\begingroup$ Thanks I see what you mean, no it should depend on the feynman parameter and $q$. I will maybe post an update to see if it works $\endgroup$ Nov 16, 2020 at 18:56
  • $\begingroup$ I'm a bit confused here due to the title. Is the integral here done only on the spatial dimension $D=d+1$ and are those 3-vectors in the denominator? If above statement are correct then it's consistent but when we do Feynman diagram we usually bunch time and space together to write the perturbed term in terms of Feynman propagator so this is a bit confusing to me. Second, is the answer known to have a closed form? $\endgroup$
    – aitfel
    Nov 22, 2020 at 14:46

2 Answers 2

1
$\begingroup$

Are you sure that you want that specific integral? It looks like you are trying to compute something like $$ \int d^d x e^{-ir(x-y)} [g(x,y)]^2 $$ where $$ g(x,y) = \int \frac{d^dk}{(2\pi)^d} \frac{ e^{ik(x-y)}}{k^2}, $$ is the massless propagator. Translation invariance makes this independent of $y$, and gives $$ \int \frac{d^dk}{(2\pi)^d} \frac{1} {(k)^2(k-r)^2}= \Pi(r) $$ without your exponential factor. I can't think of any diagram computation that would give you the integral you are trying to do.
Perhaps what you are trying to do is compute $$ \int \frac{d^dr}{(2\pi)^d} \int \frac{d^dk}{(2\pi)^d} e^{ir(x-y)} \frac{1} {(k)^2(k-r)^2}? $$ This is $$ [g(x,y)]^2= \int \frac{d^dr}{(2\pi)^d}e^{ir(x-y)} \Pi(r) $$ and gives back the bubble in $x,y$ space.

As $g(x)\propto |x|^{-2(d-2)}$, you can do any FT's of powers of $g$ by inverting $$ \int \frac{d^n k}{(2\pi)^n} e^{ik\cdot(x-x')} |k^2|^s = \frac{4^s}{\pi^{n/2}}\frac{\Gamma(s+n/2)}{\Gamma(-s)} \frac{1}{|x-x'|^{2s+n}} $$

$\endgroup$
1
  • $\begingroup$ Thanks but this is not what I need. My integral is not directly related to a Feynman graph but I thought can be done with the usual dimreg tricks. It does have the exponential factor so it's not as simple as $\Pi(r)$ since it depends on two scales, $r$ and $x$. $\endgroup$ Nov 16, 2020 at 20:51
1
$\begingroup$

The Fourier transform given by $$ F[\Delta](\vec{x})=\int \mathrm{d}^dk\, e^{i\vec{k}\cdot\vec{x}} \frac1{(\vec{k}{}^2+\Delta)^2}\,, $$ can be turned into the Hankel transform for the radial function $f(k) = 1/(k^2+\Delta)^2$ times a factor $k^{d/2-1}$. Namely $$ F[\Delta](\vec{x})= \frac{(2\pi)^{d/2}}{x^{d/2-1}} \int_0^\infty k^{d/2-1}f(k)J_{d/2-1}(kx)\,k\,\mathrm{d}k\,, $$ with $x \equiv (\vec{x}{}^2)^{1/2}$ and $J_\nu(z)$ is the Bessel function of the first kind.

In the linked Wikipedia page you can find a table with common results. If there is not what you need, some valus are also tabulated in [1].

In the aforementioned reference one finds that

$$ \int_0^\infty \frac{r^\nu}{(r^2+a^2)^{\mu+1}}\,J_\nu(xr)\,r\,\mathrm{d}r = \frac{a^{\nu-\mu} x^\mu K_{\nu-\mu}(as)}{2^\mu\Gamma(\nu+\tfrac12)}\,, $$ where $K_\alpha$ is a modified Bessel function of the second kind and $\Gamma$ the Gamma function. Now just set $\mu=1,\, \nu=d/2-1,\,a^2 = \Delta$.


[1] A. D. Poularikas, Handbook of Formulas and Tables for Signal Processing. CRC Press, 1998.

$\endgroup$
3
  • $\begingroup$ It doesn't seem mine it's a known radial function.. $\endgroup$ Nov 17, 2020 at 11:02
  • $\begingroup$ Check the reference [1], Table 17.2, the 7th identity. $\endgroup$
    – MannyC
    Nov 17, 2020 at 12:23
  • $\begingroup$ Oh nice! I missed that $\endgroup$ Nov 17, 2020 at 12:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.