The following identity is used in Peskin & Schroeder's book Eq.(19.43), page 660:

$$\int\frac{d^4k}{(2\pi)^4}\,\frac{1}{(k^2)^2}e^{ik\cdot\epsilon}=\frac{i}{(4\pi)^2}\log\frac{1}{\epsilon^2},\quad \epsilon\rightarrow 0$$

I can't figure out why it holds. Could someone provide a method to prove this? Many thanks in advace.

  • $\begingroup$ I haven't attempted the integral, but with these sort of things, spherical polars in k space are sometimes a useful approach. Have you tried that? $\endgroup$
    – twistor59
    Commented Apr 18, 2013 at 11:16
  • $\begingroup$ I have tried this approach, but I can't get it right. The trouble is how to take the integration with respect to $k^0$. $\endgroup$
    – soliton
    Commented Apr 18, 2013 at 11:24
  • $\begingroup$ It's just a loop integral which can be evaluated using the formulae given in the appendix of P&S. Then Taylor-expand the result in $\epsilon$ and you have the result. Edit: oh, just saw that Lubos already said that... $\endgroup$ Commented Apr 18, 2013 at 12:20
  • $\begingroup$ It looks to me like this can be evaluated using the integral representation of the Dirac delta: $$\delta(\epsilon) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{ik\epsilon} dk.$$ This would mean that you just have to understand the above identity, which is a standard and historical problem. $\endgroup$ Commented Apr 18, 2013 at 16:14
  • $\begingroup$ Another approach is here. $\endgroup$
    – GotchaP
    Commented Jul 17, 2017 at 3:53

4 Answers 4


That's equivalent simply to $c\int dx/x$. Switch to the Euclidean spacetime, $k_0=ik_4$ where $(k_1,\dots k_4)$ is $k_E$; i.e. analytically continue in $k_0$ (Wick rotation). The integral is $$\int \frac{i\cdot d^4 k_E}{(2\pi)^4} \frac{1}{(k_E^2)^2} \exp(ik\cdot \epsilon)$$ So it's proportional to the Fourier transform of $1/k_E^4$. The original function is $SO(4)$ symmetric, so the Fourier transform must be symmetric as well and depend on $\epsilon^2$ only. Dimensional analysis implies that the result is dimensionless i.e. it must be a combination of a constant and $\ln(\epsilon^2)$. The logarithm is there with a nonzero coefficient so the constant only determines how to take the logarithm: it should properly be written as $\ln(\epsilon^2/\epsilon_0^2)$ for some constant $\epsilon_0$ with the same dimension.

The only remaining unknown is the coefficient and one gets $4\pi^2$ from the remaining integral. It's a sort of waste of resources to compute this special integral; it's better to compute the more general integrals in appendix A.4, see especially formulae (A.44)-(A.49) on page 807, which I won't copy here because that's why Peskin and Schroeder wrote the textbook.

  • $\begingroup$ Why can we say that "it's proportional to the Fourier transform of $1/k^4_E$" since $(k^2_E)^2=(k_1^2+k_2^2+k_3^2+k_4^2)^2$ and $e^{ik\cdot\epsilon}$ is not $SO(4)$ symmetric? $\endgroup$
    – soliton
    Commented Apr 18, 2013 at 13:12
  • $\begingroup$ Dear Soliton, the $\exp(ik\cdot \epsilon)$ factor is the phase that is a part of the definition of the Fourier transform! It's nothing we have added to the function we're Fourier-transforming. $\endgroup$ Commented Apr 18, 2013 at 16:08
  • $\begingroup$ Thanks a lot. Dimensional analysis is a shortcut to obtain the result. $\endgroup$
    – soliton
    Commented Apr 19, 2013 at 12:01
  • $\begingroup$ I have found another approach by using the result of Eq.(5.2.9) in Weinberg's book (vol. 1, page 202) and the asymptotic expansion of Bessel function $K_1(x)=\frac{1}{x}+\frac{x}{2}\log\frac{x}{2}$ as $x\rightarrow 0$. $\endgroup$
    – soliton
    Commented Apr 19, 2013 at 12:10
  • 1
    $\begingroup$ When converted to 1D integrals, all these 4D Euclidean integrals may be shown to be proportional to the area of the 3-sphere, the surface of the 4-dimensional unit ball, and it's $2\pi^2$. More generally, the d-dimensional sphere may be calculated, see e.g. en.wikipedia.org/wiki/Volume_of_an_n-ball - e.g. by computing the integral of the d-dimensional Gaussian either as a power of the integral of 1D Gaussian, or as the surface of the sphere times a 1D calculable integral proportional to the Gamma function. $\endgroup$ Commented Dec 23, 2016 at 13:34

