My question comes from the textbook by Peskin & Schroeder, the integral (19.26): $$\begin{align} \int \frac{d^2 k}{(2\pi)^2}\! e^{- i k\cdot (y-z)}\frac{i \not{k}}{k^2} = -\not\partial \left(\frac{i}{4\pi}\log (y-z)^2\right) \end{align}$$

Question: how to derive the formula from the left hand side to the right hand side ?

If considering the identity (3.117) and set $m=0$, I have $$\begin{align} \int \frac{d^2k}{(2\pi)^2}\! \frac{i k\cdot\gamma}{k^2} e^{-i k\cdot (y-z)} = i \not\partial \left(D_R(y-z)\right) \end{align}$$ here $$\begin{align} D_R(y-z)= \int \frac{d^2 k}{(2\pi)^2}\frac{i}{k^2}e^{-i k\cdot (y-z)} \end{align}$$ the 2-vector: $k^\mu=(k^0,k^1)$ and owing to the massless condition:$(k^0)^2 = (k^1)^2$. set $\kappa \equiv k^1$.therefore I got $$\begin{align} &\int_{-\infty}^{+\infty}\frac{d k^1}{(2\pi)}\! \bigg[\frac{1}{2 k^0} e^{-i [k^0(y-z)_0 - k^1(y-z)_1]} + \frac{1}{-2 k^0}e^{-i[-k^0(y-z)_0 - k^1(y-z)_1]}\bigg] \\ &= - \frac{i}{4\pi}\ 2 \int_{-\infty}^{+\infty} \frac{\sin\left(\kappa (y-z)_0\right)}{\kappa}e^{i \kappa(y-z)_1}\; d\kappa \end{align}$$

But I failed to get the log-term from the above formula.

NOTE I found a related answer A four-dimensional integral in Peskin & Schroeder

  • $\begingroup$ One cannot use the on-shell condition here, since the components of $k^\mu$ are independent. $\endgroup$
    – gamebm
    Commented Feb 22, 2019 at 0:01

1 Answer 1


I found a Fourier transform can answer this question. Let me work on the Euclidean space by the Wick rotation: $k_0 \rightarrow \mathrm{i} k_2$, hence $$\begin{align} \int \frac{\mathrm{d}^2k}{(2\pi)^2}\! \frac{\mathrm{i}\not{k}}{k^2}e^{-\mathrm{i} k\cdot (y-z)} &\longrightarrow \int\frac{\mathrm{d}k_1 \mathrm{d}k_2}{(2\pi)^2}\! \frac{\mathrm{i}\not{k}}{|\vec{k}|^2} e^{\mathrm{i}\vec{k}\cdot (\vec{y}-\vec{z})} \qquad \text{set:} \quad \vec{y}-\vec{z} := \vec{r} \\ &= \mathrm{i} \not{\partial}_{\vec r}\; \frac{1}{(2\pi)^2}\,\color{blue}{\int \frac{1}{k_1^2 + k_2^2}e^{\mathrm{i}(k_1r_1 + k_2 r_2)}\; \mathrm{d}k_1 \mathrm{d}k_2} \\ &= \mathrm{i} \not{\partial}_{\vec r}\;\left( \frac{1}{(2\pi)^2} \left[\color{blue}{-2 \pi \log|\vec{r}|}\right]\right) \\ &= - \mathrm{i} \not{\partial}_{\vec r}\; \left(\frac{1}{4 \pi}\,\log(y-z)^2\right) \end{align}$$ here the blue part is a Fourier integral in 2-D, this Fourier transform is present in the Green's function of Laplace operator in 2-dimensional space.

  • $\begingroup$ $-i$ at the head of RHS of the first line is dropped. And $\log|\vec{r}|=\log |\vec{y}-\vec{z}|^{2\cdot 1/2}=\log (-(y-z)^2)^{1/2}$. So the last line is $$- \mathrm{i} \not{\partial}_{\vec r}\; \left(\frac{1}{4 \pi}\,\log(-(y-z)^2)\right).$$ Then it's in accordance with another method. The second line of (19.26) in the textbook is misprinted. $\endgroup$
    – GotchaP
    Commented Jan 24, 2017 at 1:05
  • $\begingroup$ Hi, could you please elaborate more on where did you find the proof of the blue part? $\endgroup$
    – Daren
    Commented Nov 30, 2023 at 14:57
  • $\begingroup$ Ah, it's just the solution of green function of Laplace operator in 2-d. $\endgroup$
    – Daren
    Commented Dec 1, 2023 at 1:56

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