Consider the integration in the final step:
$$= \int\frac{d^{4}k}{(2\pi)^{4}}\ \text{log}\ (-k^{2}+m^{2})$$
$$= i\int\frac{d^{4}k_{E}}{(2\pi)^{4}}\ \text{log}\ (k^{2}_{E}+m^{2})$$
$$=-i\frac{\partial}{\partial\alpha}\int \frac{d^{4}k_{E}}{(2\pi)^{4}}\frac{1}{(k_{E}^{2}+m^{2})^{\alpha}}\Bigg|_{\alpha=0}$$
$$=-i\frac{\partial}{\partial\alpha}\left(\frac{1}{(4\pi)^{d/2}} \frac{\Gamma\left(\alpha-\frac{d}{2}\right)}{\Gamma(\alpha)}\frac{1}{(m^{2})^{\alpha-d/2}}\right)\bigg|_{\alpha=0}$$
$$=-i\frac{\Gamma(-d/2)}{(4\pi)^{d/2}}\frac{1}{(m^{2})^{-d/2}}$$
where an analytic continuation via Wick rotation is performed in the second line to change variables from $k$ to $k_{E}$, where $k$ is a four-vector, and the fact that $\Gamma(\alpha)\rightarrow 1/\alpha$ as $\alpha\rightarrow 0$ is used in the final line.
How does the dependence on $d$ arise in the integral on the final line?
Can someone show the intermediate steps in the integration on the final line explicitly?