I will give another approach to this identity. First, we notice that

$$\int\frac{d^4k}{(2\pi)^4}\,\frac{1}{(k^2-m^2)^2}e^{ik\cdot\epsilon}=-i\frac{\partial}{\partial m^2}D_F(x)\big|_{x=\epsilon}$$

For space-like vector $\epsilon^2=-r^2<0$, we have


whose derivation refers to Weinberg's book vol. 1, page 202. For $r\rightarrow 0$, the following expansion holds

$$ K_1(mr)=\frac{1}{mr}+\frac{mr}{2}\log\frac{mr}{2}$$

With this conditions, we finally obtain



Further another approach.
After Wick rotation($k^0=ik_E^0,\,\epsilon^0=i\epsilon_E^0$) the integral is
$$ I_1 \equiv \int \frac{i\cdot d^4 k_E}{(2\pi)^4} \frac{1}{(k_E^2)^2} \exp(-ik_E\cdot \epsilon_E) . $$ In (19.43) we want the quntity $$ \frac{\partial}{\partial \epsilon^{\gamma}}I_1. $$ Note that $$ \frac{1}{k_E^2}=\int_0^{\infty}du e^{-k_E^2 u}, $$ $$ \frac{1}{(k_E^2)^2}=\int_0^{\infty}du\int_0^{\infty}dv e^{-k_E^2 (u+v)}, $$ $$ I_1=\frac{i}{(2\pi)^4} \int d^4k_E \int_0^{\infty}du\int_0^{\infty}dv e^{-(u+v)k_E^2-i\epsilon_E \cdot k_E}. $$ $ \displaystyle \int_{-\infty}^{\infty} dk_E^i \exp[-(u+v)(k_E^i)^2-i\epsilon_E^i k_E^i] = \sqrt{\frac{\pi}{u+v}} \exp\left[ -\frac{(\epsilon_E^i)^2}{4(u+v)} \right] $ \begin{alignat}{2} \therefore I_1&=&& \frac{i}{(2\pi)^4} \int_0^{\infty}du\int_0^{\infty}dv \frac{\pi^2}{(u+v)^2} \exp\left[-\frac{\epsilon_E^2}{4(u+v)} \right] \\ &=&& \frac{i}{16\pi^2}I_2 \left(\frac{\epsilon_E^2}{4}\right) \end{alignat} where $\displaystyle I_2(x) \equiv \int_0^{\infty}du\int_0^{\infty}dv\, \frac{1}{(u+v)^2}\exp\left(-\frac{x}{u+v}\right) $. The calculation of $I_2$ is here. \begin{alignat}{2} I_1 &=&& \frac{i}{16\pi^2}\left( -\log\left(\frac{\epsilon_E^2}{4}\right) +\gamma-1+\lim_{M\to \infty}\log M +\cal{O}(\epsilon_E^2) \right) \\ &=&& \frac{i}{16\pi^2}\left( -\log\left(-\frac{\epsilon^2}{4}\right) +\gamma-1+\lim_{M\to \infty}\log M +\cal{O}(\epsilon^2) \right) \end{alignat} After $\epsilon \to 0$, we have
$$ \frac{\partial}{\partial \epsilon^{\gamma}}I_1 = \frac{i}{16\pi^2}\frac{\partial}{\partial \epsilon^{\gamma}}\log \frac{1}{\epsilon^2}. $$


Here is yet another solution, which probably is not a physicist's way of thinking.

After Wick rotation, we may work on Euclidean space. The function $f(k_E) = |k_E|^{-4}$ is not square-integrable on $\mathbb{R}^4$, however, so its Fourier transform does not exist in ordinary sense. A moment of thought suggests that it can be realized as distribution on the space

$$ \mathcal{A} := \{ \varphi \in \mathcal{S}(\mathbb{R}^4) : \textstyle \int_{\mathbb{R}^4} \varphi(\epsilon) \, \mathrm{d}^4\epsilon = 0 \}, $$

where $\mathcal{S}(\mathbb{R}^4)$ is the Schwarz space. Then computing

$$I(\epsilon) = i \check{f}(\epsilon) = \frac{i}{(2\pi)^4} \int_{\mathbb{R}^4} |k_E|^{-4} e^{i\epsilon \cdot k_E} \, \mathrm{d}^4 k_E$$

in distribution sense amounts to identifying the following pairing

$$ \langle I, \varphi \rangle = \langle \check{f}, \varphi \rangle = \langle f, \check{\varphi} \rangle, \qquad \forall \varphi \in \mathcal{A}. $$

