I'm working on my thesis, and in an article (link below) I came up with an estimate on an integral that I can't deal with.
Let $\gamma\in(0,1),$ and take some $k\in\mathbb{Z}^3,$ and let $|k|^2$ be the square of its standard norm. The integral is the following: $$I=\int_{[-\infty,t]^2}e^{-|k|^2(2t-u-u')}\frac1{|k|^{2\gamma}|u-u'|^\gamma}\,du\,du'.$$ What the authors of the article say is that there exists a constant $C>0$ such that the following inequality holds: $$I=\int_{[-\infty,t]^2}e^{-|k|^2(2t-u-u')}\frac1{|k|^{2\gamma}|u-u'|^\gamma}\,du\,du'\le \frac{C}{|k|^4},$$ but I don't know how to justify it. Clearly, for $\gamma=0$ it is easy to obtain that bound, but I don't know why the estimate should be valid for $\gamma$ arbitrarily close to $0$. I tried to make a change of variables $v=u-u',w=u+u',$ but it doesn't work, I also tried to use the Holder's inequality, but nothing. Any help would be really appreciated.
Link: Construction of $\Phi^4_3$ diagrams for pedestrians, the formula is (4.29) at page 28.