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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.

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  • $\begingroup$ 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 Apr 18 '13 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 Apr 18 '13 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... $\endgroup$ – A friendly helper Apr 18 '13 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. $\endgroup$ – Douglas B. Staple Apr 18 '13 at 16:14
  • $\begingroup$ Another approach is here. $\endgroup$ – GotchaP Jul 17 '17 at 3:53
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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.

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  • $\begingroup$ 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 Apr 18 '13 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. $\endgroup$ – Luboš Motl Apr 18 '13 at 16:08
  • $\begingroup$ Thanks a lot. Dimensional analysis is a shortcut to obtain the result. $\endgroup$ – soliton Apr 19 '13 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 Apr 19 '13 at 12:10
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    $\begingroup$ 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. $\endgroup$ – Luboš Motl Dec 23 '16 at 13:34
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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

$$D_F(x)=\frac{m}{4\pi^2r}K_1(mr)$$

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

$$\int\frac{d^4k}{(2\pi)^4}\,\frac{1}{(k^2-m^2)^2}e^{ik\cdot\epsilon}=\frac{i}{16\pi^2}\log\frac{1}{\epsilon^2}$$

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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
$$ \frac{\partial}{\partial \epsilon^{\gamma}}I_1 = \frac{i}{16\pi^2}\frac{\partial}{\partial \epsilon^{\gamma}}\log \frac{1}{\epsilon^2}. $$

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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 $\varphi \in \mathcal{A}$.

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