Integral divergent in Peskin and Schroeder eq. (7.90) I'm working on the Eq. (7.90) in Peskin (page 252).

However, I don't understand why it diverges logarithmically. Does $\Gamma(0)$ mean logarithmically divergence? 
 A: Yes, you are right. It is general fact. In QED dimensional regularization is the best tool because it preserves gauge invariance. But for illustration of this propery (and for simplicity), we can consider $\phi^4$ theory. In this theory, let us consider two diagrams,

For the first diagram, we have
$$\sim g \Omega_d\int_{0}^{\infty}\frac{dk\,k^{d-1}}{k^2+m^2}=g\Omega_dm^{d-2}\int_{0}^{\infty}\frac{dx\,x^{d-1}}{1+x^2}=\infty, d\geq 2,$$
where $\Omega_d$ is the integral over angles and $g$ is coupling constant in our $\phi^4$ theory (also, I omit all numerical factors). For the second diagram, you can find
$$\sim g^2\Omega_dm^{d-4}\int_{0}^{\infty}\frac{dx\,x^{d-1}}{(1+x^2)^2}=\infty, d\geq 4.$$
Then let us regularize the first diagram (set $d=4$) with hard cut-off $\Lambda$. It gives that divergent piece in the limit of $\Lambda\gg m$ behaves as $\sim \Lambda^2$ and it is even divergence. For 2nd diagram with hard cut-off, you can find (I omit irrelevant numerical factors, you should keep in mind that log function has dimensionless argument!) that divergent piece in the limit of $\Lambda\gg m$ is $\sim\ln\Lambda$.
Now consider dimensional regularization. For the first diagram, we find
$$\sim g m^{d-2}\Gamma(d/2)\Gamma(d-1/2),$$
where for this result we consider that integral is convergent and perform change of variable. Then, we use $d=4-\epsilon$ and perform expansion,
$$\sim -\frac{gm^2}{\epsilon},$$
where you should take into account that 2nd power divergence of regularization with hard cut-off $\Lambda$ transforms into pole $1/\epsilon$ (coeffcients are different, in general!). Similarly, for the second diagram (using $1/(k^2+m^2)^2=-\partial(1/(k^2+m^2))/\partial m^2)$, you can find
$$\sim-\frac{g^2}{\epsilon},$$
so log (=zero power) divergence of regularization with hard cut-off transforms into pole $1/\epsilon)$. We demonstrate mentioned fact:

Dimensional regularization replaces any even divergence of diagram (this divergence can be extracted with cut-offs) by pole $1/\epsilon$

This properties (in my view) does not have any connections with definition of Gamma function.
Finally, if your integral have odd divergency, dimensional regularization is useless (see my old question).
