In Peskin&Schroeder's QFT book, one is presented with the following integral (in dimensional regularization):
$\int d^dx \log(x^2+m^2)$
here $x^2=\sum\limits_{i=1}^d x_i^2$. They evaluate it as follows:
$\displaystyle \int d^dx \log(x^2+m^2)=-\frac{\partial}{\partial\alpha}\int d^dx \frac{1}{(x^2+m^2)^\alpha}\bigg|_{\alpha=0}$
$\displaystyle=-\pi^\frac{d}{2}\frac{\partial}{\partial\alpha}\left(\frac{\Gamma(\alpha-\frac{d}{2})}{\Gamma(\alpha)}\frac{1}{(m^2)^{\alpha-d/2}}\right)\bigg|_{\alpha=0}$
How does one go from the upper to the lower line? I've tried evaluating it as follows:
$\displaystyle\int d^dx(x^2+m^2)^{-\alpha}=\int\limits_0^\infty dx\, x^{d-1}(x^2+m^2)^{-\alpha} \overbrace{\int d\Omega(d-1)}^{C(d)}\\ \displaystyle=C\int dx\, \frac{x^{d-1}}{(x^2+m^2)^\alpha}=\frac{C}{2}\int dx^2\, \frac{(x^2)^{\frac{d-2}{2}}}{(x^2+m^2)^\alpha}=|x^2=y|\\ \displaystyle=\frac{C}{2} \int dy\, \frac{y^{\frac{d-2}{2}}}{[m^2(1+\frac{y}{m^2})]^\alpha}=|y=m^2\, z|\\ \displaystyle=\frac{C}{2} (m^2)^{d-\alpha}\int\limits_0^\infty dz\, \frac{z^{\frac{d-2}{2}}}{(1+z)^\alpha}$
The integral in the last line should be (at least according to the wiki) the beta function $B(x,y)$ with $x=d$, $y=\alpha-d$, which is just $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$, so:
$\displaystyle result=\frac{C}{2}(m^2)^{d-\alpha}\frac{\Gamma(d)\Gamma(\alpha-d)}{\Gamma(\alpha)}$
but in the first formula they get stuff with $\Gamma(\alpha-\frac{d}{2})$. What am I doing wrong?