Since $\check{\varphi}(0) = 0$ and $\check{\varphi}$ has rapid decay near infinity, the pairing $\langle f, \check{\varphi} \rangle$ is realized as Lebesgue integral. Then by the Fubini's theorem,

\begin{align*} \langle f, \check{\varphi} \rangle &= i \int_{\mathbb{R}^4} \frac{1}{|k_E|^4} \check{\varphi}(k_E) \, \mathrm{d}^4 k_E = i \int_{\mathbb{R}^4} \bigg( \int_{0}^{\infty} t \mathrm{e}^{-|k_E|^2 t} \, \mathrm{d}t \bigg) \check{\varphi}(k_E) \, \mathrm{d}^4 k_E, \\ &= i \int_{0}^{\infty} t \bigg( \int_{\mathbb{R}^4} \mathrm{e}^{-t |k_E|^2} \check{\varphi}(k_E) \, \mathrm{d}^4 k_E \bigg) \mathrm{d}t. \end{align*}

Using $\langle \mathrm{e}^{-t|\cdot|^2}, \check{\varphi} \rangle = \langle (\mathrm{e}^{-t|\cdot|^2})^{\vee}, \varphi \rangle$ and the formula $\int_{\mathbb{R}} \mathrm{e}^{-tx^2}\mathrm{e}^{ix\epsilon} \, dx = \sqrt{\frac{\pi}{t}} \mathrm{e}^{-\epsilon^2/4t}$ for $t > 0$, we have

\begin{align*} \langle f, \check{\varphi} \rangle &= i \int_{0}^{\infty} t \bigg( \int_{\mathbb{R}^4} \frac{1}{(4\pi t)^2} \mathrm{e}^{-\frac{|\epsilon|^2}{4t}} \varphi(\epsilon) \, \mathrm{d}^4 \epsilon \bigg) \mathrm{d}t \\ &= \frac{i}{(4\pi)^2} \int_{0}^{\infty} \frac{1}{t} \bigg( \int_{\mathbb{R}^4} \mathrm{e}^{-\frac{|\epsilon|^2}{4t}} \varphi(\epsilon) \, \mathrm{d}^4 \epsilon \bigg) \mathrm{d}t. \end{align*}

We want to finalize the computation by switching the order of integration, but the Fubini's theorem is not applicable in this case and even the heuristic computation produces a divergent integral. Thankfully, using the fact that $\int_{\mathbb{R}} \varphi(\epsilon) \, \mathrm{d}^4\epsilon = 0$, we can regularize the inner integral so that the Fubini's theorem works:

\begin{align*} \langle f, \check{\varphi} \rangle &= \frac{i}{(4\pi)^2} \int_{0}^{\infty} \frac{1}{t} \bigg( \int_{\mathbb{R}^4} \big( \mathrm{e}^{-\frac{|\epsilon|^2}{4t}} - \mathbf{1}_{ \{ t \geq 1 \} } \big) \varphi(\epsilon) \, \mathrm{d}^4 \epsilon \bigg) \mathrm{d}t \\ &= \frac{i}{(4\pi)^2} \int_{\mathbb{R}^4} \bigg( \int_{0}^{\infty} \frac{1}{t}\big( \mathrm{e}^{-\frac{|\epsilon|^2}{4t}} - \mathbf{1}_{ \{ t \geq 1 \} } \big) \, \mathrm{d}t \bigg) \varphi(\epsilon) \, \mathrm{d}^4 \epsilon. \end{align*}

Now the inner integral can be computed using the substitution $u = |\epsilon|^2/4t$ as follows

\begin{align*} \int_{0}^{\infty} \frac{1}{t}\big( \mathrm{e}^{-\frac{|\epsilon|^2}{4t}} - \mathbf{1}_{ \{ t \geq 1 \} } \big) \, \mathrm{d}t &= \int_{0}^{\infty} \frac{1}{u} \big( \mathrm{e}^{-u} - \mathbf{1}_{ \{ u \leq |\epsilon|^2/4 \} } \big) \, \mathrm{d}u = \log\frac{4}{|\epsilon|^2} -\gamma. \end{align*}

Here, $\gamma$ is the Euler-Mascheroni constant. Therefore it follows that

$$ I(\epsilon) = \frac{i}{(4\pi)^2} \bigg( \log\frac{4}{|\epsilon|^2} -\gamma \bigg) = \frac{i}{(4\pi)^2} \log\frac{1}{\epsilon^2}. $$

The last equality follows from the fact that constants as distribution on $\mathcal{A}$ is equal to zero, i.e., for any constant $c$ we have $\langle c, \varphi \rangle = 0$ for all $\varphi \in \mathcal{A}$.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.