In a book by Jean Zinn-Justin (Phase transitions and renormalization group), there is a function $$\Delta(x,0)=\frac{2^{d-2}}{(4\pi)^{d/2}}\Gamma(d/2-1)\frac{1}{x^{d-2}}$$, when $d\to 2$, he says the function diverges as $$\Delta(x,0)=\frac{1}{2\pi(d-2)}-\frac{1}{4\pi}(\gamma+ln\pi+lnx^2)+O(d-2)$$, and he say the expansion of $\Gamma$ function is used ($\Gamma(1+s)=1-\gamma s+O(s^2)$), $\gamma$ is the Euler's constant ($\gamma=-\int^{\infty}_{0}e^{-x}lnx\,dx$).
I have no idea how to get the divergent behavior above.
$$\Delta(x,0)=\frac{1}{4} \pi ^{-d/2}\, x^{2-d}\, \Gamma \left(\frac{d}{2}-1\right)$$ Take the logarithm $$\log\big[4\Delta(x,0)\big]=-\frac d 2 \log(\pi)+(2-d)\log(x)+\log \left(\Gamma \left(\frac{d}{2}-1\right)\right)$$ Let $d=2+\epsilon$ $$\log\big[4\Delta(x,0)\big]=-\epsilon \log (x)-\frac{1}{2} (2+\epsilon ) \log (\pi )+\log \left(\Gamma \left(\frac{\epsilon }{2}\right)\right)$$ Use series expansion $$\log \left(\Gamma \left(\frac{\epsilon }{2}\right)\right)=-\log \left(\frac{\epsilon }2\right)-\frac{\gamma }{2} \epsilon+\frac{\pi ^2 }{48}\epsilon ^2+O\left(\epsilon ^3\right)$$ $$\log\big[4\Delta(x,0)\big]=\log \left(\frac{2}{\pi }\right)-\log(\epsilon)-\frac{1}{2} \left(\log \left(\pi x^2\right)+\gamma \right)\epsilon+\frac{\pi ^2 }{48}\epsilon ^2+O\left(\epsilon ^3\right)$$ Now, use $$4\Delta(x,0)=e^{\log\big[4\Delta(x,0)\big]}$$ to obtain $$\Delta(x,0)=\frac{1}{2 \pi \epsilon }-\frac{\log \left(\pi x^2\right)+\gamma }{4 \pi }+\frac{6 \left(\log \left(\pi x^2\right)+\gamma \right)^2+\pi ^2}{96 \pi } \epsilon +O\left(\epsilon ^2\right)$$