$u$, such as this one, that is associated with the running of $\alpha_s$ is called a renormalon[@thooft].

In order to understand the origins of this renormalon singularity more fully, let us re-examine the integral (\[bubble-integral\]). Applying an IR cutoff $\lambda$, we have $$(\mu^2/e^C)^u\int_\lambda^\Lambda dl\, l^{m-2u}
=(\mu^2/e^C)^u\;{\Lambda^{m+1-2u}-\lambda^{m+1-2u}\over m+1-2u}.
\label{cutoff-integral}$$ If $\Lambda$ and $\lambda$ are finite positive numbers, then the integral (\[cutoff-integral\]) is well defined, even at the point $u=(m+1)/2$. Furthermore, the inverse Borel transform is convergent, provided that $$\alpha_s(\mu^2)\beta_0\ln[\mu^2/(\lambda^2e^C)]<1,
\label{pert-cond}$$ which is identical to the condition that the integration over $l$ never passes through the Landau pole in (\[bubble-sum\]). Hence, we conclude that the renormalon singularity in the $u$ plane arises from the region of integration near $l=0$, which we have excluded by introducing the IR cutoff $\lambda$. The renormalon singularity is a signal that, because of the growth of the running coupling, the perturbation series breaks down near zero loop momentum. A renormalon that arises from the region near zero loop momentum is called an IR renormalon.

The renormalon in (\[bubble-integral\]) is on the positive real axis. Hence, the inverse Borel transform is not unique. One can obtain various values, depending on the prescription used to deform the integration contour 