I am looking at the following integral in Euclidean space:

$$I = \int_{\mathbb{R}} d\tau_3 \int_{\mathbb{R}} d\tau_4 \int_{\mathbb{R}^4} d^4 x_5 \frac{1}{x_{15}^2 x_{25}^2 x_{35}^2 x_{45}^2} \tag{1}$$

with $x_{ij} := x_i - x_j$, $x_1=(1,0,0,0)$, $x_2=(x_2^1,x_2^2,0,0)$, $x_3=(0,0,0,\tau_3)$ and finally $x_4=(0,0,0,\tau_4)$. This integral is divergent at $x_5 \sim \tau_3 \sim \tau_4$.

Using dimensional regularization $\left(\int d^4 x_5 \to \int d^{2\omega} x_5 \right)$, I could extract the divergence of $I$ by setting $x_{15} x_{25} \sim x_{13} x_{24}$, integrating the remaining integrals and expanding the resulting $\Gamma$-function. The result is:

$$I_\text{div} = \pi^4 \frac{1}{\left| x_1 \right| \left| x_2 \right|} \frac{1}{\epsilon} \tag{2}$$

with $\epsilon \to 0$ as $2\omega \to 4$.

But I am interested in the full result of the integral, i.e. divergence + finite part. My idea now is to compute the integral given in $(1)$ as far as I can in $4$d, and to try to "push" the divergence into the last one-dimensional integral. For example, I can start by performing the integrals over the $\tau$'s:

$$\int_{\mathbb{R}} d\tau_3 \frac{1}{x_{35}^2} = \int_{\mathbb{R}} d\tau_3 \frac{1}{\left[ (\tau_3 - \tau_5)^2 + \vec{x}_5^2 \right]} = \frac{\pi}{\left| \vec{x}_5 \right|}, \tag{3}$$

where I used the fact that $\vec{x}_3 = (0,0,0)$. Regarding the notation, I used here $\vec{x}^2 := \vec{x} \cdot \vec{x}$. After that, I can also perform the integral over $\tau_5$:

$$\begin{align}I &= \int_{\mathbb{R}^3} d^3 x_5 \frac{1}{\vec{x}_5^2} \int_{\mathbb{R}} d\tau_5 \frac{1}{\left(\tau_5^2 + \vec{x}_{15}^2\right)\left(\tau_5^2 + \vec{x}_{15}^2\right)} \\ &= \int_{\mathbb{R}^3} d^3 x_5 \frac{1}{\vec{x}_5^2} \frac{1}{\left| \vec{x}_{15} \right| \vec{x}_{25}^2 + \left| \vec{x}_{25} \right| \vec{x}_{15}^2}. \tag{4} \end{align}$$

Going further integrating dimension by dimension, it seems that I can actually go pretty far, maybe all the way up to the last dimension. However the computations become quite involved, and I would like to have a way to check my result in between. For example, since I know that my divergence is logarithmic, I expect to be able to extract the divergence again and get something like:

$$I_\text{div} \sim \pi^4 \left| x_1 \right| \left| x_2 \right| \int_0^\infty \frac{dr}{r}. \tag{5}$$

So I tried to do that with eq. ($5$). The divergence seems to sit at $\vec{x}_5 = (0,0,0)$, hence I try to extract it by setting $\left| \vec{x}_{15} \right| \vec{x}_{25}^2 + \left| \vec{x}_{25} \right| \vec{x}_{15}^2 \sim \left| x_1 \right| x_2^2 + \left| x_2 \right| x_1^2$, and we have:

$$\begin{align} I_\text{div} &= \pi^3 \frac{1}{\left| x_1 \right| x_2^2 + \left| x_2 \right| x_1^2} \int_{\mathbb{R}^3} d^3 x \frac{1}{\vec{x}^2} \\ &= 4 \pi^4 \tag{6} \frac{1}{\left| x_1 \right| x_2^2 + \left| x_2 \right| x_1^2} \int_0^\infty dr \end{align}$$

But I am confused: the prefactor with the $x_1$, $x_2$ is not the same as in eq. ($2$), and moreover the divergence is now linear! Am I not allowed to do that? And if not, why?

I have heard that dimensional regularization "sees" only the logarithmic divergences, maybe that could be the reason?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.