A certain regularization and renormalization scheme In a certain lecture of Witten's about some QFT in $1+1$ dimensions, I came across these two statements of regularization and renormalization, which I could not prove, 
(1) $\int ^\Lambda \frac{d^2 k}{(2\pi)^2}\frac{1}{k^2 + q_i ^2 \vert \sigma \vert ^2}  = - \frac{1}{2\pi} ln \vert q _ i \vert - \frac{1}{2\pi}ln \frac{\vert \sigma\vert}{\mu}$
(..there was an overall $\sum _i q_i$ in the above but I don't think that is germane to the point..) 
(2) $\int ^\Lambda \frac{d^2 k}{(2\pi)^2}\frac{1}{k^2 +  \vert \sigma \vert ^2} = \frac{1}{2\pi} (ln \frac{\Lambda}{\mu} - ln \frac{\vert \sigma \vert }{\mu} )$ 
I tried doing dimensional regularization and Pauli-Villar's (motivated by seeing that $\mu$ which looks like an IR cut-off) but nothing helped me reproduce the above equations.
I would glad if someone can help prove these above two equations. 
 A: Let's just look at the integral 
$$\int \frac{d^2k}{(2\pi)^2} \frac{1}{k^2+\alpha^2}.$$
The other integrals should follow from this one. 
Introduce the Pauli-Villars regulator, 
$$\begin{eqnarray*}
\int \frac{d^2k}{(2\pi)^2} \frac{1}{k^2+\alpha^2}
&\rightarrow& \int \frac{d^2k}{(2\pi)^2} \frac{1}{k^2+\alpha^2} - \int \frac{d^2k}{(2\pi)^2} \frac{1}{k^2+\Lambda^2} \\
&=& (\Lambda^2-\alpha^2)\int \frac{d^2k}{(2\pi)^2} \frac{1}{(k^2+\alpha^2)(k^2+\Lambda^2)} \\
&=& (\Lambda^2-\alpha^2)\int_0^1 dx\, \int\frac{d^2k}{(2\pi)^2} \frac{1}{(k^2 + \beta^2)^2} \\
&=& (\Lambda^2-\alpha^2)\int_0^1 dx\, \frac{1}{2} \frac{2\pi}{(2\pi)^2}
\int_0^\infty dk^2\,\frac{1}{(k^2 + \beta^2)^2} \\
&=& (\Lambda^2-\alpha^2) \frac{1}{4\pi}
\int_0^1 dx\, \frac{1}{\beta^2} \\
&=& (\Lambda^2-\alpha^2) \frac{1}{4\pi} 
\int_0^1 dx\, \frac{1}{\Lambda^2 - x(\Lambda^2-\alpha^2)} \\
&=& -\frac{1}{2\pi} \ln \frac{|\alpha|}{\Lambda}
\end{eqnarray*}$$
Where we have combined denominators with the Feynman parameter $x$, with the intermediate variable $\beta^2 = \Lambda^2 - x(\Lambda^2-\alpha^2)$.
Of course, this could also be approached with dimensional regularization with the same result. 
Addendum: After regularization we must renormalize. 
Using the minimal subtraction prescription we find 
$$\int \frac{d^2k}{(2\pi)^2} \frac{1}{k^2+\alpha^2} 
\rightarrow -\frac{1}{2\pi} \ln \frac{|\alpha|}{\mu},$$
as required. 
A: perhaps since your itnegral is logarithmic divergent you could do the following
$$ \int_{0}^{\infty}\frac{kdk}{k^{2}+a^{2}}\to \int_{0}^{\infty}\frac{kdk}{k^{2}+a^{2}}- \int_{0}^{\infty}\frac{dx}{x+b}+\int_{0}^{\infty}\frac{dx}{x+b} $$
then the integral $$ A=\int_{0}^{\infty}\frac{kdk}{k^{2}+a^{2}}- \int_{0}^{\infty}\frac{dk}{k+b}$$ is convergent so we must now regularize
$ \int_{0}^{\infty}\frac{dx}{x+b} $ which can be made by using Ramanujan's resummation to get $ \sum_{n=0}^{\infty} \frac{1}{n+b}= -\Psi (b) $ now use the Euler-Maclaurin summation formula
